solve the initial value problem with using laplace Y'+3Y=6e^{-3t}cos 6t+24 Y(0)=0

Question
Laplace transform
asked 2020-12-16
solve the initial value problem with using laplace
\(Y'+3Y=6e^{-3t}\cos 6t+24\)
\(Y(0)=0\)

Answers (1)

2020-12-17
Step 1
The given IVP is as follows.
\(Y'+3Y=6e^{-3t}\cos 6t+24\)
\(Y(0)=0\)
Apply Laplace transform on both sides of the given differential equation as follows.
\(L\left\{Y'+3Y\right\}=L\left\{6e^{-3t}\cos 6t+24\right\}\)
\(L\left\{Y'\right\}+\left\{3Y\right\}=L\left\{6e^{-3t}\cos 6t\right\}+L\left\{24\right\}\)
\(sL\left\{Y\right\}-Y(0)+3L\left\{Y\right\}=6L\left\{e^{-3t}\cos 6t\right\}+L\left\{24\right\}\)
\((s+3)L\left\{Y\right\}=6\frac{s+3}{(s+3)^{2}+6^{2}}+\frac{24}{s}\)
\(L\left\{Y\right\}=\frac{6}{(s+3)^{2}+6^{2}}+\frac{24}{s(s+3)}\)
\(L\left\{Y\right\}=\frac{6}{(s+3)^{2}+6^{2}}+\frac{8}{s}-\frac{8}{(s+3)}\)
Step 2
Apply inverse Laplace transform on both sides as follows.
\(L^{-1}\left\{L\left\{Y\right\}\right\}=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}+\frac{8}{s}-\frac{8}{(s+3)}\right\}\)
\(Y(t)=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}\right\}+8L^{-1}\left\{\frac{1}{s}\right\}-8L^{-1}\left\{\frac{1}{(s+3)}\right\}\)
\(Y(t)=e^{-3t}\sin 6t+8-8L^{-1}\left\{\frac{1}{(s+3)}\right\}\)
0

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