2.1.48 Problem 48
Internal
problem
ID
[9119]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
48
Date
solved
:
Friday, February 21, 2025 at 09:15:53 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} y {y^{\prime \prime }}^{4}+{y^{\prime }}^{2}&=0 \end{align*}
Solved as second order missing x ode
Time used: 15.905 (sec)
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using
\begin{align*} y' &= p \end{align*}
Then
\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}
Hence the ode becomes
\begin{align*} y p \left (y \right )^{4} \left (\frac {d}{d y}p \left (y \right )\right )^{4}+p \left (y \right )^{2} = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Factoring the ode gives these factors
\begin{align*}
\tag{1} p^{2} &= 0 \\
\tag{2} {p^{\prime }}^{4} p^{2} y +1 &= 0 \\
\end{align*}
Now each of the above equations is solved in turn.
Solving equation (1)
Solving for \(p\) from
\begin{align*} p^{2} = 0 \end{align*}
Solving gives \(p = 0\)
Solving equation (2)
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} p^{\prime }&=\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\
\tag{2} p^{\prime }&=\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\
\tag{3} p^{\prime }&=-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\
\tag{4} p^{\prime }&=-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}p} &= \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(y,p) &= -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied
\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= -\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=-\frac {p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}
Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)
\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}
Integrating (2) w.r.t. \(p\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\
\tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y)
\end{equation}
But equation (1) says that \(\frac {\partial \phi }{\partial y} = -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes
\begin{equation}
\tag{5} -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}
\]
Integrating the above w.r.t \(y\) gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= -\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_1 \\
\end{align*}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = \frac {2 p^{{3}/{2}}}{3}-\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_1
\]
But since \(\phi \) itself is a constant function, then
let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = \frac {2 p^{{3}/{2}}}{3}-\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}
\]
Solving Eq. (2)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}p} &= \left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(y,p) &= -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied
\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= -\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=\frac {i p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}
Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)
\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}
Integrating (2) w.r.t. \(p\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\
\tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y)
\end{equation}
But equation (1) says that \(\frac {\partial \phi }{\partial y} = -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes
\begin{equation}
\tag{5} -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}
\]
Integrating the above w.r.t \(y\) gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= -\frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_2 \\
\end{align*}
Where \(c_2\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = \frac {2 p^{{3}/{2}}}{3}-\frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_2
\]
But since \(\phi \) itself is a constant function, then
let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_2\) and \(c_2\) constants into the constant \(c_2\) gives the solution as
\[
c_2 = \frac {2 p^{{3}/{2}}}{3}-\frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}
\]
Solving Eq. (3)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}p} &= \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(y,p) &= \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied
\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= \frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=\frac {p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}
Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)
\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}
Integrating (2) w.r.t. \(p\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\
\tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y)
\end{equation}
But equation (1) says that \(\frac {\partial \phi }{\partial y} = \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}
\]
Integrating the above w.r.t \(y\) gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_3 \\
\end{align*}
Where \(c_3\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = \frac {2 p^{{3}/{2}}}{3}+\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_3
\]
But since \(\phi \) itself is a constant function, then
let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_3\) and \(c_2\) constants into the constant \(c_3\) gives the solution as
\[
c_3 = \frac {2 p^{{3}/{2}}}{3}+\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}
\]
Solving Eq. (4)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}p} &= \left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(y,p) &= \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied
\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= \frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}
Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=-\frac {i p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}
Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)
\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}
Integrating (2) w.r.t. \(p\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\
\int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\
\tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y)
\end{equation}
But equation (1) says that \(\frac {\partial \phi }{\partial y} = \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}
\]
Integrating the above w.r.t \(y\) gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_4 \\
\end{align*}
Where \(c_4\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)
\[
\phi = \frac {2 p^{{3}/{2}}}{3}+\frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_4
\]
But since \(\phi \) itself is a constant function, then
let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_4\) and \(c_2\) constants into the constant \(c_4\) gives the solution as
\[
c_4 = \frac {2 p^{{3}/{2}}}{3}+\frac {4 i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}
\]
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}-\frac {4 \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_1 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{3} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{4} y^{\prime }&=\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\
\tag{5} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{6} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{7} y^{\prime }&=\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\
\tag{8} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{9} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{10} y^{\prime }&=\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\
\tag{11} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{12} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +c_5 \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_6 \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +c_7 \]
Solving Eq. (4)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +c_8 \]
Solving Eq. (5)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_9 \]
Solving Eq. (6)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C10} \]
Solving Eq. (7)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C11} \]
Solving Eq. (8)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C12} \]
Solving Eq. (9)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C13} \]
Solving Eq. (10)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C14} \]
Solving Eq. (11)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C15} \]
Solving Eq. (12)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C16} \]
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}+\frac {4 \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_3 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{3} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{4} y^{\prime }&=\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}{4} \\
\tag{5} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{6} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{7} y^{\prime }&=\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}{4} \\
\tag{8} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{9} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{10} y^{\prime }&=\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}{4} \\
