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Question

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

Solution

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Answered 10 months ago
Answered 10 months ago
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First find solution of homogeneous problem.

r22r3=0r1,2=1,3r^2-2r-3=0 \rightarrow r_{1,2}=-1,3 \rightarrow using quadratic formula

Homogeneous solution:

yc=c1et+c2e3ty_c=c_1e^{-t}+c_2e^{3t}

Let Y=Ae2tY=Ae^{2t} - because g(t)=3e2tg(t)=3e^{2t}

Plug YY into starting equation to find particular solution:

(Ae2t)2(Ae2t)3Ae2t=3e2t4Ae2t4Ae2t3Ae2t=3e2t3Ae2t=3e2tA=1(Ae^{2t})''-2(Ae^{2t})'-3Ae^{2t}=3e^{2t}\\\\ 4Ae^{2t}-4Ae^{2t}-3Ae^{2t}=3e^{2t} \\\\ -3Ae^{2t}=3e^{2t}\\\\ A=-1

Y=e2tY=-e^{2t}

Solution: y=yc+Y=c1et+c2e3te2ty=y_c+Y=c_1e^{-t}+c_2e^{3t}-e^{2t}

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