 # Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+5y=2t^2-9 , y(0)=0 , y'(0)=-3 ankarskogC 2021-09-17 Answered

Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$y"+5y=2{t}^{2}-9,y\left(0\right)=0,{y}^{\prime }\left(0\right)=-3$

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$y"+5y=2{t}^{2}-9$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=-3$
$L\left\{y"\right\}+sL\left\{{y}^{\prime }\right\}=L\left\{2{t}^{2}-9\right\}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+5Y\left(s\right)=L\left\{2{t}^{2}\right\}-L\left\{9\right\}$
$=\frac{4}{{s}^{3}}-\frac{9}{s}$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)+3+5Y\left(s\right)=\frac{4}{{s}^{3}}-\frac{9}{s}$
$⇒Y\left(s\right)\left[{s}^{2}+5\right]=\frac{4}{{s}^{3}}-\frac{9}{s}-3$
$⇒Y\left(s\right)=\frac{\frac{4}{{s}^{3}}-\frac{9}{s}-3}{{s}^{2}+5}=\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}$
$⇒y\left(t\right)={L}^{-1}\left\{\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}\right\}$
$=\frac{2{t}^{2}}{5}-\frac{3\sqrt{5}\mathrm{sin}\left(\sqrt{5}t\right)}{5}+\frac{49\mathrm{cos}\left(\sqrt{5}t\right)}{25}-\frac{49}{25}$
$\left[\because {L}^{-1}\left\{\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}\right\}=\frac{4}{5{s}^{3}}-\frac{15}{s\left({s}^{2}+25\right)}+\frac{49s}{25\left(25+{5}^{2}\right)}-\frac{49}{25{s}^{2}}\right]$
$\therefore \frac{2{t}^{2}}{5}-\frac{49}{25}-\frac{3\sqrt{5}}{25}\mathrm{sin}\left(\sqrt{5}t\right)+\frac{49}{25}\mathrm{cos}\left(\sqrt{5}t\right)$