2.1.16 Problem 16

Internal problem ID [9087]
Book : Second order enumerated odes
Section : section 1
Problem number : 16
Date solved : Monday, January 27, 2025 at 05:32:24 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

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

Solved as second order missing x ode

Time used: 276.831 (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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 1 \end{align*}

Which is now solved as first order ode for \(p(y)\).

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

\begin{align*} p^{2} p^{2}+p = 1 \end{align*}

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

\begin{align*} \tag{1} p &= \frac {-1+\sqrt {4 p^{2}+1}}{2 p^{2}} \\ \tag{2} p &= -\frac {1+\sqrt {4 p^{2}+1}}{2 p^{2}} \\ \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 &= 0\\ g &= \frac {-1+\sqrt {4 p^{2}+1}}{2 p^{2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {1}{p^{3}}-\frac {\sqrt {4 p^{2}+1}}{p^{3}}+\frac {2}{p \sqrt {4 p^{2}+1}}\right ) p^{\prime }\left (y \right ) \end{equation}

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

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} p = 1 \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 {p \left (y \right )}{\frac {1}{p \left (y \right )^{3}}-\frac {\sqrt {4 p \left (y \right )^{2}+1}}{p \left (y \right )^{3}}+\frac {2}{p \left (y \right ) \sqrt {4 p \left (y \right )^{2}+1}}} \end{equation}

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

Integrating gives

\begin{align*} \int -\frac {2 p^{2}-\sqrt {4 p^{2}+1}+1}{p^{4} \sqrt {4 p^{2}+1}}d p &= dy\\ \frac {-2 \sqrt {4 p^{2}+1}\, p^{2}+\sqrt {4 p^{2}+1}-1}{3 p^{3}}&= y +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {p^{4} \sqrt {4 p^{2}+1}}{2 p^{2}-\sqrt {4 p^{2}+1}+1}&= 0 \end{align*}

for \(p \left (y \right )\). 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*} p \left (y \right ) = -\frac {i}{2}\\ p \left (y \right ) = \frac {i}{2} \end{align*}

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

\begin{align*} p = \frac {-1+\sqrt {4 {\left (\frac {{\left (\left (-3 y -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y +486 c_1^{2} y^{2}+324 c_1 \,y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y -144 y^{2}+64}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}-\frac {4}{{\left (\left (-3 y -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y +486 c_1^{2} y^{2}+324 c_1 \,y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y -144 y^{2}+64}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}}{2 {\left (\frac {{\left (\left (-3 y -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y +486 c_1^{2} y^{2}+324 c_1 \,y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y -144 y^{2}+64}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}-\frac {4}{{\left (\left (-3 y -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y +486 c_1^{2} y^{2}+324 c_1 \,y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y -144 y^{2}+64}{9 c_1^{2}+18 c_1 y +9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}}\\ p = 2\\ p = 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 &= 0\\ g &= \frac {-1-\sqrt {4 p^{2}+1}}{2 p^{2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (-\frac {2}{p \sqrt {4 p^{2}+1}}+\frac {1}{p^{3}}+\frac {\sqrt {4 p^{2}+1}}{p^{3}}\right ) p^{\prime }\left (y \right ) \end{equation}

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

\begin{align*} p = 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 {p \left (y \right )}{-\frac {2}{p \left (y \right ) \sqrt {4 p \left (y \right )^{2}+1}}+\frac {1}{p \left (y \right )^{3}}+\frac {\sqrt {4 p \left (y \right )^{2}+1}}{p \left (y \right )^{3}}} \end{equation}

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

Integrating gives

\begin{align*} \int \frac {2 p^{2}+\sqrt {4 p^{2}+1}+1}{p^{4} \sqrt {4 p^{2}+1}}d p &= dy\\ \frac {2 \sqrt {4 p^{2}+1}\, p^{2}-\sqrt {4 p^{2}+1}-1}{3 p^{3}}&= y +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {p^{4} \sqrt {4 p^{2}+1}}{2 p^{2}+\sqrt {4 p^{2}+1}+1}&= 0 \end{align*}

for \(p \left (y \right )\). 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*} p \left (y \right ) = 0\\ p \left (y \right ) = -\frac {i}{2}\\ p \left (y \right ) = \frac {i}{2} \end{align*}

