# The Laplace transform L {e^(-t^2)} exists, but without finding it solve the initial-value problem y''+9y=3e^(-t^2), y(0)=0,y'(0)=0

The Laplace transform L $\left\{{e}^{-{t}^{2}}\right\}$ exists, but without finding it solve the initial-value problem $y{}^{″}+9y=3{e}^{-{t}^{2}},y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$
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Yusuf Keller
Apply the Laplace transform to both sides of the equation:
$L\left(y{}^{″}+9y\right)=L\left(3{e}^{-{t}^{2}}\right)$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+9Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)$
$⇒\left({s}^{2}+9\right)Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)$
$⇒Y\left(s\right)=\frac{L\left(3{e}^{-{t}^{2}}\right)}{{s}^{2}+9}=\frac{1}{3}L\left(3{e}^{-{t}^{2}}\right)\frac{3}{{s}^{2}+9}$
$⇒Y\left(s\right)=L\left({e}^{-{t}^{2}}\right)xL\left(\mathrm{sin}\left(3t\right)\right)$
Using the property of convolution:
$f\left(t\right)\cdot g\left(t\right)←L\to F\left(s\right)G\left(s\right)$
$⇒y\left(t\right)={e}^{-{t}^{2}}\cdot \mathrm{sin}\left(3t\right)$
$⇒y\left(t\right)={\int }_{0}^{t}{e}^{-{\tau }^{2}}\sim \left(3\left(t-\tau \right)\right)d\tau$
Result would be:
$y\left(t\right)={\int }_{0}^{t}{e}^{-{\tau }^{2}}\sim \left(3\left(t-\tau \right)\right)d\tau$