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

BenoguigoliB 2021-02-09 Answered
Solve y(t)+20te2(tτ)y(τ)dτ=e2t
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Expert Answer

BleabyinfibiaG
Answered 2021-02-10 Author has 118 answers

Given equation is :
y(t)+20te2(tτ)y(τ)dτ=e2t
To solve this equation use Laplace transform. Apply Laplace transform on both sides:
L{y(t)}+2L{0te2(tτ)y(τ)dτ}=L{e2t}
The standard Laplace transform gives that L{0te2(tτ)y(τ)dτ}=L{e2t}L{y(t)}
Let the Laplace transform of y(t) be Y(s)
Then the equation turns into:
Y(s)+2L{e2t}Y(s)=L{e2t}
Use the standard Laplace transform of eat, which is 1s+a
Y(s)+2s+2Y(s)=1(s+2)
[(s+2+2)(s+2)]Y(s)=1(s+2)
Y(s)=1(s+4)
Step 2
Now, apply the inverse Laplace transform on the obtained equation above.
L1{Y(s)}=L1{1(s+4)}
Since the Laplace transform of y(t) is Y(s), inverse Laplace of Y(s) is y(t)
Similarly, when L{e(at)}=1(s+a), the inverse Laplace is: L1{1(s+a)}=eat
Then, the resultant equation is:
y(t)=e4t
Therefore, the solution for the given equation is

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