# Solve y(t)+2int_0^t e^{-2(t-tau)}y(tau)d tau=e^{-2t}

Question
Laplace transform
Solve $$y(t)+2\int_0^t e^{-2(t-\tau)}y(\tau)d \tau=e^{-2t}$$

2021-02-10
Given equation is :
$$y(t)+2\int_0^t e^{-2(t-\tau)}y(\tau)d \tau=e^{-2t}$$
To solve this equation use Laplace transform. Apply Laplace transform on both sides:
$$L\left\{y(t)\right\}+2L\left\{\int_0^te^{-2(t-\tau)}y(\tau)d \tau\right\}=L\left\{e^{-2t}\right\}$$
The standard Laplace transform gives that $$L\left\{\int_0^te^{-2(t-\tau)}y(\tau)d \tau\right\}=L\left\{e^{-2t}\right\}\cdot L\left\{y(t)\right\}$$
Let the Laplace transform of y(t) be Y(s)
Then the equation turns into:
$$Y(s)+2L\left\{e^{-2t}\right\}\cdot Y(s)=L\left\{e^{-2t}\right\}$$
Use the standard Laplace transform of $$e^{-at}$$, which is $$\frac{1}{s+a}$$
$$\Rightarrow Y(s)+\frac{2}{s+2}Y(s)=\frac{1}{(s+2)}$$
$$\Rightarrow \left[\frac{(s+2+2)}{(s+2)}\right]Y(s)=\frac{1}{(s+2)}$$
$$\Rightarrow Y(s)=\frac{1}{(s+4)}$$
Step 2
Now, apply the inverse Laplace transform on the obtained equation above.
$$L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{1}{(s+4)}\right\}$$
Since the Laplace transform of y(t) is Y(s), inverse Laplace of Y(s) is y(t)
Similarly, when $$L\left\{e^(-at)}\right\}=\frac{1}{(s+a)}$$, the inverse Laplace is: $$L^{-1}\left\{\frac{1}{(s+a)}\right\}=e^{-at}$$
Then, the resultant equation is:
$$y(t)=e^{-4t}$$
Therefore, the solution for the given equation is

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