nagasenaz

2021-01-08

Solve Using Laplace transform
$\frac{dy}{dt}=y+10{u}_{4}\left(t\right)\mathrm{sin}\left(2\left(t-4\right)\right),y\left(0\right)=2$

Roosevelt Houghton

Step 1
The given IVP is as follows.
$\frac{dy}{dt}=y+10{u}_{4}\left(t\right)\mathrm{sin}\left(2\left(t-4\right)\right),y\left(0\right)=2$
Apply Laplace transform on both sides of the above equations as follows.
$L\left\{\frac{dy}{dt}\right\}=L\left\{y+10{u}_{4}\left(t\right)\mathrm{sin}\left(2\left(t-4\right)\right)\right\}$
$L\left\{\frac{dy}{dt}\right\}=L\left\{y\right\}+10L\left\{{u}_{4}\left(t\right)\mathrm{sin}\left(2\left(t-4\right)\right)\right\}$
$sL\left\{y\right\}-y\left(0\right)=L\left\{y\right\}+10{e}^{-4s}\frac{2}{{s}^{2}+4}$
$sL\left\{y\right\}-2=L\left\{y\right\}+\frac{20{e}^{-4s}}{{s}^{2}+4}$
$L\left\{y\right\}\left(s-1\right)=2+\frac{20{e}^{-4s}}{{s}^{2}+4}$
$L\left\{y\right\}=\frac{2}{s-1}+\frac{20{e}^{-4s}}{\left(s-1\right)\left({s}^{2}+4\right)}$
Step 2
Apply inverse Laplace transform on both sides as follows.
$L\left\{y\right\}=\frac{2}{s-1}+\frac{20{e}^{-4s}}{\left(s-1\right)\left({s}^{2}+4\right)}$
${L}^{-1}\left\{L\left\{y\right\}\right\}={L}^{-1}\left\{\frac{2}{s-1}+\frac{20{e}^{-4s}}{\left(s-1\right)\left({s}^{2}+4\right)}\right\}$
$y\left(t\right)=2{L}^{-1}\left\{\frac{1}{s-1}\right\}+{L}^{-1}\left\{{e}^{-4s}\frac{20}{\left(s-1\right)\left({s}^{2}+4\right)}\right\}$
$y\left(t\right)=2{L}^{-1}\left\{\frac{1}{s-1}\right\}+{L}^{-1}\left\{{e}^{-4s}\left[\frac{4}{s-1}-\frac{4s}{{s}^{2}+4}-\frac{4}{{s}^{2}+4}\right]\right\}$
$y\left(t\right)=2{e}^{t}+u\left(t-4\right)\left[4{e}^{t-4}-4\mathrm{cos}\left(2\left(t-4\right)\right)-2\mathrm{sin}\left(2\left(t-4\right)\right)\right]$

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