Laplace Transform of A Second Order ODE

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Homework Help Overview

The discussion revolves around the Laplace Transform of a second-order ordinary differential equation (ODE) given by y'' + 2y' + y = 4e-t, with initial conditions y(0) = 2 and y'(0) = -1. Participants are exploring the transformation process and the subsequent algebraic manipulation required to solve for Y(s).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the Laplace Transform and the resulting expression for Y(s). There is an exploration of partial fraction decomposition as a method to simplify the expression, with some questioning the correctness of their approaches and the structure of the denominator.

Discussion Status

Several participants have provided insights into the partial fraction decomposition process, with some suggesting methods and others expressing uncertainty about their previous attempts. There is an ongoing examination of the algebraic steps involved, particularly regarding the handling of repeated linear factors in the denominator.

Contextual Notes

Participants note the complexity of the algebraic manipulation and the potential for confusion regarding the setup of the partial fractions. There is a recognition of the need for clarity in the decomposition process, especially when dealing with repeated factors.

iismitch55
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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!
 
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Have you tried completing the square for the numerator?
 
iismitch55 said:
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!
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.
 
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Mark44 said:
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.

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?
 
iismitch55 said:
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
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.
iismitch55 said:
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?
 
Mark44 said:
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.

Thanks, I will give this a shot. Could you tell me why the denominator breaks up this way rather than (s + 1) each denominator?
 
iismitch55 said:
Thanks, I will give this a shot. Could you tell me why the denominator breaks up this way rather than (s + 1) each denominator?
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).
 
Mark44 said:
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).

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

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