\tag{11} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\tag{12} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{1}/{3}}}{4}\right )}^{2} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C17} \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C18} \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C19} \]
Solving Eq. (4)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C20} \]
Solving Eq. (5)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C21} \]
Solving Eq. (6)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C22} \]
Solving Eq. (7)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C23} \]
Solving Eq. (8)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C24} \]
Solving Eq. (9)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C25} \]
Solving Eq. (10)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C26} \]
Solving Eq. (11)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C27} \]
Solving Eq. (12)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C28} \]
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}-\frac {4 i \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_2 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {{\left (12 c_2 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{3} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{4} y^{\prime }&=\frac {{\left (12 c_2 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{5} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{6} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{7} y^{\prime }&=\frac {{\left (12 c_2 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{8} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{9} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{10} y^{\prime }&=\frac {{\left (12 c_2 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{11} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{12} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_2 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C29} \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C30} \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C31} \]
Solving Eq. (4)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C32} \]
Solving Eq. (5)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C33} \]
Solving Eq. (6)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C34} \]
Solving Eq. (7)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C35} \]
Solving Eq. (8)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C36} \]
Solving Eq. (9)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C37} \]
Solving Eq. (10)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C38} \]
Solving Eq. (11)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C39} \]
Solving Eq. (12)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C40} \]
For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}+\frac {4 i \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_4 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {{\left (12 c_4 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{3} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 -16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{4} y^{\prime }&=\frac {{\left (12 c_4 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{5} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{6} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 -16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{7} y^{\prime }&=\frac {{\left (12 c_4 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{8} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{9} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 +16 \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{10} y^{\prime }&=\frac {{\left (12 c_4 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{11} y^{\prime }&=\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\tag{12} y^{\prime }&=\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\left (12 c_4 +16 i \left (-y^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}{4} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C41} \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C42} \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C43} \]
Solving Eq. (4)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C44} \]
Solving Eq. (5)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C45} \]
Solving Eq. (6)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C46} \]
Solving Eq. (7)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C47} \]
Solving Eq. (8)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C48} \]
Solving Eq. (9)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C49} \]
Solving Eq. (10)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {4}{{\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C50} \]
Solving Eq. (11)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C51} \]
Solving Eq. (12)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau = x +\textit {\_C52} \]
For solution (5) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + \textit {\_C53} \\ y &= \textit {\_C53} \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {4}{{\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C29} \\
\int _{}^{y}\frac {4}{{\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C35} \\
\int _{}^{y}\frac {4}{{\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C32} \\
\int _{}^{y}\frac {4}{{\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C38} \\
\int _{}^{y}\frac {4}{{\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C47} \\
\int _{}^{y}\frac {4}{{\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C41} \\
\int _{}^{y}\frac {4}{{\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C44} \\
\int _{}^{y}\frac {4}{{\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C50} \\
\int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C23} \\
\int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C17} \\
\int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C11} \\
\int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau &= x +c_5 \\
\int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C14} \\
\int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C26} \\
\int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau &= x +c_8 \\
\int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}}}d \tau &= x +\textit {\_C20} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C37} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C31} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C34} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C40} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C49} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C43} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C46} \\
\int _{}^{y}\frac {16}{\left (1+i \sqrt {3}\right )^{2} {\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C52} \\
\int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C12} \\
\int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C13} \\
\int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C25} \\
\int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +c_6 \\
\int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +c_7 \\
\int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C18} \\
\int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C19} \\
\int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C24} \\
\int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C15} \\
\int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C28} \\
\int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C27} \\
\int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C16} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C30} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C36} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C33} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_2 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C39} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 -16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C42} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 +16 \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C48} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 -16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C45} \\
\int _{}^{y}\frac {16}{\left (i \sqrt {3}-1\right )^{2} {\left (12 c_4 +16 i \left (-\tau ^{3}\right )^{{1}/{4}}\right )}^{{2}/{3}}}d \tau &= x +\textit {\_C51} \\
\int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +\textit {\_C21} \\
\int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_3 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C22} \\
\int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +c_9 \\
\int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +\textit {\_C10} \\
y &= \textit {\_C53} \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
*** Sublevel 2 ***
Methods for second order ODEs:
Successful isolation of d^2y/dx^2: 4 solutions were found. Trying to solve each resulting ODE.
