Find the solution of the following differential equation by Laplace transforms: y'''- 5y" + 7y’-3y =20sin(t) , y(0)=y'(0)=0 , y"(0)=-2

Question
Laplace transform
asked 2020-12-15
Find the solution of the following differential equation by Laplace transforms:
\(y'''- 5y" + 7y’-3y =20\sin(t) , y(0)=y'(0)=0 , y"(0)=-2\)

Answers (1)

2020-12-16
Step 1
Consider the following initial value problem and apply the Laplace transform on both sides:
\(y'''-5y"+7y'-3y=20\sin t\)
\(y(0)=y'(0)=0 , y"(0)=-2\)
\(s^3L\left\{y(t)\right\}-s^2y(0)-sy'(0)-y"(0)-5\left\{s^2L\left\{y(t)\right\}-sy(0)-y'(0)\right\}+7\left\{sL\left\{y(t)\right\}-y(0)\right\}-3L\left\{y(t)\right\}=\frac{20}{(s^2+1)}\)
\(s^3Y\left\{y(t)\right\}+2-5\left\{s^2L\left\{y(t)\right\}\right\}+7\left\{sL\left\{y(t)\right\}\right\}-3L\left\{y(t)\right\}=\frac{20}{(s^2+1)}\)
\((s^3-5s^2+7s-3)L\left\{y(t)\right\}+2=\frac{20}{(s^2+1)}\)
\((s^3-5s^2+7s-3)L\left\{y(t)\right\}=\frac{20}{(s^2+1)}-2\)
\(L\left\{y(t)\right\}=\frac{20}{(s^2+1)(s^3-5s^2+7s-3)}-\frac{2}{(s^3-5s^2+7s-3)}\)
\(=\frac{1-3s}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}\)
\(=\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}\)
Step 2
Apply the inverse Laplace transform:
\(L\left\{y(t)\right\}=\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}\)
\(\left\{y(t)\right\}=L^{-1}\left[\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}\right]\)
\(y(t)=\sin t +3e^t-4te^t-3\cos t\)
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