Second Order ODE Using Laplace Transforms

In summary: The important thing is to have a clear and concise solution that can be easily understood and applied.
  • #1
jazznaz
23
0
Little homework problem that I'm beating myself up over...

Solve:

[tex]xy'' - 2y' + xy = -2\cos x[/tex]

Using the method of Laplace transforms...

So I do some jiggling and get to:

[tex](1+s^{2})\frac{dF}{ds}+4sF(s) = \frac{2s}{s^{2} + 1}[/tex]

To which I find the following solution:

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}}{(1 + s^{2})^{2}}[/tex]

But I'm supposed to get to:

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

But the method I'm using to solve the first order ODE (variation of parameters/constants) won't seem to lead to the answer I need. It almost looks as though I should include an aribtrary constant where they're expecting me to get -1, but it seems that any constant value will be a solution to the first order DE obtained after applying the transform. So I don't really understand how they've deduced that it should be -1...

We've been given the answer to the problem,

[tex]y = \frac{C}{2}(\sin x - x\cos x) + x\cos x[/tex]

Where the result is obtained easily using a table of Laplace Transforms.
 
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  • #2
jazznaz said:
[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}}{(1 + s^{2})^{2}}[/tex]

But I'm supposed to get to:

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

Those are equivalent. To see this, start from the second expression.

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

[tex]F(s) = \frac{C+s^2-1}{(1 + s^{2})^{2}}[/tex]

[tex]F(s) = \frac{s^2+(C-1)}{(1 + s^{2})^{2}}[/tex]

Since C is an arbitrary constant, so is C-1.
 
  • #3
So the justification is altering the constant to give a nice form for the inverse Laplace transform?

Thanks very much!
 
  • #4
Yes, you can use either C or C-1. It doesn't matter.
 

Related to Second Order ODE Using Laplace Transforms

1. What is a Second Order ODE?

A Second Order ODE (Ordinary Differential Equation) is a mathematical equation that involves the second derivative of a dependent variable with respect to an independent variable. It is commonly used to model physical systems and phenomena in science and engineering.

2. What is the Laplace Transform?

The Laplace Transform is a mathematical operation that converts a function of time into a function of complex frequency. It is commonly used to solve differential equations, including Second Order ODEs, by transforming the problem into an algebraic equation.

3. Why is the Laplace Transform useful for solving Second Order ODEs?

The Laplace Transform allows for the direct and efficient solution of Second Order ODEs by converting them into simpler algebraic equations. This method is often preferred over traditional methods such as separation of variables or integration, as it can handle a wider range of problems and does not require advanced integration techniques.

4. What are the steps for solving a Second Order ODE using Laplace Transforms?

The general steps for solving a Second Order ODE using Laplace Transforms are: 1. Take the Laplace Transform of both sides of the equation, 2. Solve the resulting algebraic equation for the transformed dependent variable, 3. Take the inverse Laplace Transform to get the solution in terms of the original dependent variable, and 4. Apply any initial or boundary conditions to determine the specific solution.

5. What are some common applications of Second Order ODEs solved using Laplace Transforms?

Second Order ODEs solved using Laplace Transforms have a wide range of applications in science and engineering, including electrical circuits, mechanical systems, heat transfer, fluid dynamics, and more. They are also commonly used in control theory and signal processing.

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