2.1.51 Problem 51

Internal problem ID [9122]
Book : Second order enumerated odes
Section : section 1
Problem number : 51
Date solved : Monday, January 27, 2025 at 05:43:58 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y {y^{\prime \prime }}^{3}+y^{3} y^{\prime }&=0 \end{align*}

Factoring the ode gives these factors

\begin{align*} \tag{1} y &= 0 \\ \tag{2} {y^{\prime \prime }}^{3}+y^{\prime } y^{2} &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(y\) from

\begin{align*} y = 0 \end{align*}

Solving gives \(y = 0\)

Solving equation (2)

Solved as second order missing x ode

Time used: 84.304 (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*} p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3}+p \left (y \right ) y^{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 &= 0 \\ \tag{2} {p^{\prime }}^{3} p^{2}+y^{2} &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(p\) from

\begin{align*} p = 0 \end{align*}

Solving gives \(p = 0\)

Solving equation (2)

Let \(p=p^{\prime }\) the ode becomes

\begin{align*} p^{3} p^{2}+y^{2} = 0 \end{align*}

Solving for \(p\) from the above results in

\begin{align*} \tag{1} p &= -\frac {y}{\sqrt {-p}\, p} \\ \tag{2} p &= \frac {y}{\sqrt {-p}\, p} \\ \end{align*}

This has the form

\begin{align*} p=yf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=p'(y)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(y\) gives

\begin{align*} p &= f+(y f'+g') \frac {dp}{dy}\\ p-f &= (y f'+g') \frac {dp}{dy}\tag {2} \end{align*}

Comparing the form \(p=y f + g\) to (1A) shows that

\begin{align*} f &= \frac {1}{\left (-p \right )^{{3}/{2}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {1}{\left (-p \right )^{{3}/{2}}} = \frac {3 y p^{\prime }\left (y \right )}{2 \left (-p \right )^{{5}/{2}}} \end{equation}

The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives

\begin{align*} p -\frac {1}{\left (-p \right )^{{3}/{2}}} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (y \right ) = \frac {2 \left (p \left (y \right )-\frac {1}{\left (-p \left (y \right )\right )^{{3}/{2}}}\right ) \left (-p \left (y \right )\right )^{{5}/{2}}}{3 y} \end{equation}

This ODE is now solved for \(p \left (y \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (y \right ) = \frac {2 \left (\left (-p \left (y \right )\right )^{{5}/{2}}+1\right ) p \left (y \right )}{3 y} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (y \right )&= \frac {2 \left (\left (-p \left (y \right )\right )^{{5}/{2}}+1\right ) p \left (y \right )}{3 y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= \frac {2}{3 y}\\ g(p) &= \left (\left (-p \right )^{{5}/{2}}+1\right ) p \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\ \int { \frac {1}{\left (\left (-p \right )^{{5}/{2}}+1\right ) p}\,dp} &= \int { \frac {2}{3 y} \,dy} \\ \end{align*}

\[ -\frac {2 \ln \left (p \left (y \right )^{2}-\left (-p \left (y \right )\right )^{{3}/{2}}-p \left (y \right )-\sqrt {-p \left (y \right )}+1\right )}{5}+\ln \left (-p \left (y \right )\right )-\frac {2 \ln \left (\sqrt {-p \left (y \right )}+1\right )}{5}=\ln \left (y^{{2}/{3}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \left (\left (-p \right )^{{5}/{2}}+1\right ) p=0 \]

for \(p \left (y \right )\) gives

\begin{align*} p \left (y \right )&=0\\ p \left (y \right )&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\\ p \left (y \right )&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =4\right )^{2} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

The solution \(-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\) will not be used

The solution \(-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =4\right )^{2}\) will not be used

