Question

# find the general solution of the given differential equation. y” + 2y' + 5y = 3 sin 2t

Solution

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5 (15 ratings)
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First find solution of homogeneous problem.

$r^2+2r+5=0 \rightarrow r_{1,2}=-1\pm 2i$ $\rightarrow$ using quadratic formula

Homogeneous solution:

$y_c=c_1e^{-t}\cos 2t+c_2e^{-t}\sin 2t$

Let $Y=A\cos 2t+B\sin 2t$ - because $g(t)=3\sin 2t$

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

$(A\cos 2t+B\sin 2t)''+2(A\cos 2t+B\sin 2t)'+5(A\cos 2t+B\sin 2t)=3\sin 2t\\\\ -4A\cos 2t-4B\sin 2t-4A\sin 2t+4B\cos 2t+5A\cos 2t+5B\sin 2t=3\sin 2t\\\\ \cos 2t(A+4B)+\sin 2t(B-4A)=3\sin 2t$

We obtain system:

$A+4B=0$

$B-4A=3$ - multiply first equation with 4 and add to second

$B-4A+4A+16B=3 \rightarrow B=\dfrac{3}{17} \rightarrow A=-\dfrac{12}{17}$

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