Answer to Question #105437 in Differential Equations for Zoya

Question

Solve the following differential equations changing the independent variables
X^2 d^2y/dx^2 - dy/dx - 4x^3 y= 8x^3 sinx^2 ,X>0.

Answer 1

Solve

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

Divide by $x$

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

Compare with

$$y''+Py'+Qy=R$$

We have

$$P=-{1 \over x},Q=-4x^2,R=8x^2\sin(x^2)$$

Chooze $z$ such that $(dz/dx)^2=4x^2$ or $dz/dx=2x$ so that $z=x^2$

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

$${d^2y \over dx^2}=2{dy \over dz}+4x^2{d^2y \over dz^2}$$

$$2{dy \over dz}+4x^2{d^2y \over dz^2}-{1 \over x}(2x{dy \over dz})-4x^2y(z)=8x^2\sin(x^2)$$

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

We will get

$$(D_1^2-1)y=2\sin z$$

$$C.F.=c_1e^z+c_2e^{-z}$$

$$P.I.={1 \over D_1^2-1}(2\sin z)=2{1 \over -1^2-1}\cdot\sin z=-\sin z$$

The required solution is

$$y(z)=c_1e^z+c_2e^{-z}-\sin z$$

$$y=c_1e^{x^2}+c_2e^{-x^2}-\sin (x^2)$$