Laplace Transform of A Second Order ODE

In summary, the given equation can be written in the Laplace domain as Y(s) = (2s2 + 5s + 7)/[(s+1)(s+1)(s+1)] and can be solved using partial fractions decomposition. The solution for Y(s) will have three terms, each with a denominator of (s+1), which is a result of the repeated linear factors in the original equation. The coefficients for each term can be found by equating the numerators on both sides of the equation.
  • #1
iismitch55
11
0
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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  • #2
Have you tried completing the square for the numerator?
 
  • #3
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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  • #4
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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
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).
 
  • #8
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!
 

Related to Laplace Transform of A Second Order ODE

1. What is the Laplace Transform of a Second Order ODE?

The Laplace Transform of a Second Order ODE is a mathematical tool used to convert a second order ordinary differential equation (ODE) into an algebraic equation. This transformation allows for easier solving of the original ODE and is particularly useful for solving initial value problems.

2. How is the Laplace Transform of a Second Order ODE calculated?

The Laplace Transform of a Second Order ODE is calculated by taking the Laplace Transform of each term in the original ODE and then substituting the transformed terms into the algebraic equation. The initial values of the ODE are also taken into account when solving for the constants in the transformed equation.

3. What types of second order ODEs can be solved using the Laplace Transform?

The Laplace Transform can be applied to a wide range of second order ODEs, including linear, non-linear, and variable coefficient ODEs. However, it is most commonly used for linear ODEs with constant coefficients.

4. What are the advantages of using the Laplace Transform to solve second order ODEs?

One major advantage of using the Laplace Transform to solve second order ODEs is that it simplifies the solving process by transforming the ODE into an algebraic equation. This can be especially helpful when dealing with complex or difficult ODEs. Additionally, the Laplace Transform can be used to solve initial value problems, which is not always possible with other methods.

5. Are there any limitations to using the Laplace Transform for second order ODEs?

While the Laplace Transform is a powerful tool for solving second order ODEs, it does have some limitations. It may not be suitable for solving highly non-linear or variable coefficient ODEs. Additionally, the Laplace Transform cannot be used to solve boundary value problems, as it requires initial values to be known.

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