Answer to Question #120663 in Differential Equations for Kunal

Question

Solve the following differential equation by changing the independent variable:
x.d^2y/dx^2+(4x^2-1)dy/dx+4x^3=2x^3

Answer 1

$$x{d^2y\over dx^2}+(4x^2-1){dy\over dx}+4x^3y=2x^3$$ $${d^2y\over dx^2}+(4x-{1\over x}){dy\over dx}+4x^2y=2x^2,x\not=0$$

Here

$P=4x-\dfrac{1}{x},$

$Q=4x^2$

$R=2x^2$

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

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

$$P_1=\dfrac{{d^2z\over dx^2}+P{dz\over dx}}{({dz\over dx})^2}=\dfrac{2+(4x-\dfrac{1}{x})(2x)}{(2x)^2}=2$$

$$R_1=\dfrac{R}{({dz\over dx})^2}=\dfrac{2x^2}{(2x)^2}={1\over 2}$$

$${d^2y\over dz^2}+2{dy\over dz}+y={1\over 2}$$

Method of undetermined coefficients

Homogeneous second order differentional equation

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

Characteristic equation

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

The general solution of the homogeneous differential equation is

$$y_0=C_1ze^{-z}+C_2e^{-z}$$

Let $Y=A,$ then

$${dY\over dz}=0, {d^2Y\over dz^2}=0$$

$$Y={1\over 2}$$

The general solution of the nonhomogeneous equation

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

$$y(z)=C_1ze^{-z}+C_2e^{-z}+{1\over 2}$$

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

$$y(z)=C_1x^2e^{-x^2}+C_2e^{-x^2}+{1\over 2}$$