"Solve y(t)+2∫0te−2(t−τ)y(τ)dτ=e−2t" was answered by students like you

Name: Solve y(t)+2∫0te−2(t−τ)y(τ)dτ=e−2t
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve y(t)+2∫0te−2(t−τ)y(τ)dτ=e−2t

BenoguigoliB

Answered question

2021-02-09

Solve

y (t) + 2 \int_{0}^{t} e^{- 2 (t - τ)} y (τ) d τ = e^{- 2 t}

Answer & Explanation

BleabyinfibiaG

Skilled2021-02-10Added 118 answers

Given equation is :
$y (t) + 2 \int_{0}^{t} e^{- 2 (t - τ)} y (τ) d τ = e^{- 2 t}$
To solve this equation use Laplace transform. Apply Laplace transform on both sides:
$L {y (t)} + 2 L {\int_{0}^{t} e^{- 2 (t - τ)} y (τ) d τ} = L {e^{- 2 t}}$
The standard Laplace transform gives that $L {\int_{0}^{t} e^{- 2 (t - τ)} y (τ) d τ} = L {e^{- 2 t}} \cdot L {y (t)}$
Let the Laplace transform of y(t) be Y(s)
Then the equation turns into:
$Y (s) + 2 L {e^{- 2 t}} \cdot Y (s) = L {e^{- 2 t}}$
Use the standard Laplace transform of $e^{- a t}$ , which is $\frac{1}{s + a}$
$\Rightarrow Y (s) + \frac{2}{s + 2} Y (s) = \frac{1}{(s + 2)}$
$\Rightarrow [\frac{(s + 2 + 2)}{(s + 2)}] Y (s) = \frac{1}{(s + 2)}$
$\Rightarrow Y (s) = \frac{1}{(s + 4)}$
Step 2
Now, apply the inverse Laplace transform on the obtained equation above.
$L^{- 1} {Y (s)} = L^{- 1} {\frac{1}{(s + 4)}}$
Since the Laplace transform of y(t) is Y(s), inverse Laplace of Y(s) is y(t)
Similarly, when $L {e^{(- a t)}} = \frac{1}{(s + a)}$ , the inverse Laplace is: $L^{- 1} {\frac{1}{(s + a)}} = e^{- a t}$
Then, the resultant equation is:
$y (t) = e^{- 4 t}$
Therefore, the solution for the given equation is

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?