# y"-8y'+41y=0 y(0)=0 y'(0)=5 First , using Y for the Laplace transform of y(t), i.e., Y=Lleft{y(t)right} find the equation you get by taking the Laplace transform of the differential equation

$y"-8{y}^{\prime }+41y=0$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=5$First , using Y for the Laplace transform of y(t), i.e., $Y=L\left\{y\left(t\right)\right\}$find the equation you get by taking the Laplace transform of the differential equation
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Step 1
Given
$y"-8{y}^{\prime }+41y=0$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=5$
Step 2
Calculation
Apply Laplace transform.
$L\left\{y"\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-y\left(0\right)$
$L\left\{y\right\}=Y\left(s\right)$
$L\left\{c\right\}=\frac{c}{s}$
$L\left\{y"\right\}-8L\left\{{y}^{\prime }\right\}+41L\left\{y\right\}=L\left\{0\right\}$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-8\left[sY\left(s\right)-y\left(0\right)\right]+41Y\left(s\right)=\frac{0}{s}$
$⇒{s}^{2}Y\left(s\right)-0-5-8\left[sY\left(s\right)-0\right]+41Y\left(s\right)=0$
Hence the equation is
${s}^{2}Y\left(s\right)-0-5-8\left[sY\left(s\right)-0\right]+41Y\left(s\right)=0$
Now,
${s}^{2}Y\left(s\right)-0-5-8\left[sY\left(s\right)-0\right]+41Y\left(s\right)=0$
$⇒Y\left(s\right)\left[{s}^{2}-8s+41\right]=5$
$⇒Y\left(s\right)=\frac{5}{\left({s}^{2}-8s+41\right)}$
Hence the value of $Y\left(s\right)=\frac{5}{\left({s}^{2}-8s+41\right)}$
Apply inverse Laplace transform
$Y\left(s\right)=\frac{5}{\left({s}^{2}-8s+41\right)}$
$=\frac{5}{\left({s}^{2}-8s+16+25\right)}$
$=\frac{5}{\left(s-4{\right)}^{2}+{5}^{2}}$
$y\left(t\right)={L}^{-1}\left(\frac{5}{\left(s-4{\right)}^{2}+{5}^{2}}\right)$
$⇒y\left(t\right)={e}^{4t}\mathrm{sin}\left(5t\right)$
Hence the value of $y\left(t\right)={e}^{4t}\mathrm{sin}\left(5t\right)$