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Question

# find the general solution of the given differential equation. y” − 2y' − 3y = 3e2t

Solution

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Step 1
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First find solution of homogeneous problem.

$r^2-2r-3=0 \rightarrow r_{1,2}=-1,3$ $\rightarrow$ using quadratic formula

Homogeneous solution:

$y_c=c_1e^{-t}+c_2e^{3t}$

Let $Y=Ae^{2t}$ - because $g(t)=3e^{2t}$

Plug $Y$ into starting equation to find particular solution:

$(Ae^{2t})''-2(Ae^{2t})'-3Ae^{2t}=3e^{2t}\\\\ 4Ae^{2t}-4Ae^{2t}-3Ae^{2t}=3e^{2t} \\\\ -3Ae^{2t}=3e^{2t}\\\\ A=-1$

$Y=-e^{2t}$

Solution: $y=y_c+Y=c_1e^{-t}+c_2e^{3t}-e^{2t}$

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