Internal problem ID [10869]
Internal file name [OUTPUT/10126_Sunday_December_24_2023_05_13_38_PM_911658/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-2 Equation of form
\(y''+f(x)y'+g(x)y=0\)
Problem number: 46.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_1", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+2 a \,x^{n} y^{\prime }+a \left (a \,x^{2 n}+n \,x^{n -1}\right ) y=0} \]
The ode satisfies this form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \] Where \( p(x) = 2 a \,x^{n}\). Therefore, there is an integrating factor given by \begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int 2 a \,x^{n} \, dx} \\ &= {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} \end {align*}
Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential \begin{align*} \left ( M(x) y \right )'' &= 0 \\ \left ( {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} y \right )'' &= 0 \\ \end{align*} Integrating once gives \[ \left ( {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} y \right )' = c_{1} \] Integrating again gives \[ \left ( {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} y \right ) = c_{1} x +c_{2} \] Hence the solution is \begin{align*} y &= \frac {c_{1} x +c_{2}}{{\mathrm e}^{\frac {a \,x^{n +1}}{n +1}}} \\ \end{align*} Or \[ y = c_{1} x \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}+c_{2} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}+c_{2} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}+c_{2} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \] Verified OK.
In normal form the given ode is written as \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=2 a \,x^{n}\\ q \left (x \right )&=x^{2 n} a^{2}+\frac {x^{n} a n}{x} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= x^{2 n} a^{2}+\frac {x^{n} a n}{x} - \frac {\left (2 a \,x^{n}\right )'}{2}- \frac {\left (2 a \,x^{n}\right )^2}{4} \\ &= x^{2 n} a^{2}+\frac {x^{n} a n}{x} - \frac {\left (\frac {2 x^{n} a n}{x}\right )}{2}- \frac {\left (4 x^{2 n} a^{2}\right )}{4} \\ &= x^{2 n} a^{2}+\frac {x^{n} a n}{x} - \left (\frac {x^{n} a n}{x}\right )-x^{2 n} a^{2}\\ &= 0 \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {2 a \,x^{n}}{2} }\\ &= {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} v^{\prime \prime }\left (x \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} = 0 \end {align*}
Which is now solved for \(v \left (x \right )\) Integrating twice gives the solution \[ v \left (x \right )= c_{1} x + c_{2} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} x +c_{2}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} x +c_{2} \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} x +c_{2} \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} x +c_{2} \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 24
dsolve(diff(y(x),x$2)+2*a*x^n*diff(y(x),x)+a*(a*x^(2*n)+n*x^(n-1))*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \left (c_{2} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.112 (sec). Leaf size: 28
DSolve[y''[x]+2*a*x^n*y'[x]+a*(a*x^(2*n)+n*x^(n-1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (c_2 x+c_1) e^{-\frac {a x^{n+1}}{n+1}} \]