28.5 problem 65

Internal problem ID [10888]
Internal file name [OUTPUT/10145_Sunday_December_24_2023_05_15_08_PM_11268123/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 2.1.2-3 Equation of form \((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 65.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"

Maple gives the following as the ode type

[[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

\[ \boxed {x y^{\prime \prime }+n y^{\prime }+b \,x^{1-2 n} y=0} \]

28.5.1 Solving as second order change of variable on x method 2 ode

In normal form the ode \begin {align*} x y^{\prime \prime }+n y^{\prime }+b \,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 {n}{x}\\ q \left (x \right )&=b \,x^{-2 n} \end {align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \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 \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}

This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {n}{x}d x \right )}d x\\ &= \int e^{-n \ln \left (x \right )} \,dx\\ &= \int x^{-n}d x\\ &= -\frac {x^{-n +1}}{n -1}\tag {6} \end {align*}

Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {b \,x^{-2 n}}{x^{-2 n}}\\ &= b\tag {7} \end {align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+b y \left (\tau \right )&=0 \end {align*}

The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=b\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+b \,{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+b = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=b\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (b\right )}\\ &= \pm \sqrt {-b} \end {align*}

Hence \begin{align*} \lambda _1 &= + \sqrt {-b} \\ \lambda _2 &= - \sqrt {-b} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \sqrt {-b} \\ \lambda _2 &= -\sqrt {-b} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (\sqrt {-b}\right )\tau } +c_{2} e^{\left (-\sqrt {-b}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\sqrt {-b}\, \tau }+c_{2} {\mathrm e}^{-\sqrt {-b}\, \tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-b}\, x^{-n +1}}{n -1}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b}\, x^{-n +1}}{n -1}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-b}\, x^{-n +1}}{n -1}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b}\, x^{-n +1}}{n -1}} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{-\frac {\sqrt {-b}\, x^{-n +1}}{n -1}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b}\, x^{-n +1}}{n -1}} \] Verified OK.

28.5.2 Solving as second order change of variable on x method 1 ode

In normal form the ode \begin {align*} x y^{\prime \prime }+n y^{\prime }+b \,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 {n}{x}\\ q \left (x \right )&=b \,x^{-2 n} \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 {b \,x^{-2 n}}}{c}\tag {6} \\ \tau '' &= -\frac {b \,x^{-2 n} n}{c \sqrt {b \,x^{-2 n}}\, x} \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 {b \,x^{-2 n} n}{c \sqrt {b \,x^{-2 n}}\, x}+\frac {n}{x}\frac {\sqrt {b \,x^{-2 n}}}{c}}{\left (\frac {\sqrt {b \,x^{-2 n}}}{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 {b \,x^{-2 n}}d x}{c}\\ &= -\frac {x \sqrt {b \,x^{-2 n}}}{c \left (n -1\right )} \end {align*}

Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right )-c_{2} \sin \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right )-c_{2} \sin \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} \cos \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right )-c_{2} \sin \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right ) \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 43

dsolve(x*diff(y(x),x$2)+n*diff(y(x),x)+b*x^(1-2*n)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \sin \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right )+c_{2} \cos \left (\frac {x^{-n +1} \sqrt {b}}{n -1}\right ) \]

✓ Solution by Mathematica

Time used: 0.072 (sec). Leaf size: 52

DSolve[x*y''[x]+n*y'[x]+b*x^(1-2*n)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 \cos \left (\frac {\sqrt {b} x^{1-n}}{n-1}\right )+c_2 \sin \left (\frac {\sqrt {b} x^{1-n}}{1-n}\right ) \]