Question

(D^2 - 1) y = x^2 sinx

Answer 1

$(D^2-1)y=x^2\sin x$

First consider the homogeneous part,

$(D^2-1)y=\dfrac{\mathrm d^2y}{\mathrm dx^2}-y=0$

which has characteristic equation

$r^2-1=(r-1)(r+1)=0$

This has roots $r=\pm1$, so the characteristic solution is

$y_c=C_1e^x+C_2e^{-x}$

For the nonhomogeneous part, consider a solution of the form

$y_p=(a_2x^2+a_1x+a_0)\sin x+(b_2x^2+b_1x+b_0)\cos x$

which has second derivative

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=(-b_2x^2+(4a_2x-b_1)x+2a_1-b_0+2b_2)\cos x+(a_2x^2+(a_1+4b_2)x+a_0-2a_2+2b_1)\sin x$

Substituting into the ODE gives

$(-2b_2x^2+(4a_2-2b_1)x+2a_1-2b_0+2b_2)\cos x+(-2a_2x^2+(-2a_2-4b_2)x-2a_0-2b_1+2a_2)\sin x=x^2\sin x$

Matching up coefficients gives the system

$\begin{cases}-2b_2=0\\4a_2-2b_1=0\\2a_1-2b_0+2b_2=0\\-2a_2=1\\-2a_1-4b_2=0\\-2a_0-2b_1+2a_2=0\end{cases}$

which has solutions

$a_2=-\dfrac12,a_1=0,a_0=\dfrac12,b_2=0,b_1=-1,b_0=0$

So the particular solution is

$y_p=\left(-\dfrac12x^2+\dfrac12\right)\sin x-x\cos x$

Therefore the general solution is

$y=y_c+y_p$
$y=C_1e^x+C_2e^{-x}+\left(-\dfrac12x^2+\dfrac12\right)\sin x-x\cos x$