# find the solution of the initial-value problem-second order

Q.

$$\text{ Three solutions of a second order linear equation L[y]=g(t) are }$$

$$\psi_1(t)=3e^t+e^{t^2}$$

$$\psi_2(t)=7e^t+e^{t^2}$$

$$\psi_3(t)=5e^t+e^{-t^3}+e^{t^2}$$

$$\text{ find the solution of the initial-value problem }$$

$$L[y]=g, y(0)=1, y'(0)=2$$

I first found the general equation:

$\psi_2-\psi_1=4e^t$

$\psi_3-\psi_2=-2e^t+e^{-t^3}$

so the general solution is:

$$y(t)=c_14e^t+c_2(-2e^t+e^{-t^3})+\psi(t)$$

but I do not know how to find the solution of the initial value problem, what I am thinking is to pick $\psi_1(t)$ and use that as my particular solution. Plug in t=0 and solve for 1 and find the derivative of y'(t), plug in 0 and solve for 2 to obtain the other constant.

Does that sound like I am on the right path?

• You are on the right way. Just use $y(t)=c_14e^t+c_2(-2e^t+e^{-t^3}).$ – mfl Mar 28 '15 at 20:09
• @mfl, I just solved it out and I got the answer. so yay! – kero Mar 28 '15 at 20:10