Toronto Math Forum
MAT2442014F => MAT244 MathLectures => Topic started by: Chang Peng (Eddie) Liu on December 06, 2014, 01:34:45 PM

$$y^{(4)}âˆ’3yâ€³âˆ’4y=\sin(t)+8t.$$
apparently the particular solution is $Y = At\cos(t) + Btsin(t) + Ct + D$
I know why it's $At\cos(t) + Bt\sin(t)$ , but I have no idea where the $Ct + D$ is coming from. Can someone please explain this?

Observe that $r_{1,2}=\pm 2$, $r_{3,4}=\pm i$.
Because righthand expression is $f_1+f_2$ with $f_1=\sin(t)$ and $f_2=8t$. Then we need $Y=Y_1+Y_2$ with $Y_1= (A\cos(t)+B\sin (t))t $ and $Y_2= Ct+D$.

That makes sense. Thank you professor.

Prof, why you make the particular solution Y1=(Acos(t)+Bsin(t))t^2?
I think for this question, the particular solution multiplies by t is enough, i.e Y1= Atcos(t)+Btsin(t).
Thank you.

Yes, it was a misprint. Corrected