# 2nd-ODE problem

1. Mar 3, 2012

### karlmartin

Hey ya'll!

This is the equation under discussion:

y'' - 2y' - 3y = x + 2

I'm asked to use the method of variation of parameters to determine a solution for this differential equation, but I reach a point where my the equations just look too ridiculous to continue.

The point I have in mind is where I finally get the second equation for the system of equations that is supposed to give me meaningful functions for the derivatives of the two variable functions.

These are my equations(the variable functions are u1 and u2):

u1'e3x + u2'e-x = 0
3u1'e3x - u2'e-x = x + 2

From these I am unable to extract equations for the variables that I would be able to integrate without resorting to some sort of external aid. Also, it seems unlikely that these overly difficult equations would really have be the result of the necessary work, considering this is from a beginner calculus course(Early Transcendentals, 6th international edition, chapter 17.3, problem 20).

I need to master this technique, but every problem I try ends up with me looking at an insane integral, puzzled beyond the twilight zone. Could someone please point out my mistake, or at least confirm that the equations are indeed correct and perhaps point out a method of simplification that doesn't result in an integral too hard to process?

Thank you very much for your time.

2. Mar 3, 2012

### karlmartin

I have to apologize in advance in case I violated some rule, since this is a first post. Also, it is 5 AM here and it is quite possible that I'm dreaming of posting this instead of actually doing it. Good night.

3. Mar 3, 2012

### Dick

Very poetic. But you don't need to do variation of parameters. You just need to find a particular solution for your ODE. Try substituting y=ax+b and try to find a and b,

4. Mar 4, 2012

### karlmartin

Thanks for the alternative, Dick, but I know that I could use the method of undetermined coefficients, I'm just asked not to. My goal here is to correctly use the variation of parameters and to know that technique too.

5. Mar 4, 2012

### NewtonianAlch

Dreaming of posting instead of actually posting, that made me fall off my chair LOL.

6. Mar 4, 2012

### NewtonianAlch

y'' - 2y' - 3y = x + 2

Like Dick said, you need to find a particular solution to it.

y$_{p}$ = ax + b

However, you should first find y$_{h}$; the homogeneous solution:

y'' - 2y' - 3y = 0

Where you should get, y$_{h}$ = A*e^ etc. etc. you know what I mean?

7. Mar 4, 2012

### karlmartin

I know what you mean and I've already done that. As I said, I know how to find the particular solution using the method of undetermined coefficients, as you suggested. However, my quarrel is with the method of variation of parameters. I don't care for the solution as much as I want to use the method of variation of parameters properly.

8. Mar 4, 2012

9. Mar 4, 2012

### Dickfore

Solve the linear system of equations w.r.t. u'1, and u'2. Integrate. Substitute in the general form of the solution.

10. Mar 4, 2012

### karlmartin

But I can't integrate the solutions. The integrals are too hard and I don't see how they would become any simpler. The solution itself according to the undetermined coefficients method is pretty simple. I feel like flipping the table, opening the windows and yelling prophecies in tongues to the people on the streets.

11. Mar 4, 2012

### NewtonianAlch

You are one funny character, lol.

12. Mar 4, 2012

### Dick

Variation of parameters is definitely the hard way to do this. But it can be done.

13. Mar 5, 2012

### karlmartin

Thank you all for the help, I think I have the technique on my toolbelt now and it's properly calibrated. It just seems the necessary integration is mostly very difficult.

NewtonianAlch, jokes are all I can offer in return. Also, manbreasts. But those aren't very noteworthy so I'll stick to the jokes.

14. Mar 5, 2012

### Dickfore

$$\int{x \, e^{a \, x} \, dx} = \frac{d}{d a} \int{e^{a \, x} \, dx} = \frac{d}{d a} \frac{e^{a \, x}}{a} = \frac{e^{a \, x}}{a^2} (a \, x - 1)$$

15. Mar 5, 2012

### HallsofIvy

Staff Emeritus
In general, you can always integrate something like $\int e^{ax}x^n dx$ (even more generally, $\int f(x)x^ndx$, for f any repeatedly integrable function) by using integration by parts n times, each time taking $dv= e^{ax}dx$, $u= x^j$ so that you are repeatedly reducing the power of x until you just have $\int e^{ax}dx$.