# Differential equation

1. Aug 30, 2010

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

If my solution is right please tell me why I should omit $$2xe^{-x}+0.5e^{2x}$$

#### Attached Files:

• ###### Differential equation.jpg
File size:
67.2 KB
Views:
154
2. Aug 30, 2010

### vela

Staff Emeritus
You shouldn't omit the particular solution. The initial conditions apply to the complete solution, not just the homogeneous solution.

3. Aug 30, 2010

So the solution of my book is wrong as I expected!

In this case we have
$$c_{1}=4 c_{2}=-4.5 c_{3}=0$$

4. Aug 30, 2010

### vela

Staff Emeritus
They can't all be 0 otherwise you'd have y(0)=1/2.

5. Aug 30, 2010

### gomunkul51

Do you know how to solve ODEs with Laplace Transform? it is perfectly suited for constant coefficient ODEs with starting conditions at x=0.

6. Aug 30, 2010

### Je m'appelle

This is a very good advice. I'd take it, in case you know Laplace Transforms.

7. Aug 30, 2010

### vela

Staff Emeritus
Really? I wouldn't. I find using Laplace transforms for a problem like this is usually more tedious than solving it using the method of undetermined coefficients.

8. Aug 30, 2010

### Coto

Depends if you want to solve the inverse LT by hand. :)

9. Aug 30, 2010

### gomunkul51

But then you have to find the 3 constants that solve the IVP.
and if you solve by Laplace Transform you will find them along the way.

Nevertheless, both ways are good !
and you better know how to solve if by different methods.

Also learn Lagrange's Variation of Parameters, it's a little longer but it's ingenious ! :)

10. Aug 30, 2010

### vela

Staff Emeritus
But with the Laplace transform, you'll need to do a partial fractions decomposition, so you end up having to solve a system of linear equations anyway.

I agree it's good to know both ways. It's kind of neat to see it all work out with Laplace transforms, but once you do it a few times, the novelty wears off. ;)

11. Aug 31, 2010

Thank you very much for your kind help. I know Laplace but I have to revise it now.

I actually solved it again and I got the same results:
$$c_{1}=4 c_{2}=-4.5 c_{3}=0$$ I mean again $$c_{3}=0$$

12. Aug 31, 2010

### vela

Staff Emeritus
Obviously, you're doing something wrong. Why don't you post your work so we can see where the problem is?

13. Aug 31, 2010

#### Attached Files:

• ###### My work.jpg
File size:
43.3 KB
Views:
86
14. Aug 31, 2010

### vela

Staff Emeritus
Oh, you did get the right answer. It's just that your equations ran together, so it looked like you got c1=4c2=4.5c3=0 so that all of the coefficients were 0.

If you're going to use LaTeX for something like that, you should put the equations on separate lines. :)

15. Aug 31, 2010