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

Laplace transform
asked 2021-02-09
Solve \(y(t)+2\int_0^t e^{-2(t-\tau)}y(\tau)d \tau=e^{-2t}\)

Answers (1)

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.
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:
Therefore, the solution for the given equation is

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