Therefore the solutions found are

\begin{align*} -\frac {2 \ln \left (p \left (y \right )^{2}-\left (-p \left (y \right )\right )^{{3}/{2}}-p \left (y \right )-\sqrt {-p \left (y \right )}+1\right )}{5}+\ln \left (-p \left (y \right )\right )-\frac {2 \ln \left (\sqrt {-p \left (y \right )}+1\right )}{5} &= \ln \left (y^{{2}/{3}}\right )+c_1 \\ p \left (y \right ) &= 0 \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} p = \frac {y}{{\left ({\left ({\operatorname {RootOf}\left (-1+\left (y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1\right ) \textit {\_Z}^{75}+\left (-15 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-15\right ) \textit {\_Z}^{70}+\left (105 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+105\right ) \textit {\_Z}^{65}+\left (-455 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-455\right ) \textit {\_Z}^{60}+\left (1365 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1365\right ) \textit {\_Z}^{55}+\left (-3000 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-3003\right ) \textit {\_Z}^{50}+\left (4975 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+5005\right ) \textit {\_Z}^{45}+\left (-6300 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-6435\right ) \textit {\_Z}^{40}+\left (6075 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+6435\right ) \textit {\_Z}^{35}+\left (-4375 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-5005\right ) \textit {\_Z}^{30}+\left (2250 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+3003\right ) \textit {\_Z}^{25}+\left (-750 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-1365\right ) \textit {\_Z}^{20}+\left (125 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+455\right ) \textit {\_Z}^{15}-105 \textit {\_Z}^{10}+15 \textit {\_Z}^{5}\right )}^{5}-1\right )}^{2}\right )}^{{3}/{2}}}\\ \end{align*}

Solving ode 2A

Taking derivative of (*) w.r.t. \(y\) gives

\begin{align*} p &= f+(y f'+g') \frac {dp}{dy}\\ p-f &= (y f'+g') \frac {dp}{dy}\tag {2} \end{align*}

Comparing the form \(p=y f + g\) to (1A) shows that

\begin{align*} f &= -\frac {1}{\left (-p \right )^{{3}/{2}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p +\frac {1}{\left (-p \right )^{{3}/{2}}} = -\frac {3 y p^{\prime }\left (y \right )}{2 \left (-p \right )^{{5}/{2}}} \end{equation}

The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives

\begin{align*} p +\frac {1}{\left (-p \right )^{{3}/{2}}} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-1 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} p = -y \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (y \right ) = -\frac {2 \left (p \left (y \right )+\frac {1}{\left (-p \left (y \right )\right )^{{3}/{2}}}\right ) \left (-p \left (y \right )\right )^{{5}/{2}}}{3 y} \end{equation}

This ODE is now solved for \(p \left (y \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (y \right ) = -\frac {2 \left (\left (-p \left (y \right )\right )^{{5}/{2}}-1\right ) p \left (y \right )}{3 y} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (y \right )&= -\frac {2 \left (\left (-p \left (y \right )\right )^{{5}/{2}}-1\right ) p \left (y \right )}{3 y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {2}{3 y}\\ g(p) &= \left (\left (-p \right )^{{5}/{2}}-1\right ) p \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\ \int { \frac {1}{\left (\left (-p \right )^{{5}/{2}}-1\right ) p}\,dp} &= \int { -\frac {2}{3 y} \,dy} \\ \end{align*}

\[ \frac {2 \ln \left (p \left (y \right )^{2}+\left (-p \left (y \right )\right )^{{3}/{2}}-p \left (y \right )+\sqrt {-p \left (y \right )}+1\right )}{5}-\ln \left (-p \left (y \right )\right )+\frac {2 \ln \left (\sqrt {-p \left (y \right )}-1\right )}{5}=\ln \left (\frac {1}{y^{{2}/{3}}}\right )+c_2 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \left (\left (-p \right )^{{5}/{2}}-1\right ) p=0 \]

for \(p \left (y \right )\) gives

\begin{align*} p \left (y \right )&=-1\\ p \left (y \right )&=0\\ p \left (y \right )&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\\ p \left (y \right )&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =4\right )^{2} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

The solution \(-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\) will not be used

The solution \(-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =4\right )^{2}\) will not be used

