# Diff eq. question

1. Oct 4, 2008

### Unassuming

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

x'=ax +f(t) where a & x are elements of the reals, a is a constant
Let x_p be any particular solution of the equation. Show that the general solution is
x(t)=x_p(t) +ce^at

2. Relevant equations

Where should I start with this? I know integrating x'=ax gives the ce^at part but how do I show that the general solution has the entire form asked for?
3. The attempt at a solution

2. Oct 4, 2008

### gabbagabbahey

If $x_p(t)$ is the particular solution, then $x'_p(t)=ax_p(t)+f(t)$ What do you get when you substitute the proposed general solution into the DE?

3. Oct 6, 2008

### Unassuming

I am confused. Which of these is the proposed general solution and which is the DE?

4. Oct 6, 2008

### gabbagabbahey

DE stands for differential equation. The proposed general solution is $x(t)=x_p(t)+Ce^{at}$ What do you get when you substitute this into the differential equation?

5. Oct 6, 2008

### Unassuming

Do I take the proposed general solution and solve for x_p(t) and then differentiate it? Then set it equal ot the DE?

6. Oct 6, 2008

### gabbagabbahey

Why not just differentiate $x(t)$?

7. Oct 6, 2008

### Unassuming

I just differentiated x(t) and got x'(t)=x'_p(t) + cae^at. Do I then solve this for x'_p(t) and then substitute this into the DE?

8. Oct 6, 2008

### gabbagabbahey

You already know that $x'_p(t)=ax_p(t)+f(t)$ since the particular solution must satisfy the differential equation, so why not substitute this into your above result?

9. Oct 6, 2008

### Unassuming

thank you. I substituted x'_p(t) into what is above and got x'(t)=axp(t) +f(t)+cae^at. To show the gen. sol is x(t)=x_p(t)+ce^at, do I need to integrate what I got?

10. Oct 6, 2008

### gabbagabbahey

No, just rearrange what you've got: $x'(t)=a(x_p(t)+ce^{at})+f(t)$ but what is $a(x_p(t)+ce^{at})$? ;)

11. Oct 6, 2008

### Unassuming

thank you. I see that after rearranging, x_p(t)+ce^at is x! I don't really understand how what I did works to slove the problem. Is it possible for you to explain what you have helped me to do?!

12. Oct 6, 2008

### gabbagabbahey

Well, you've just shown that $x(t)=x_p(t)+Ce^{at}$ satisfies the DE $x'(t)=ax(t)+f(t)$....Doesn't that mean that x(t) IS the general solution?

13. Oct 6, 2008

### Unassuming

I guess so! The problem had a hint suggesting I look at the difference x - x_p. What did this mean?

14. Oct 6, 2008

### gabbagabbahey

Well, the general solution is the sum of the particular solution and the complimentary solution (i.e. $x(t)=x_p(t)+x_c(t)$. So, $x(t)-x_p(t)=x_c(t)$. The complimentary solution must satisfy the complimentary, homogeneous DE $x'_c(t)=ax_c(t)$. Does $x(t)-x_p(t)$ satisfy the homogeneous DE?