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

I suppose you meant to have the second derivative as the first term:[tex]2y''+3y'+y=2+3\sin t[/tex]The corresponding homogeneous equation[tex]2y''+3y'+y=0[/tex]has characteristic equation[tex]2r^2+3r+1=(2r+1)(r+1)=0[/tex]with roots at [tex]r=-\dfrac12[/tex] and [tex]r=-1[/tex], giving the characteristic solution[tex]y_c=C_1e^{-t/2}+C_2e^{-t}[/tex]For the nonhomogeneous equation, assume a solution of the form[tex]y_p=a+b\sin t+c\cos t[/tex]with derivatives[tex]{y_p}'=b\cos t-c\sin t[/tex][tex]{y_p}''=-b\sin t-c\cos t[/tex]Substituting these into the ODE gives[tex]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[/tex][tex]-(b+3c)\sin t+(3b-c)\cos t+a=2+3\sin t[/tex][tex]\implies a=2,b=-\dfrac3{10},c=-\dfrac9{10}[/tex]Then the ODE has solution[tex]\boxed{y(t)=C_1e^{-t/2}+C_2e^{-t}+2-\dfrac3{10}\sin t-\dfrac9{10}\cos t}[/tex]