Therefore the solutions found are

\begin{align*} \frac {2 \ln \left (p \left (y \right )^{2}+\left (-p \left (y \right )\right )^{{3}/{2}}-p \left (y \right )+\sqrt {-p \left (y \right )}+1\right )}{5}-\ln \left (-p \left (y \right )\right )+\frac {2 \ln \left (\sqrt {-p \left (y \right )}-1\right )}{5} &= \ln \left (\frac {1}{y^{{2}/{3}}}\right )+c_2 \\ p \left (y \right ) &= -1 \\ p \left (y \right ) &= 0 \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} p = -\frac {y}{{\left ({\left ({\operatorname {RootOf}\left (\left (y^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{75}+\left (15 y^{5}+15 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{70}+\left (105 y^{5}+105 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{65}+\left (455 y^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{60}+\left (1365 y^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{55}+\left (3000 y^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{50}+\left (4975 y^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{45}+\left (6300 y^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{40}+\left (6075 y^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{35}+\left (4375 y^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{30}+\left (2250 y^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{25}+\left (750 y^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{20}+\left (125 y^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{15}+105 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{10}+15 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right )}^{5}+1\right )}^{2}\right )}^{{3}/{2}}}\\ p = -y\\ \end{align*}

For solution (1) 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} + c_3 \\ y &= c_3 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \frac {y}{{\left ({\left ({\operatorname {RootOf}\left (-1+\left (y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1\right ) \textit {\_Z}^{75}+\left (-15 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-15\right ) \textit {\_Z}^{70}+\left (105 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+105\right ) \textit {\_Z}^{65}+\left (-455 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-455\right ) \textit {\_Z}^{60}+\left (1365 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1365\right ) \textit {\_Z}^{55}+\left (-3000 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-3003\right ) \textit {\_Z}^{50}+\left (4975 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+5005\right ) \textit {\_Z}^{45}+\left (-6300 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-6435\right ) \textit {\_Z}^{40}+\left (6075 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+6435\right ) \textit {\_Z}^{35}+\left (-4375 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-5005\right ) \textit {\_Z}^{30}+\left (2250 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+3003\right ) \textit {\_Z}^{25}+\left (-750 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}-1365\right ) \textit {\_Z}^{20}+\left (125 y^{5} {\mathrm e}^{\frac {15 c_1}{2}}+455\right ) \textit {\_Z}^{15}-105 \textit {\_Z}^{10}+15 \textit {\_Z}^{5}\right )}^{5}-1\right )}^{2}\right )}^{{3}/{2}}} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {{\left ({\left ({\operatorname {RootOf}\left (-1+\left (\tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1\right ) \textit {\_Z}^{75}+\left (-15 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-15\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+105\right ) \textit {\_Z}^{65}+\left (-455 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-455\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1365\right ) \textit {\_Z}^{55}+\left (-3000 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-3003\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+5005\right ) \textit {\_Z}^{45}+\left (-6300 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-6435\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+6435\right ) \textit {\_Z}^{35}+\left (-4375 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-5005\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+3003\right ) \textit {\_Z}^{25}+\left (-750 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-1365\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+455\right ) \textit {\_Z}^{15}-105 \textit {\_Z}^{10}+15 \textit {\_Z}^{5}\right )}^{5}-1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau = x +c_4 \]

Solving for \(y\) gives

\begin{align*} y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {{\left ({\left ({\operatorname {RootOf}\left (-1+\left (\tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1\right ) \textit {\_Z}^{75}+\left (-15 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-15\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+105\right ) \textit {\_Z}^{65}+\left (-455 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-455\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1365\right ) \textit {\_Z}^{55}+\left (-3000 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-3003\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+5005\right ) \textit {\_Z}^{45}+\left (-6300 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-6435\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+6435\right ) \textit {\_Z}^{35}+\left (-4375 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-5005\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+3003\right ) \textit {\_Z}^{25}+\left (-750 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-1365\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+455\right ) \textit {\_Z}^{15}-105 \textit {\_Z}^{10}+15 \textit {\_Z}^{5}\right )}^{5}-1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau +x +c_4 \right ) \\ \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -y \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{y}d y &= dx\\ -\ln \left (y \right )&= x +c_5 \end{align*}

