Question

Find the general solution of the given differential equation 2y"' + 3y' +y= 2 + 3 sint

Answer 1

I suppose you meant to have the second derivative as the first term:

$2y''+3y'+y=2+3\sin t$

The corresponding homogeneous equation

$2y''+3y'+y=0$

has characteristic equation

$2r^2+3r+1=(2r+1)(r+1)=0$

with roots at $r=-\dfrac12$ and $r=-1$, giving the characteristic solution

$y_c=C_1e^{-t/2}+C_2e^{-t}$

For the nonhomogeneous equation, assume a solution of the form

$y_p=a+b\sin t+c\cos t$

with derivatives

${y_p}'=b\cos t-c\sin t$

${y_p}''=-b\sin t-c\cos t$

Substituting these into the ODE gives

$2(-b\sin t-c\cos t)+3(b\cos t-c\sin t)+(a+b\sin t+c\cos t)=2+3\sin t$

$-(b+3c)\sin t+(3b-c)\cos t+a=2+3\sin t$

$\implies a=2,b=-\dfrac3{10},c=-\dfrac9{10}$

Then the ODE has solution

$\boxed{y(t)=C_1e^{-t/2}+C_2e^{-t}+2-\dfrac3{10}\sin t-\dfrac9{10}\cos t}$