# solve the initial value problem with using laplace Y'+3Y=6e^{-3t}cos 6t+24 Y(0)=0 Question
Laplace transform solve the initial value problem with using laplace
$$Y'+3Y=6e^{-3t}\cos 6t+24$$
$$Y(0)=0$$ 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.
$$L^{-1}\left\{L\left\{Y\right\}\right\}=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}+\frac{8}{s}-\frac{8}{(s+3)}\right\}$$
$$Y(t)=L^{-1}\left\{\frac{6}{(s+3)^{2}+6^{2}}\right\}+8L^{-1}\left\{\frac{1}{s}\right\}-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$$
$$Y(t)=e^{-3t}\sin 6t+8-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$$

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