2.9 problem 10

Internal problem ID [8098]
Book : Second order enumerated odes
Section : section 2
Problem number : 10
Date solved : Monday, October 21, 2024 at 04:52:12 PM
CAS classification : [_Liouville, [_2nd_order, _reducible, _mu_xy]]

Solve

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

2.9.1 Solved as second nonlinear ode solved by Mainardi Lioville method

Time used: 0.200 (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) &= x\\ g(y) &= y \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 \(g=y\) and \(f=x\), then

\begin{align*} \int -g d y &= \int -y d y\\ &= -\frac {y^{2}}{2}\\ \int -f d x &= \int -x d x\\ &= -\frac {x^{2}}{2} \end{align*}

Substituting the above into Eq(6A) gives

\[ y^{\prime } = c_2 \,{\mathrm e}^{-\frac {y^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{2}} \]

Which is now solved as first order separable ode. The ode \(y^{\prime } = c_2 \,{\mathrm e}^{-\frac {y^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{2}}\) is separable as it can be written as

\begin{align*} y^{\prime }&= c_2 \,{\mathrm e}^{-\frac {y^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{2}}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= {\mathrm e}^{-\frac {x^{2}}{2}} c_2\\ g(y) &= {\mathrm e}^{-\frac {y^{2}}{2}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { {\mathrm e}^{\frac {y^{2}}{2}}\,dy} &= \int { {\mathrm e}^{-\frac {x^{2}}{2}} c_2 \,dx}\\ -\frac {i \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, y}{2}\right )}{2}&=\frac {c_2 \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {2}\, x}{2}\right )}{2}+c_3 \end{align*}

Will add steps showing solving for IC soon.

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

\begin{align*} y = -i \operatorname {RootOf}\left (i \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {2}\, x}{2}\right ) c_2 +i \sqrt {2}\, c_3 -\operatorname {erf}\left (\textit {\_Z} \right ) \sqrt {\pi }\right ) \sqrt {2} \end{align*}

2.9.2 Maple step by step solution

2.9.3 Maple trace

Methods for second order ODEs:
 

2.9.4 Maple dsolve solution

Solving time : 0.010 (sec)
Leaf size : 37

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

\[ y = -i \operatorname {RootOf}\left (i \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {2}\, x}{2}\right ) c_1 +i \sqrt {2}\, c_2 -\operatorname {erf}\left (\textit {\_Z} \right ) \sqrt {\pi }\right ) \sqrt {2} \]

2.9.5 Mathematica DSolve solution

Solving time : 0.128 (sec)
Leaf size : 44

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

\[ y(x)\to -i \sqrt {2} \text {erf}^{-1}\left (i \left (\sqrt {\frac {2}{\pi }} c_2-c_1 \text {erf}\left (\frac {x}{\sqrt {2}}\right )\right )\right ) \]