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

\begin{align*} p = \frac {-1-\sqrt {4 {\left (\frac {{\left (\left (-3 y -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y +486 c_2^{2} y^{2}+324 c_2 \,y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y -144 y^{2}+64}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}-\frac {4}{{\left (\left (-3 y -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y +486 c_2^{2} y^{2}+324 c_2 \,y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y -144 y^{2}+64}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}}{2 {\left (\frac {{\left (\left (-3 y -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y +486 c_2^{2} y^{2}+324 c_2 \,y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y -144 y^{2}+64}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}-\frac {4}{{\left (\left (-3 y -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y +486 c_2^{2} y^{2}+324 c_2 \,y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y -144 y^{2}+64}{9 c_2^{2}+18 c_2 y +9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y +9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}}\\ p = 2\\ p = 2\\ \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 } = 1 \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 {1\, dx}\\ y &= x + 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 } = 2 \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 {2\, dx}\\ y &= 2 x + c_4 \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 } = \frac {-1-\sqrt {4 {\left (\frac {{\left (\left (-3 y-3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y+486 c_2^{2} y^{2}+324 c_2 y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y-144 y^{2}+64}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}-\frac {4}{{\left (\left (-3 y-3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y+486 c_2^{2} y^{2}+324 c_2 y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y-144 y^{2}+64}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}}{2 {\left (\frac {{\left (\left (-3 y-3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y+486 c_2^{2} y^{2}+324 c_2 y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y-144 y^{2}+64}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}-\frac {4}{{\left (\left (-3 y-3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} y+486 c_2^{2} y^{2}+324 c_2 y^{3}+81 y^{4}-144 c_2^{2}-288 c_2 y-144 y^{2}+64}{9 c_2^{2}+18 c_2 y+9 y^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}} \end{align*}

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

\[ \int _{}^{y}-\frac {2 \left (36 \tau ^{2}+72 c_2 \tau -{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}+36 c_2^{2}-64\right )^{2}}{\left (1+\sqrt {4 {\left (\frac {{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}-\frac {4}{{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2} {\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}}d \tau = x +c_5 \]

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 {-1+\sqrt {4 {\left (\frac {{\left (\left (-3 y-3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y+486 c_1^{2} y^{2}+324 c_1 y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y-144 y^{2}+64}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}-\frac {4}{{\left (\left (-3 y-3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y+486 c_1^{2} y^{2}+324 c_1 y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y-144 y^{2}+64}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}}{2 {\left (\frac {{\left (\left (-3 y-3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y+486 c_1^{2} y^{2}+324 c_1 y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y-144 y^{2}+64}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}-\frac {4}{{\left (\left (-3 y-3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} y+486 c_1^{2} y^{2}+324 c_1 y^{3}+81 y^{4}-144 c_1^{2}-288 c_1 y-144 y^{2}+64}{9 c_1^{2}+18 c_1 y+9 y^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 y+9 y^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}} \end{align*}

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

\[ \int _{}^{y}\frac {2 \left (36 \tau ^{2}+72 c_1 \tau -{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}+36 c_1^{2}-64\right )^{2}}{\left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2} {\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}} \left (-1+\sqrt {4 {\left (\frac {{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}-\frac {4}{{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}\right )}d \tau = x +c_6 \]

Will add steps showing solving for IC soon.

The solution

\[ \int _{}^{y}-\frac {2 \left (36 \tau ^{2}+72 c_2 \tau -{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}+36 c_2^{2}-64\right )^{2}}{\left (1+\sqrt {4 {\left (\frac {{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}-\frac {4}{{\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2} {\left (\left (-3 \tau -3 c_2 +\sqrt {\frac {81 c_2^{4}+324 c_2^{3} \tau +486 c_2^{2} \tau ^{2}+324 c_2 \,\tau ^{3}+81 \tau ^{4}-144 c_2^{2}-288 c_2 \tau -144 \tau ^{2}+64}{9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_2^{2}+18 c_2 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}}d \tau = x +c_5 \]

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

\[ \int _{}^{y}\frac {2 \left (36 \tau ^{2}+72 c_1 \tau -{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}}+36 c_1^{2}-64\right )^{2}}{\left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2} {\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{2}/{3}} \left (-1+\sqrt {4 {\left (\frac {{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}-\frac {4}{{\left (\left (-3 \tau -3 c_1 +\sqrt {\frac {81 c_1^{4}+324 c_1^{3} \tau +486 c_1^{2} \tau ^{2}+324 c_1 \,\tau ^{3}+81 \tau ^{4}-144 c_1^{2}-288 c_1 \tau -144 \tau ^{2}+64}{9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16}}\right ) \left (9 c_1^{2}+18 c_1 \tau +9 \tau ^{2}-16\right )^{2}\right )}^{{1}/{3}}}\right )}^{2}+1}\right )}d \tau = x +c_6 \]

