# Laplace Transform of A Second Order ODE

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1. Apr 18, 2015

### iismitch55

23.) y'' + 2y' + y = 4e-t; y(0) = 2, y'(0) = -1

Y(s) = [(as + b) y(0) + a y'(0) + F(s)]/(as2 + bs + c)

My attempt:

a = 1, b = 2, c = 1

F(s) = 4 L{ e-t } = 4/(s+1) (From Laplace Transform Table)

Plugging and simplifying:

Y(s) = (2s2 + 5s + 7)/[(s + 1)(s2 + 2s + 1)

Here is where I get stuck. I've tried Partial Fraction Decomposition a couple times, with no real luck. The numerator also doesn't factor. I need to get it to match something in my table (sorry I have no way to post it). I do find it peculiar that the denominator is a perfect cube. Any help would be greatly appreciated!

2. Apr 18, 2015

### SteamKing

Staff Emeritus
Have you tried completing the square for the numerator?

3. Apr 18, 2015

### Staff: Mentor

Assuming your work is correct (I didn't check) the denominator of Y(s) is (s + 1)3. The partial fractions decomposition of $\frac 1 {(s + 1)^3}$ is $\frac A {s + 1} + \frac B {(s + 1)^2} + \frac C {(s + 1)^3}$.
Edit: Maybe you figured this out already, but the way you wrote the denominator suggested to me that you didn't realize it.

Edit2: Partial fractions is definitely the way to go, and comes out with nice numbers. Show me what you did for the partial fractions work and I'll take a look at it and steer you in the right direction if you have an error.

Last edited: Apr 18, 2015
4. Apr 18, 2015

### iismitch55

So, which would be the correct way to proceed with partial fractions:

A(s + 1)^2(s + 1)^3 + B(s +1)(s + 1)^3 + C(s + 1)(s + 1)^2 = 2s^2 + 5s + 7

Or

A(s + 1)(s + 1) + B(s +1)(s + 1) + C(s + 1)(s + 1) = 2s^2 + 5s + 7

My PFD is a bit rusty sorry. Method 2 is what I tried before, which yielded the following:

A + B + C = 2
2A + 2B +2C = 5
A + B + C = 7

Which makes no sense. It sounds like method 1 is right, but since I'm rusty I don't know why it would be right. Could you exxplain?

5. Apr 18, 2015

### Staff: Mentor

Neither of these.

You have $$\frac {2s^2 + 5s + 7}{(s + 1)^3} = \frac A {s + 1} + \frac B {(s + 1)^2} + \frac C {(s + 1)^3}$$

Multiply both sides of this equation by (s + 1)3. You should end up with a quadratic on each side of the equation.

6. Apr 18, 2015

### iismitch55

Thanks, I will give this a shot. Could you tell me why the denominator breaks up this way rather than (s + 1) each denominator?

7. Apr 18, 2015

### Staff: Mentor

That's just the way it works when you have repeated linear factors. Off the top of my head I don't know why that works, but I know that writing A/(s + 1) + B/(s + 1) + C/(s + 1) doesn't work. That would be the same as (A + B + C)/(s + 1), which you could simplify to D/(s + 1).

8. Apr 18, 2015

### iismitch55

Can confirm, tried, doesn't work. Thank you for all your help!