Answer to Question #105027 in Differential Equations for Vaishali

Question

solve the following equation by changing the independent variable x^2 d^2 y/dx - dy/dx -4x^3 y = 8x^3 sin x^2 , x>0

Answer 1

$$x{d^2y\over dx^2}-{dy\over dx}-4x^3y=8x^3\sin x^2,x>0$$

$${d^2y\over dx^2}-{1\over x}\cdot{dy\over dx}-4x^2y=8x^2\sin x^2$$

Here

$P=-\dfrac{1}{x},$

$Q=-4x^2,$

$R=8x^2\sin x^2$

Changing the independent variable from $x$ to $z$

$${d^2y\over dz^2}+P_1{dy\over dz}+Q_1y=R_1$$

where

$$P_1=\dfrac{{d^2z\over dx^2}+P{dz\over dx}}{({dz\over dx})^2}$$

$$Q_1=\dfrac{Q}{({dz\over dx})^2}$$

$$R_1=\dfrac{R}{({dz\over dx})^2}$$

We choose $z$ such that

$$Q_1=\dfrac{Q}{({dz\over dx})^2}=constant$$

Let

$$Q_1=\dfrac{-4x^2}{({dz\over dx})^2}=-1$$

Then

$${dz\over dx}=2x$$

integrating

$$z=x^2$$

We have

$${d^2z\over dx^2}=2$$

$$P_1=\dfrac{{d^2z\over dx^2}+P{dz\over dx}}{({dz\over dx})^2}=\dfrac{2+(-\dfrac{1}{x})(2x)}{(2)^2}=0$$

$$R_1=\dfrac{R}{({dz\over dx})^2}=\dfrac{8x^2\sin x^2}{(2x)^2}=2\sin z$$

$${d^2y\over dz^2}-y=2\sin z$$

Method of undetermined coefficients

Homogeneous second order differentional equation

$${d^2y\over dz^2}-y=0$$

Characteristic equation

$$\lambda^2-1=0$$ $$\lambda_1=-1,\lambda_2=1$$

The general solution of the homogeneous differential equation is

$$y_0=C_1e^{-z}+C_2e^{z}$$

Let $Y=A\cos z+B\sin z,$ then

$${dY\over dz}=-A\sin z+B\cos z,$$

$${d^2Y\over dz^2}=-A\cos z-B\sin z$$

$$A\cos z+B\sin z+A\cos z+B\sin z=2\sin z$$

$A=0$

$B=1$

$$Y=\sin z$$

The general solution of the nonhomogeneous equation

$$y(z)=y_0+Y$$

$$y(z)=C_1e^{-z}+C_2e^{z}+\sin z$$

Substitute $z=x^2$ and obtain the general solution

$$y(z)=C_1e^{-x^2}+C_2e^{x^2}+\sin x^2$$