Rui Baldwin
2021-09-22
Answered

The Laplace transform L $\left\{{e}^{-{t}^{2}}\right\}$ exists, but without finding it solve the initial-value problem $y{}^{\u2033}+9y=3{e}^{-{t}^{2}},y\left(0\right)=0,{y}^{\prime}\left(0\right)=0$

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Yusuf Keller

Answered 2021-09-23
Author has **90** answers

Apply the Laplace transform to both sides of the equation:

$L(y{}^{\u2033}+9y)=L\left(3{e}^{-{t}^{2}}\right)$

$\Rightarrow {s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime}\left(0\right)+9Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)$

$\Rightarrow ({s}^{2}+9)Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)$

$\Rightarrow Y\left(s\right)=\frac{L\left(3{e}^{-{t}^{2}}\right)}{{s}^{2}+9}=\frac{1}{3}L\left(3{e}^{-{t}^{2}}\right)\frac{3}{{s}^{2}+9}$

$\Rightarrow Y\left(s\right)=L\left({e}^{-{t}^{2}}\right)xL\left(\mathrm{sin}\left(3t\right)\right)$

Using the property of convolution:

$f\left(t\right)\cdot g\left(t\right)\leftarrow L\to F\left(s\right)G\left(s\right)$

$\Rightarrow y\left(t\right)={e}^{-{t}^{2}}\cdot \mathrm{sin}\left(3t\right)$

$\Rightarrow y\left(t\right)={\int}_{0}^{t}{e}^{-{\tau}^{2}}\sim \left(3(t-\tau )\right)d\tau$

Result would be:

$y\left(t\right)={\int}_{0}^{t}{e}^{-{\tau}^{2}}\sim \left(3(t-\tau )\right)d\tau$

Using the property of convolution:

Result would be:

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