Laplace Transform of A Second Order ODE

  • #1
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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!
 

Answers and Replies

  • #2
SteamKing
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Have you tried completing the square for the numerator?
 
  • #3
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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!
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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  • #4
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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?
 
  • #5
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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?
 
  • #6
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0
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?
 
  • #7
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7,140
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).
 
  • #8
11
0
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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