Internal problem ID [11075]
Internal file name [OUTPUT/10332_Wednesday_January_24_2024_10_07_30_PM_78114833/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 2.1.2-8. Other equations.
Problem number: 252.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_1"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {x \left (x^{2 n}+a \right ) y^{\prime \prime }+\left (x^{2 n}+a -a n \right ) y^{\prime }-b^{2} x^{-1+2 n} y=0} \]
In normal form the ode \begin {align*} x \left (x^{2 n}+a \right ) y^{\prime \prime }+\left (x^{2 n}+a -a n \right ) y^{\prime }-b^{2} x^{-1+2 n} y&=0 \tag {1} \end {align*}
Becomes \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 )&=\frac {x^{2 n}+a -a n}{x \left (x^{2 n}+a \right )}\\ q \left (x \right )&=-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}}{c}\tag {6} \\ \tau '' &= \frac {-\frac {b^{2} x^{2 n -2} \left (2 n -2\right )}{x \left (x^{2 n}+a \right )}+\frac {2 b^{2} x^{2 n -2} x^{2 n} n}{\left (x^{2 n}+a \right )^{2} x}}{2 c \sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {\frac {-\frac {b^{2} x^{2 n -2} \left (2 n -2\right )}{x \left (x^{2 n}+a \right )}+\frac {2 b^{2} x^{2 n -2} x^{2 n} n}{\left (x^{2 n}+a \right )^{2} x}}{2 c \sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}}+\frac {x^{2 n}+a -a n}{x \left (x^{2 n}+a \right )}\frac {\sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}}{c}}{\left (\frac {\sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}}{c}\right )^2} \\ &=0 \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}d x}{c}\\ &= \frac {\int \sqrt {-\frac {b^{2} x^{2 n -2}}{x^{2 n}+a}}d x}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = c_{1} \cosh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right )+i c_{2} \sinh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cosh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right )+i c_{2} \sinh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cosh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right )+i c_{2} \sinh \left (\frac {b \,\operatorname {arcsinh}\left (\frac {x^{n}}{\sqrt {a}}\right )}{n}\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] Solution is available but has integrals. Trying a simpler solution using Kovacics algorithm... Solution via Kovacic is not simpler. Returning default solution <- linear_1 successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 61
dsolve(x*(x^(2*n)+a)*diff(y(x),x$2)+(x^(2*n)+a-a*n)*diff(y(x),x)-b^2*x^(2*n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{i b \left (\int x^{n -1} \sqrt {-\frac {1}{x^{2 n}+a}}d x \right )}+c_{2} {\mathrm e}^{-i b \left (\int x^{n -1} \sqrt {-\frac {1}{x^{2 n}+a}}d x \right )} \]
✓ Solution by Mathematica
Time used: 0.458 (sec). Leaf size: 47
DSolve[x*(x^(2*n)+a)*y''[x]+(x^(2*n)+a-a*n)*y'[x]-b^2*x^(2*n-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \cosh \left (\frac {b \text {arcsinh}\left (\frac {x^n}{\sqrt {a}}\right )}{n}\right )+i c_2 \sinh \left (\frac {b \text {arcsinh}\left (\frac {x^n}{\sqrt {a}}\right )}{n}\right ) \]