Question

Solve the differential equation :
(x^2)y'' - y' - 4(x^3)y = 8(x^3)sin(x^2)

Answer 1

We need to solve : $(x^2)y'' - y' - 4(x^3)y = 8(x^3)sin(x^2)$

Substituting $x^2=t \implies 2xdx=dt$

$\implies dy/dx=2xdy/dt$ ---(1)

Differentiating this again, we get;

$d^2y/dx^2=4xd^2y/dt^2-(2/x)(dy/dt)$ ---(2)

Using (1) and (2) in the given equation, we get

$4x^3(y''-y)=8x^3sint$

$\implies y''-y=2sint$

Solving this, we get;

$(D^2-1)y=2sint$

Complimentary function : $c_1e^t+c_2e^{-t}$ (Since, roots of auxiliary equation are 1, -1)

Particular solution : $y=2sint/(D^2-1)=2sint/(-(1)^2-1)=-sint$

$=-sint$

Thus, final solution is $y=c_1e^t+c_2e^{-t}-sint$

Substituting $t=x^2$ back, we get;

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