Question

Y" - 4y = (x2 - 3) sin 2x

Answer 1

`y''-4y=0`

has characteristic equation


`r^2-4=0`

which has roots at
`r=\pm2`, giving the characteristic solution


`y_c=C_1e^(2x)+C_2e^(-2x)`

For the nonhomogeneous part of the ODE, let
`y_p=(a_2x^2+a_1x+a_0)\sin2x+(b_2x^2+b_1x+b_0)\cos2x`. Then


`{y_p}''=(-4b_2x^2+(8a_2-b_1)x+4a_1-4b_0+2b_2)\cos2x+(-4a_2x^2+(-4a_1-8b_2)x-4a_0+2a_2-4b_1)\sin2x`

Substituting into the ODE gives


`(-8b_2x^2+(8a_2-b_1)x+4a_1-8b_0+2b_2)\cos2x+(-8a_2x^2+(-8a_1-8b_2)x-8a_0+2a_2-4b_1)\sin2x=(x^2-3)\sin2x`

It follows that


`\begin{cases}-8b_2=0\\8a_2-8b_1=0\\4a_1-8b_0+2b_2=0\\-8a_2=1\\-8a_1-8b_2=0\\-8a_0+2a_2-4b_1=-3\end{cases}\implies\begin{cases}a_2=-\frac18\\\\a_1=0\\\\a_0=(13)/(32)\\\\b_2=0\\\\b_1=-\frac18\\\\b_0=0\end{cases}`

which yields the particular solution


`y_p=-\frac18x^2\sin2x+(13)/(32)\sin2x-\frac18x\cos2x`

So the general solution is


`y=y_c+y_p`

`y=C_1e^(2x)+C_2e^(-2x)-\frac18x^2\sin2x+(13)/(32)\sin2x-\frac18x\cos2x`