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

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

2020-11-15
Step 1
Given a differntial equation,
$$y′′+2y′+y =\delta(t-4)$$
with condition
$$y(0)=0$$
$$y′(0)=0$$
Step 2
Appling laplace transform on both sides,
$$L\left\{y′′+2y′+y\right\}=L\left\{\delta(t-4)\right\}$$
$$L\left\{y"\right\}+2L\left\{y'\right\}+L\left\{y\right\}=L\left\{\delta(t-4)\right\}$$
Since,
$$L\left\{y"\right\}=s^2Y(s)-sy(0)-y'(0)$$
$$L\left\{y"\right\}=s^2Y(s)-s(0)-(0)=s^2Y(s)$$ and $$L\left\{y'\right\}=sY(s)-y(0)$$
$$L\left\{y'\right\}=sY(s)-(0)=sY(s)$$
Step 3
Substituting the values,
$$s^2Y(s)+2sY(s)+Y(s)=e^{-4s}$$
$$(s^2+2s+1)Y(s)=e^{-4s}$$ Thus , $$Y(s)=\frac{e^{-4s}}{s^2+2s+1}$$
Step 4
Now, applying the inverse Laplace transform,
$$L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{e^{-4s}}{s^2+2s+1}\right\}$$
$$y(t)=L^{-1}\left\{\frac{e^{-4s}}{(s+1)^2}\right\}$$ Since , $$L^{-1}\left\{\frac{1}{(s+1)^2}\right\}=te^{-t}$$
Now, if inverse laplace transform of F(s) is f(t) then
$$L^{-1}\left\{e^{-at}F(s)\right\}=H(t-a)f(t-a)$$
Thus,
$$y(t)=H(t-4)(t-4)e^{-(t-4)}$$
Where H(t-4) is a Heaviside step function.
Step5
Thus, the solution of the given initial value problem is
$$y(t)=H(t-4)e^{-(t-4)}(t-4)$$

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