# Use the Laplace transform to solve the given initial-value problem y′′+2y′+y =delta(t-4) y(0)=0 y′(0)=0

Use the Laplace transform to solve the given initial-value problem
$y\prime \prime +2y\prime +y=\delta \left(t-4\right)$
$y\left(0\right)=0$
$y\prime \left(0\right)=0$
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Step 1
Given a differntial equation,
$y\prime \prime +2y\prime +y=\delta \left(t-4\right)$
with condition
$y\left(0\right)=0$
$y\prime \left(0\right)=0$
Step 2
Appling laplace transform on both sides,
$L\left\{y\prime \prime +2y\prime +y\right\}=L\left\{\delta \left(t-4\right)\right\}$
$L\left\{y"\right\}+2L\left\{{y}^{\prime }\right\}+L\left\{y\right\}=L\left\{\delta \left(t-4\right)\right\}$
Since,
$L\left\{y"\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{y"\right\}={s}^{2}Y\left(s\right)-s\left(0\right)-\left(0\right)={s}^{2}Y\left(s\right)$and$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-y\left(0\right)$
$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-\left(0\right)=sY\left(s\right)$
Step 3
Substituting the values,
${s}^{2}Y\left(s\right)+2sY\left(s\right)+Y\left(s\right)={e}^{-4s}$
$\left({s}^{2}+2s+1\right)Y\left(s\right)={e}^{-4s}$Thus ,$Y\left(s\right)=\frac{{e}^{-4s}}{{s}^{2}+2s+1}$
Step 4
Now, applying the inverse Laplace transform,
${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\frac{{e}^{-4s}}{{s}^{2}+2s+1}\right\}$
$y\left(t\right)={L}^{-1}\left\{\frac{{e}^{-4s}}{\left(s+1{\right)}^{2}}\right\}$Since , ${L}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}}\right\}=t{e}^{-t}$
Now, if inverse laplace transform of F(s) is f(t) then
${L}^{-1}\left\{{e}^{-at}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right)$
Thus,
$y\left(t\right)=H\left(t-4\right)\left(t-4\right){e}^{-\left(t-4\right)}$
Where H(t-4) is a Heaviside step function.
Step5
Thus, the solution of the given initial value problem is
$y\left(t\right)=H\left(t-4\right){e}^{-\left(t-4\right)}\left(t-4\right)$