Singular solutions are found by solving

\begin{align*} -y&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = 0 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{-x -c_5} \\ \end{align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {y}{{\left ({\left ({\operatorname {RootOf}\left (\left (y^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{75}+\left (15 y^{5}+15 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{70}+\left (105 y^{5}+105 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{65}+\left (455 y^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{60}+\left (1365 y^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{55}+\left (3000 y^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{50}+\left (4975 y^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{45}+\left (6300 y^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{40}+\left (6075 y^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{35}+\left (4375 y^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{30}+\left (2250 y^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{25}+\left (750 y^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{20}+\left (125 y^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{15}+105 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{10}+15 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right )}^{5}+1\right )}^{2}\right )}^{{3}/{2}}} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {{\left ({\left ({\operatorname {RootOf}\left (\left (\tau ^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{75}+\left (15 \tau ^{5}+15 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5}+105 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{65}+\left (455 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{55}+\left (3000 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{45}+\left (6300 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{35}+\left (4375 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{25}+\left (750 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{15}+105 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{10}+15 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right )}^{5}+1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau = x +c_6 \]

Solving for \(y\) gives

\begin{align*} y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {{\left ({\left ({\operatorname {RootOf}\left (\left (\tau ^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{75}+\left (15 \tau ^{5}+15 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5}+105 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{65}+\left (455 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{55}+\left (3000 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{45}+\left (6300 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{35}+\left (4375 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{25}+\left (750 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{15}+105 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{10}+15 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right )}^{5}+1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau +x +c_6 \right ) \\ \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {{\left ({\left ({\operatorname {RootOf}\left (-1+\left (\tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1\right ) \textit {\_Z}^{75}+\left (-15 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-15\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+105\right ) \textit {\_Z}^{65}+\left (-455 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-455\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+1365\right ) \textit {\_Z}^{55}+\left (-3000 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-3003\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+5005\right ) \textit {\_Z}^{45}+\left (-6300 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-6435\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+6435\right ) \textit {\_Z}^{35}+\left (-4375 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-5005\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+3003\right ) \textit {\_Z}^{25}+\left (-750 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}-1365\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5} {\mathrm e}^{\frac {15 c_1}{2}}+455\right ) \textit {\_Z}^{15}-105 \textit {\_Z}^{10}+15 \textit {\_Z}^{5}\right )}^{5}-1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau +x +c_4 \right ) \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {{\left ({\left ({\operatorname {RootOf}\left (\left (\tau ^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{75}+\left (15 \tau ^{5}+15 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{70}+\left (105 \tau ^{5}+105 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{65}+\left (455 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{60}+\left (1365 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{55}+\left (3000 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{50}+\left (4975 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{45}+\left (6300 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{40}+\left (6075 \tau ^{5}+6435 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{35}+\left (4375 \tau ^{5}+5005 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{30}+\left (2250 \tau ^{5}+3003 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{25}+\left (750 \tau ^{5}+1365 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{20}+\left (125 \tau ^{5}+455 \,{\mathrm e}^{\frac {15 c_2}{2}}\right ) \textit {\_Z}^{15}+105 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{10}+15 \,{\mathrm e}^{\frac {15 c_2}{2}} \textit {\_Z}^{5}+{\mathrm e}^{\frac {15 c_2}{2}}\right )}^{5}+1\right )}^{2}\right )}^{{3}/{2}}}{\tau }d \tau +x +c_6 \right ) \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} y &= 0 \\ y &= c_3 \\ y &= {\mathrm e}^{-x -c_5} \\ \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: 3 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 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -_b(_a)^2+(-_b(_a))^(1/3), _b(_a), explicit, HINT = [[1, 0]]`         *** 
         symmetry methods on request 
      `, `1st order, trying reduction of order with given symmetries:`[1, 0]
 

Maple dsolve solution

Solving time : 0.184 (sec)
Leaf size : 126

dsolve(y(x)*diff(diff(y(x),x),x)^3+diff(y(x),x)*y(x)^3 = 0,y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= c_{1} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -\int _{}^{\textit {\_Z}}-\frac {1}{\textit {\_f}^{2}-\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} +c_{1} \right )d x +c_{2}} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x +2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \left (-\textit {\_f} \right )^{{1}/{3}} \sqrt {3}+2 \textit {\_f}^{2}+\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \left (-\textit {\_f} \right )^{{1}/{3}} \sqrt {3}-2 \textit {\_f}^{2}-\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ \end{align*}

Mathematica DSolve solution

Solving time : 2.742 (sec)
Leaf size : 800

DSolve[{y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]==0,{}},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5}{3} (-1) c_1\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ \end{align*}