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

\[ y = 2 x +c_4 \]

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

Summary of solutions found

\begin{align*} y &= x +c_3 \\ \end{align*}

Solved as second order missing y ode

Time used: 3.957 (sec)

This is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} {u^{\prime }\left (x \right )}^{2}+u \left (x \right )-1 = 0 \end{align*}

Which is now solved for \(u(x)\) as first order ode.

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} u^{\prime }\left (x \right )&=\sqrt {1-u \left (x \right )} \\ \tag{2} u^{\prime }\left (x \right )&=-\sqrt {1-u \left (x \right )} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {1-u}}d u &= dx\\ -2 \sqrt {1-u}&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {1-u}&= 0 \end{align*}

for \(u \left (x \right )\). 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*} u \left (x \right ) = 1 \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {1-u}}d u &= dx\\ 2 \sqrt {1-u}&= x +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {1-u}&= 0 \end{align*}

for \(u \left (x \right )\). 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*} u \left (x \right ) = 1 \end{align*}

In summary, these are the solution found for \(u(x)\)

\begin{align*} -2 \sqrt {1-u \left (x \right )} &= x +c_1 \\ 2 \sqrt {1-u \left (x \right )} &= x +c_2 \\ u \left (x \right ) &= 1 \\ \end{align*}

For solution \(-2 \sqrt {1-u \left (x \right )} = x +c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} -2 \sqrt {1-y^{\prime }} = x +c_1 \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 {-\frac {1}{4} c_1^{2}-\frac {1}{2} c_1 x -\frac {1}{4} x^{2}+1\, dx}\\ y &= -\frac {x^{3}}{12}-\frac {c_1 \,x^{2}}{4}-\frac {\left (c_1 +2\right ) \left (c_1 -2\right ) x}{4} + c_3 \end{align*}

\begin{align*} y&= -\frac {1}{12} x^{3}-\frac {1}{4} c_1 \,x^{2}-\frac {1}{4} c_1^{2} x +x +c_3 \end{align*}

For solution \(2 \sqrt {1-u \left (x \right )} = x +c_2\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} 2 \sqrt {1-y^{\prime }} = x +c_2 \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 {-\frac {1}{4} c_2^{2}-\frac {1}{2} c_2 x -\frac {1}{4} x^{2}+1\, dx}\\ y &= -\frac {x^{3}}{12}-\frac {c_2 \,x^{2}}{4}-\frac {\left (c_2 +2\right ) \left (c_2 -2\right ) x}{4} + c_4 \end{align*}

\begin{align*} y&= -\frac {1}{12} x^{3}-\frac {1}{4} c_2 \,x^{2}-\frac {1}{4} c_2^{2} x +x +c_4 \end{align*}

For solution \(u \left (x \right ) = 1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 1 \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 {1\, dx}\\ y &= x + c_5 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_1 \,x^{2}-\frac {1}{4} c_1^{2} x +x +c_3 \\ y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_2 \,x^{2}-\frac {1}{4} c_2^{2} x +x +c_4 \\ y &= x +c_5 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= x +c_5 \\ y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_1 \,x^{2}-\frac {1}{4} c_1^{2} x +x +c_3 \\ y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_2 \,x^{2}-\frac {1}{4} c_2^{2} x +x +c_4 \\ \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: 2 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 
      <- 2nd order ODE linearizable_by_differentiation successful 
   ------------------- 
   * Tackling next 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 
      <- 2nd order ODE linearizable_by_differentiation successful 
-> Calling odsolve with the ODE`, diff(y(x), x) = 1, y(x), singsol = none`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful`
 

Maple dsolve solution

Solving time : 0.195 (sec)
Leaf size : 30

dsolve(diff(diff(y(x),x),x)^2+diff(y(x),x) = 1,y(x),singsol=all)
 

\begin{align*} y &= x +c_{1} \\ y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +x +c_{2} \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.025 (sec)
Leaf size : 67

DSolve[{(D[y[x],{x,2}])^2+D[y[x],x]==1,{}},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -\frac {x^3}{12}-\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2 \\ y(x)\to -\frac {x^3}{12}+\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2 \\ \end{align*}