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Let $g(t)$ be $1$ on the interval $[1,2]$ and zero otherwise.

Consider the differential equation $y''+2y'+2y=g(t)$ with initial conditions $y(0)=1, y'(0)=-2$.

The solution is:

A

$e^{-t}\cos t+u_1(t)h(t-1)-u_2(t)h(t-2)$ where $h(t)=\frac{1}{2}-\frac{1}{2}e^{-t}\cos(t)$

B

$e^{-t}\cos t-e^{-t}\sin t+u_1(t)h(t-1)-u_2(t)h(t-2)$ where $h=\frac{1}{2}-\frac{1}{2}e^{-t}\cos(t)-\frac{1}{2}e^{-t}\sin(t)$

C

$e^{-t}\cos t+u_2(t)h(t-2)-u_1(t)h(t-1)$ where $h(t)=\frac{1}{2}-\frac{1}{2}e^{-t}\cos(t)$

D

$e^{-t}\cos t-e^{-t}\sin t+u_2(t)h(t-2)-u_1(t)h(t-1)$ where $h=\frac{1}{2}-\frac{1}{2}e^{-t}\cos(t)-\frac{1}{2}e^{-t}\sin(t)$

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