Use the Laplace transform to solve the following initial...

Use the Laplace transform to solve the following initial value problem: y'' - 4y' + 20y = 0 y(0) = 1, y'(0) = 2 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = By completing the square in the denominator and inverting the transform, find y(t) = .

Transcript

00:01 For this question, we're asked to solve this initial value problem using the laplace transform.

00:05 So i'm going to take the laplace transform of this differential equation, thereby turning it into an algebraic equation, which is much easier to solve.

00:13 So i'll have the following.

00:19 All right, now we're told that y of 0, this is equal to 0.

00:22 So this term is 0 and this term is 0 as well.

00:28 And then i can factor out y of s from a couple of terms.

00:33 So i'll have s squared minus 8s plus 32.

00:42 And then we're told that y prime of 0, this is equal to 4.

00:45 So this is minus 4.

00:48 I'll put that to the other side.

00:50 I get 4 and i get that y of s is equal to this.

00:56 So on most laplace transform tables, you probably will not see this function.

01:01 So the question is, how do we algebraically manipulate it to something that is on a laplace transform table? and if we use partial fractions, then complex numbers will be involved.

01:11 So we don't want to do that.

01:12 And just real quick, that's because the discriminant of this polynomial.

01:19 So it's 8 squared minus 4 times 32.

01:27 So this number in here, this is less than 0.

01:31 So the roots of this polynomial are complex numbers.

01:35 So we're not going to use partial fraction decomposition.

01:39 Instead, we will complete the square.

01:45 So actually, let me switch to blue for some scratch work.

01:50 So s squared minus 8s plus 32...

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