2.12 problem 13

Internal problem ID [8101]
Book : Second order enumerated odes
Section : section 2
Problem number : 13
Date solved : Monday, October 21, 2024 at 04:52:16 PM
CAS classification : [_Liouville, [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

Solve

\begin{align*} 10 y^{\prime \prime }+x^{2} y^{\prime }+\frac {3 {y^{\prime }}^{2}}{y}&=0 \end{align*}

2.12.1 Solved as second nonlinear ode solved by Mainardi Lioville method

Time used: 0.204 (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 {x^{2}}{10}\\ g(y) &= \frac {3}{10 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 \(g=\frac {3}{10 y}\) and \(f=\frac {x^{2}}{10}\), then

\begin{align*} \int -g d y &= \int -\frac {3}{10 y}d y\\ &= -\frac {3 \ln \left (y\right )}{10}\\ \int -f d x &= \int -\frac {x^{2}}{10}d x\\ &= -\frac {x^{3}}{30} \end{align*}

Substituting the above into Eq(6A) gives

\[ y^{\prime } = \frac {c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}}}{y^{{3}/{10}}} \]

Which is now solved as first order separable ode. The ode \(y^{\prime } = \frac {c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}}}{y^{{3}/{10}}}\) is separable as it can be written as

\begin{align*} y^{\prime }&= \frac {c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}}}{y^{{3}/{10}}}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}}\\ g(y) &= \frac {1}{y^{{3}/{10}}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { y^{{3}/{10}}\,dy} &= \int { c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}} \,dx}\\ \frac {10 y^{{13}/{10}}}{13}&=\int c_2 \,{\mathrm e}^{-\frac {x^{3}}{30}}d x +2 c_3 \end{align*}

Will add steps showing solving for IC soon.

2.12.2 Maple step by step solution

2.12.3 Maple trace

Methods for second order ODEs:
 

2.12.4 Maple dsolve solution

Solving time : 0.013 (sec)
Leaf size : 55

dsolve(10*diff(diff(y(x),x),x)+x^2*diff(y(x),x)+3/y(x)*diff(y(x),x)^2 = 0, 
       y(x),singsol=all)
 

\[ -\frac {\frac {3 c_1 \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right ) x \,{\mathrm e}^{-\frac {x^{3}}{60}} 30^{{1}/{6}}}{4}+\left (c_1 x \,{\mathrm e}^{-\frac {x^{3}}{30}}+c_2 -\frac {10 y^{{13}/{10}}}{13}\right ) \left (x^{3}\right )^{{1}/{6}}}{\left (x^{3}\right )^{{1}/{6}}} = 0 \]

2.12.5 Mathematica DSolve solution

Solving time : 66.802 (sec)
Leaf size : 73

DSolve[{10*D[y[x],{x,2}]+x^2*D[y[x],x]+3/y[x]*(D[y[x],x])^2==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_2 \exp \left (\int _1^x\frac {30 e^{-\frac {1}{30} K[1]^3} \sqrt [3]{K[1]^3}}{30 c_1 \sqrt [3]{K[1]^3}-13 \sqrt [3]{30} \Gamma \left (\frac {1}{3},\frac {K[1]^3}{30}\right ) K[1]}dK[1]\right ) \]