3.26 problem 26

Internal problem ID [7864]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 26
Date solved : Monday, October 21, 2024 at 04:30:23 PM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

Solve

\begin{align*} y^{\prime \prime }+\sin \left (y\right ) {y^{\prime }}^{2}&=0 \end{align*}

3.26.1 Solved as second order missing x ode

Time used: 0.250 (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 ) \left (\frac {d}{d y}p \left (y \right )\right )+\sin \left (y \right ) p \left (y \right )^{2} = 0 \end{align*}

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

In canonical form a linear first order is

\begin{align*} p^{\prime } + q(y)p &= p(y) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(y) &=\sin \left (y \right )\\ p(y) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dy}}\\ &= {\mathrm e}^{\int \sin \left (y \right )d y}\\ &= {\mathrm e}^{-\cos \left (y \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (p \,{\mathrm e}^{-\cos \left (y \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} p \,{\mathrm e}^{-\cos \left (y \right )}&= \int {0 \,dy} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-\cos \left (y \right )}\) gives the final solution

\[ p = {\mathrm e}^{\cos \left (y \right )} c_1 \]

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 } = {\mathrm e}^{\cos \left (y\right )} c_1 \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

\[ \int _{}^{y}\frac {{\mathrm e}^{-\cos \left (\tau \right )}}{c_1}d \tau = x +c_2 \]

Will add steps showing solving for IC soon.

3.26.2 Maple step by step solution

3.26.3 Maple trace

Methods for second order ODEs:
 

3.26.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 21

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

\[ \int _{}^{y}{\mathrm e}^{-\cos \left (\textit {\_a} \right )}d \textit {\_a} -c_1 x -c_2 = 0 \]

3.26.5 Mathematica DSolve solution

Solving time : 1.302 (sec)
Leaf size : 111

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

\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {e^{\sin (K[1])-\cos (K[1]) K[1]}}{c_1}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {e^{\sin (K[1])-\cos (K[1]) K[1]}}{c_1}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {e^{\sin (K[1])-\cos (K[1]) K[1]}}{c_1}dK[1]\&\right ][x+c_2] \\ \end{align*}