Question

(x^2)d^2y/dx^2 –dy/dx –4x^3y =8x^3 sinx^2 ,x>0

Answer 1

Firstly the question is,

$xd^2y/dx^2 –dy/dx –4x^3y =8x^3 sin(x^2)$

Substituting $x^2=t, 2xdx/dy=dt/dy \implies dy/dx=2xdy/dt$ , we differentiate again to get;

$d^2y/dx^2=2dy/dt+4x^2d^2y/dt^2$

Substituting these values of $dy/dx, d^2y/dx^2$ back in the given equation, we get;

$4x^3(d^2y/dt^2-y)=8x^3sin(t)$

Dividing throughout by $4x^3$ ;

$\implies d^2y/dt^2-y=2sin(t)$

Solving this for complimentary function and particular solution we get;

Complimentary function : $c_1e^t+c_2e^{-t}$

Particular solution : $-sin(t)$

Thus, solution is $y(t)=c_1e^t+c_2e^{-t}-sin(t)$

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