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

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

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

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

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

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

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:


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?

  • 1
    $\begingroup$ You are on the right way. Just use $y(t)=c_14e^t+c_2(-2e^t+e^{-t^3}).$ $\endgroup$ – mfl Mar 28 '15 at 20:09
  • $\begingroup$ @mfl, I just solved it out and I got the answer. so yay! $\endgroup$ – kero Mar 28 '15 at 20:10

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