 # Solve y(t)+2int_0^t e^{-2(t-tau)}y(tau)d tau=e^{-2t} BenoguigoliB 2021-02-09 Answered
Solve $y\left(t\right)+2{\int }_{0}^{t}{e}^{-2\left(t-\tau \right)}y\left(\tau \right)d\tau ={e}^{-2t}$
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Given equation is :
$y\left(t\right)+2{\int }_{0}^{t}{e}^{-2\left(t-\tau \right)}y\left(\tau \right)d\tau ={e}^{-2t}$
To solve this equation use Laplace transform. Apply Laplace transform on both sides:
$L\left\{y\left(t\right)\right\}+2L\left\{{\int }_{0}^{t}{e}^{-2\left(t-\tau \right)}y\left(\tau \right)d\tau \right\}=L\left\{{e}^{-2t}\right\}$
The standard Laplace transform gives that $L\left\{{\int }_{0}^{t}{e}^{-2\left(t-\tau \right)}y\left(\tau \right)d\tau \right\}=L\left\{{e}^{-2t}\right\}\cdot L\left\{y\left(t\right)\right\}$
Let the Laplace transform of y(t) be Y(s)
Then the equation turns into:
$Y\left(s\right)+2L\left\{{e}^{-2t}\right\}\cdot Y\left(s\right)=L\left\{{e}^{-2t}\right\}$
Use the standard Laplace transform of ${e}^{-at}$, which is $\frac{1}{s+a}$
$⇒Y\left(s\right)+\frac{2}{s+2}Y\left(s\right)=\frac{1}{\left(s+2\right)}$
$⇒\left[\frac{\left(s+2+2\right)}{\left(s+2\right)}\right]Y\left(s\right)=\frac{1}{\left(s+2\right)}$
$⇒Y\left(s\right)=\frac{1}{\left(s+4\right)}$
Step 2
Now, apply the inverse Laplace transform on the obtained equation above.
${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{\left(s+4\right)}\right\}$
Since the Laplace transform of y(t) is Y(s), inverse Laplace of Y(s) is y(t)
Similarly, when $L\left\{{e}^{\left(-at\right)}\right\}=\frac{1}{\left(s+a\right)}$, the inverse Laplace is: ${L}^{-1}\left\{\frac{1}{\left(s+a\right)}\right\}={e}^{-at}$
Then, the resultant equation is:
$y\left(t\right)={e}^{-4t}$
Therefore, the solution for the given equation is