1.47 problem 47

Internal problem ID [8084]
Book : Second order enumerated odes
Section : section 1
Problem number : 47
Date solved : Monday, October 21, 2024 at 04:48:02 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

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

1.47.1 Solved as second order missing x ode

Time used: 5.078 (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^{2} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end{align*}

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

The ode \(p^{\prime } = -\frac {1}{\sqrt {-p}\, y}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {1}{\sqrt {-p}\, y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {1}{y}\\ g(p) &= \frac {1}{\sqrt {-p}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { \sqrt {-p}\,dp} &= \int { -\frac {1}{y} \,dy}\\ -\frac {2 \left (-p\right )^{{3}/{2}}}{3}&=\ln \left (\frac {1}{y}\right )+c_1 \end{align*}

Solving for \(p\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p = -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \end{align*}

Solving Eq. (2)

The ode \(p^{\prime } = \frac {1}{\sqrt {-p}\, y}\) is separable as it can be written as

\begin{align*} p^{\prime }&= \frac {1}{\sqrt {-p}\, y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= \frac {1}{y}\\ g(p) &= \frac {1}{\sqrt {-p}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { \sqrt {-p}\,dp} &= \int { \frac {1}{y} \,dy}\\ -\frac {2 \left (-p\right )^{{3}/{2}}}{3}&=\ln \left (y \right )+c_2 \end{align*}

Solving for \(p\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p = -\frac {\left (-12 \ln \left (y \right )-12 c_2 \right )^{{2}/{3}}}{4} \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 } = -\frac {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{2}/{3}}}{4} \end{align*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-12 \ln \left (y \right )-12 c_2 \right )^{{2}/{3}}}{4}&= 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 = {\mathrm e}^{-c_2} \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 {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \end{align*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4}&= 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 = {\mathrm e}^{c_1} \end{align*}

Will add steps showing solving for IC soon.

1.47.2 Maple step by step solution

1.47.3 Maple trace

Methods for second order ODEs:
 

1.47.4 Maple dsolve solution

Solving time : 0.069 (sec)
Leaf size : 225

dsolve(y(x)^2*diff(diff(y(x),x),x)^2+diff(y(x),x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= c_1 \\ y &= 0 \\ -4 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -4 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )-2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )-2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_2 \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \end{align*}

1.47.5 Mathematica DSolve solution

Solving time : 3.841 (sec)
Leaf size : 449

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

\begin{align*} y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i (-c_1)} (-\log (\text {$\#$1})-i (-1) c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i (-1) c_1-\log (\text {$\#$1})\right )}{(-i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i (-c_1)} (-\log (\text {$\#$1})+i (-c_1)){}^{2/3} \Gamma \left (\frac {1}{3},i (-c_1)-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ \end{align*}