TITLE: How to solve $\ddot{y} = t^2$
QUESTION [6 upvotes]: I am having trouble to solve $\ddot{y} = t^2$.  
Step 1: Find the homogenous solution: (this part is simple)
$$y_H = c_1+c_2t$$ 
Step 2: Find the particular solution:  Since the nonhomogenous part is a polynomial of degree $2$, so $$y_P = At^2+Bt+C$$  
Step 3:  $y = y_H+y_P$ and then plug in:  
We have $$2A = t^2$$
which is not correct.  
How should I modify this?  please advise, thanks!

REPLY [4 votes]: $$y''=t^2$$
Integrate,
$$y'=\int t^2dt=\frac { t^3}{3}+K$$
Integrate again,
$$y=\int \frac { t^3}{3}+Kdt=\frac { t^4}{12}+K_1t+K_2$$