How to solve $\ddot{y} = t^2$

Question

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!

Answer 1

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

Answer 2

Hint: $y_p=\frac {t^4}{12}+At+B$