General Solution for x'=ax+f(t) in Reals with Particular Solution x_p(t)

[Unassuming]

Homework Statement

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

Homework 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?

The Attempt at a Solution

 

[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?

 

[Unassuming]

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

 

[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?

 

[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?

 

[gabbagabbahey]

Why not just differentiate $x(t)$?

 

[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?

 

[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?

 

[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?

 

[gabbagabbahey]

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

 

[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?!

 

[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?

 

[Unassuming]

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

 

[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?