*** Sublevel 3 ***
Methods for second order ODEs:
--- Trying classification methods ---
trying 2nd order Liouville
trying 2nd order WeierstrassP
trying 2nd order JacobiSN
differential order: 2; trying a linearization to 3rd order
trying 2nd order ODE linearizable_by_differentiation
trying 2nd order, 2 integrating factors of the form mu(x,y)
trying differential order: 2; missing variables
`, `-> Computing symmetries using: way = 3
-> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(-_a^3*_b(_a)^2)^(1/4)/_a = 0, _b(_a), HINT = [[_a, (1/2)*_b]]`
symmetry methods on request
`, `1st order, trying reduction of order with given symmetries:`[_a, 1/2*_b]
Maple dsolve solution
Solving time : 13.932 (sec)
Leaf size : 2829
dsolve(y(x)*diff(diff(y(x),x),x)^4+diff(y(x),x)^2 = 0,y(x),singsol=all)
\begin{align*}
y &= c_{1} \\
y &= 0 \\
\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} -\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (1+i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a}^{3} {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-2 \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right ) \left (\textit {\_a} +\frac {\left (c_{1} \textit {\_a} \right )^{{1}/{4}}}{2}\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} -\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (1-i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (1-i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \left (1+i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a}^{3} {\left (\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-2 \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right ) \left (\textit {\_a} +\frac {\left (c_{1} \textit {\_a} \right )^{{1}/{4}}}{2}\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\textit {\_a} \right )-2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_f}}{2 i \left (-\textit {\_f}^{2}\right )^{{1}/{4}}+\textit {\_f}^{2}}d \textit {\_f} \right )+c_{1} \right ) \sqrt {\textit {\_a}}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (2 i \textit {\_a} -\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-i \textit {\_a} {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {\textit {\_a}}{\sqrt {i \textit {\_a} {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (-i+\sqrt {3}\right ) \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (\sqrt {3}+i\right ) \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\sqrt {3}+i\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (2 i \textit {\_a} -\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-i \textit {\_a} {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {\textit {\_a}}{\sqrt {i \textit {\_a} {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-i+\sqrt {3}\right ) \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-i+\sqrt {3}\right ) \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (-i+\sqrt {3}\right ) \left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}+2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (\sqrt {3}+i\right ) \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
-\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i \left (c_{1} \textit {\_a} \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (c_{1} \textit {\_a} \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\sqrt {3}+i\right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\
\int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\textit {\_a} \right )-2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_f}}{\textit {\_f}^{2}+2 \left (-\textit {\_f}^{2}\right )^{{1}/{4}}}d \textit {\_f} \right )+c_{1} \right ) \sqrt {\textit {\_a}}}d \textit {\_a} -x -c_{2} &= 0 \\
\end{align*}
✓Mathematica DSolve solution
Solving time : 3.846 (sec)
Leaf size : 1237
DSolve[{y[x]*D[y[x],{x,2}]^4+D[y[x],x]^2==0,{}},y[x],x,IncludeSingularSolutions->True]
Too large to display
✗Sympy solution
Solving time : 0.000 (sec)
Leaf size : 0
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(y(x)*Derivative(y(x), (x, 2))**4 + Derivative(y(x), x)**2,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -sqrt(-y(x))*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x) cannot be solved by the factorable group method