2.11 problem 12

Internal problem ID [8100]
Book : Second order enumerated odes
Section : section 2
Problem number : 12
Date solved : Monday, October 21, 2024 at 04:52:14 PM
CAS classification : [_Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

Solve

\begin{align*} 3 y^{\prime \prime }+\cos \left (x \right ) y^{\prime }+\sin \left (y\right ) {y^{\prime }}^{2}&=0 \end{align*}

2.11.1 Solved as second nonlinear ode solved by Mainardi Lioville method

Time used: 0.338 (sec)

The ode has the Liouville form given by

\begin{align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end{align*}

Where in this problem

\begin{align*} f(x) &= \frac {\cos \left (x \right )}{3}\\ g(y) &= \frac {\sin \left (y \right )}{3} \end{align*}

Dividing through by \(y^{\prime }\) then Eq (1A) becomes

\begin{align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end{align*}

But the first term in Eq (2A) can be written as

\begin{align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end{align*}

And the last term in Eq (2A) can be written as

\begin{align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end{align*}

Substituting (3A,4A) back into (2A) gives

\begin{align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end{align*}

Integrating the above w.r.t. \(x\) gives

\begin{align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_1 \end{align*}

Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives

\begin{align*} y^{\prime } &= c_2 e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end{align*}

Where \(c_2\) is a new arbitrary constant. But since \(g=\frac {\sin \left (y \right )}{3}\) and \(f=\frac {\cos \left (x \right )}{3}\), then

\begin{align*} \int -g d y &= \int -\frac {\sin \left (y \right )}{3}d y\\ &= \frac {\cos \left (y\right )}{3}\\ \int -f d x &= \int -\frac {\cos \left (x \right )}{3}d x\\ &= -\frac {\sin \left (x \right )}{3} \end{align*}

Substituting the above into Eq(6A) gives

\[ y^{\prime } = c_2 \,{\mathrm e}^{\frac {\cos \left (y\right )}{3}} {\mathrm e}^{-\frac {\sin \left (x \right )}{3}} \]

Which is now solved as first order separable ode. The ode \(y^{\prime } = c_2 \,{\mathrm e}^{\frac {\cos \left (y\right )}{3}} {\mathrm e}^{-\frac {\sin \left (x \right )}{3}}\) is separable as it can be written as

\begin{align*} y^{\prime }&= c_2 \,{\mathrm e}^{\frac {\cos \left (y\right )}{3}} {\mathrm e}^{-\frac {\sin \left (x \right )}{3}}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= {\mathrm e}^{-\frac {\sin \left (x \right )}{3}} c_2\\ g(y) &= {\mathrm e}^{\frac {\cos \left (y \right )}{3}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { {\mathrm e}^{-\frac {\cos \left (y \right )}{3}}\,dy} &= \int { {\mathrm e}^{-\frac {\sin \left (x \right )}{3}} c_2 \,dx}\\ \int _{}^{y}{\mathrm e}^{-\frac {\cos \left (\tau \right )}{3}}d \tau = \int {\mathrm e}^{-\frac {\sin \left (x \right )}{3}} c_2 d x +2 c_3 \end{align*}

Will add steps showing solving for IC soon.

2.11.2 Maple step by step solution

2.11.3 Maple trace

Methods for second order ODEs:
 

2.11.4 Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 27

dsolve(3*diff(diff(y(x),x),x)+cos(x)*diff(y(x),x)+sin(y(x))*diff(y(x),x)^2 = 0, 
       y(x),singsol=all)
 

\[ \int _{}^{y}{\mathrm e}^{-\frac {\cos \left (\textit {\_a} \right )}{3}}d \textit {\_a} -c_1 \left (\int {\mathrm e}^{-\frac {\sin \left (x \right )}{3}}d x \right )-c_2 = 0 \]

2.11.5 Mathematica DSolve solution

Solving time : 0.855 (sec)
Leaf size : 47

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

\[ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}e^{-\frac {1}{3} \cos (K[1])}dK[1]\&\right ]\left [\int _1^x-e^{-\frac {1}{3} \sin (K[2])} c_1dK[2]+c_2\right ] \]