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

Yulia 2020-12-16 Answered
solve the initial value problem with using laplace
${Y}^{\prime }+3Y=6{e}^{-3t}\mathrm{cos}6t+24$
$Y\left(0\right)=0$
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## Expert Answer

casincal
Answered 2020-12-17 Author has 82 answers

Step 1
The given IVP is as follows.
${Y}^{\prime }+3Y=6{e}^{-3t}\mathrm{cos}6t+24$
$Y\left(0\right)=0$
Apply Laplace transform on both sides of the given differential equation as follows.
$L{Y}^{\prime }+3Y=L6{e}^{-3t}\mathrm{cos}6t+24$
$L{Y}^{\prime }+3Y=L6{e}^{-3t}\mathrm{cos}6t+L24$
$sLY-Y\left(0\right)+3LY=6L{e}^{-3t}\mathrm{cos}6t+L24$
$\left(s+3\right)LY=6\frac{s+3}{\left(s+3{\right)}^{2}+{6}^{2}}+\frac{24}{s}$
$LY=\frac{6}{\left(s+3{\right)}^{2}+{6}^{2}}+\frac{24}{s\left(s+3\right)}$
$LY=\frac{6}{\left(s+3{\right)}^{2}+{6}^{2}}+\frac{8}{s}-\frac{8}{\left(s+3\right)}$
Step 2
Apply inverse Laplace transform on both sides as follows.
${L}^{-1}\left\{L\left\{Y\right\}\right\}={L}^{-1}\left\{\frac{6}{\left(s+3{\right)}^{2}+{6}^{2}}+\frac{8}{s}-\frac{8}{\left(s+3\right)}\right\}$
$Y\left(t\right)={L}^{-1}\left\{\frac{6}{\left(s+3{\right)}^{2}+{6}^{2}}\right\}+8{L}^{-1}\left\{\frac{1}{s}\right\}-8{L}^{-1}\left\{\frac{1}{\left(s+3\right)}\right\}$
$Y\left(t\right)={e}^{-3t}\mathrm{sin}6t+8-8{L}^{-1}\left\{\frac{1}{\left(s+3\right)}\right\}$

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