Internal problem ID [7453]
Internal file name [OUTPUT/6420_Sunday_June_05_2022_04_51_42_PM_54690149/index.tex]
Book: Second order enumerated odes
Section: section 2
Problem number: 13.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[_Liouville, [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {10 y^{\prime \prime }+x^{2} y^{\prime }+\frac {3 {y^{\prime }}^{2}}{y}=0} \]
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 \[ y^{\prime } = \frac {c_{2} {\mathrm e}^{-\frac {x^{3}}{30}}}{y^{\frac {3}{10}}} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{2} {\mathrm e}^{-\frac {x^{3}}{30}}}{y^{\frac {3}{10}}} \end {align*}
Where \(f(x)=c_{2} {\mathrm e}^{-\frac {x^{3}}{30}}\) and \(g(y)=\frac {1}{y^{\frac {3}{10}}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y^{\frac {3}{10}}}} \,dy &= c_{2} {\mathrm e}^{-\frac {x^{3}}{30}} \,d x \\ \int { \frac {1}{\frac {1}{y^{\frac {3}{10}}}} \,dy} &= \int {c_{2} {\mathrm e}^{-\frac {x^{3}}{30}} \,d x} \\ \frac {10 y^{\frac {13}{10}}}{13}&=\frac {10^{\frac {1}{3}} 9^{\frac {2}{3}} c_{2} \left (\frac {3 \,243^{\frac {1}{6}} 10^{\frac {5}{6}} x \,{\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{40 \left (x^{3}\right )^{\frac {1}{6}}}+\frac {3 \,30^{\frac {5}{6}} {\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{x^{2} \left (x^{3}\right )^{\frac {1}{6}}}\right )}{9}+c_{3} \\ \end{align*} The solution is \[ \frac {10 y^{\frac {13}{10}}}{13}-\frac {10^{\frac {1}{3}} 9^{\frac {2}{3}} c_{2} \left (\frac {3 \,243^{\frac {1}{6}} 10^{\frac {5}{6}} x \,{\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{40 \left (x^{3}\right )^{\frac {1}{6}}}+\frac {3 \,30^{\frac {5}{6}} {\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{x^{2} \left (x^{3}\right )^{\frac {1}{6}}}\right )}{9}-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {10 y^{\frac {13}{10}}}{13}-\frac {10^{\frac {1}{3}} 9^{\frac {2}{3}} c_{2} \left (\frac {3 \,243^{\frac {1}{6}} 10^{\frac {5}{6}} x \,{\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{40 \left (x^{3}\right )^{\frac {1}{6}}}+\frac {3 \,30^{\frac {5}{6}} {\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{x^{2} \left (x^{3}\right )^{\frac {1}{6}}}\right )}{9}-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {10 y^{\frac {13}{10}}}{13}-\frac {10^{\frac {1}{3}} 9^{\frac {2}{3}} c_{2} \left (\frac {3 \,243^{\frac {1}{6}} 10^{\frac {5}{6}} x \,{\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{40 \left (x^{3}\right )^{\frac {1}{6}}}+\frac {3 \,30^{\frac {5}{6}} {\mathrm e}^{-\frac {x^{3}}{60}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right )}{x^{2} \left (x^{3}\right )^{\frac {1}{6}}}\right )}{9}-c_{3} = 0 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 55
dsolve(10*diff(y(x),x$2)+x^2*diff(y(x),x)+3/y(x)*(diff(y(x),x))^2=0,y(x), singsol=all)
\[ -\frac {3 \left (c_{1} x \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {x^{3}}{30}\right ) {\mathrm e}^{-\frac {x^{3}}{60}} 30^{\frac {1}{6}}+\frac {4 \left (x^{3}\right )^{\frac {1}{6}} \left (c_{1} x \,{\mathrm e}^{-\frac {x^{3}}{30}}+c_{2} -\frac {10 y \left (x \right )^{\frac {13}{10}}}{13}\right )}{3}\right )}{4 \left (x^{3}\right )^{\frac {1}{6}}} = 0 \]
✓ Solution by Mathematica
Time used: 66.444 (sec). Leaf size: 73
DSolve[10*y''[x]+x^2*y'[x]+3/y[x]*(y'[x])^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \exp \left (\int _1^x\frac {30 e^{-\frac {1}{30} K[1]^3} \sqrt [3]{K[1]^3}}{30 c_1 \sqrt [3]{K[1]^3}-13 \sqrt [3]{30} \Gamma \left (\frac {1}{3},\frac {K[1]^3}{30}\right ) K[1]}dK[1]\right ) \]