(20 points) In an integro-differential equation, the unknown...

(20 points) In an integro-differential equation, the unknown dependent variable $y$ appears within an integral, and its derivative $\frac{dy}{dt}$ also appears. Consider the following initial value problem, defined for $t \ge 0$: $\frac{dy}{dt} + 4 \int_0^t y(t - w)e^{-4w} dw = 8$, $y(0) = 0$. a. Use convolution and Laplace transforms to find the Laplace transform of the solution. $Y(s) = \mathcal{L} \{y(t)\} =$ b. Obtain the solution $y(t)$. $y(t) = $

Transcript

00:01 In this problem, we are given an integral differential equation and we are going to solve it using the method of laplace transforms.

00:10 So this is the equation, dy over dt plus 4 times integral from 0 to t, y of t minus w times exponential minus 4 times w, dw equal to 1.

00:29 And the initial condition is y0 is equal to 0.

00:35 So in the first part, we are going to find the laplace transform of the solution, namely y of s.

00:44 And in the second part, we are going to obtain the actual solution y of t.

00:51 Okay, so let us start by rewriting this integral differential equation in a more suggestive form.

00:59 So we have the derivative term plus, we have 4 times the convolution of this function y with the other function exponential minus 40.

01:12 This is the definition of the laplace transform.

01:16 So i am just rewriting it in a more suggestive manner.

01:21 So this is equal to 1.

01:23 Now let us take the laplace transform of both sides.

01:27 We start with the unknown function and we just make it a capital letter.

01:35 As for the derivative, we have s minus capital y minus y0.

01:43 So y0 is equal to 0.

01:45 So we have simply s times capital y.

01:49 So what about the convolution? we have y convolution exponential minus 40.

01:58 Using the convolution theorem, we know that this is equal to the laplace transform of y times the laplace transform of exponential minus 40, namely the other function.

02:13 Okay, we know that the laplace transform of y is capital y.

02:17 And using a table of laplace transforms for elementary functions, we know that the second term is just 1 over s plus 4.

02:28 Okay, now we have this one final term, namely the right -hand side.

02:34 The laplace transform of 1 is just 1 over s.

02:37 Okay, now let us bring everything together.

02:41 In this differential equation, we have s times capital y plus 4 times capital y times 1 over s plus 4 equal to 1 over s.

02:55 If you solve this equation for y and simplify, you will obtain s plus 4 over s times s plus 2 squared.

03:08 This is the laplace transform of this unknown function.

03:15 Now, we are going to invert this.

03:19 We will compute the inverse laplace transform...

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