Related questions with answers

Consider the equation y”−y′−2y=0y”−y'−2y=0. (a) Show that y1(t)=e−ty_1(t)=e^{−t} and y2(t)=e2ty_2(t)=e^{2t} form a fundamental set of solutions. (b) Let y3(t)=−2e2ty_3(t)=−2e^{2t}, y4(t)=y1(t)+2y2(t)y_4(t)=y_1(t)+2_{y2}(t), and y5(t)=2y1(t)−2y3(t)y_5(t)=2_{y1}(t)−2_{y3}(t). Are y3(t)y_3(t), y4(t)y_4(t), and y5(t)y_5(t) also solutions of the given differential equation? (c) Determine whether each of the following pairs forms a fundamental set of solutions: [y1(t),y3(t)];[y2(t),y3(t)];[y1(t),y4(t)];[y4(t),y5(t)][y_1(t),y_3(t)];[y_2(t),y_3(t)];[y_1(t),y_4(t)];[y_4(t),y_5(t)].

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.

r2+2r+5=0→r1,2=−1±2ir^2+2r+5=0 \rightarrow r_{1,2}=-1\pm 2i →\rightarrow using quadratic formula

Homogeneous solution:

yc=c1e−tcos⁡2t+c2e−tsin⁡2ty_c=c_1e^{-t}\cos 2t+c_2e^{-t}\sin 2t

Let Y=Acos⁡2t+Bsin⁡2tY=A\cos 2t+B\sin 2t - because g(t)=3sin⁡2tg(t)=3\sin 2t

Plug YY into starting equation to find particular solution:

(Acos⁡2t+Bsin⁡2t)′′+2(Acos⁡2t+Bsin⁡2t)′+5(Acos⁡2t+Bsin⁡2t)=3sin⁡2t−4Acos⁡2t−4Bsin⁡2t−4Asin⁡2t+4Bcos⁡2t+5Acos⁡2t+5Bsin⁡2t=3sin⁡2tcos⁡2t(A+4B)+sin⁡2t(B−4A)=3sin⁡2t(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=0A+4B=0

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

B−4A+4A+16B=3→B=317→A=−1217B-4A+4A+16B=3 \rightarrow B=\dfrac{3}{17} \rightarrow A=-\dfrac{12}{17}

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