Diff eq. question

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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

 

Answers and Replies

  • #2
gabbagabbahey
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If [itex]x_p(t)[/itex] is the particular solution, then [itex]x'_p(t)=ax_p(t)+f(t)[/itex] What do you get when you substitute the proposed general solution into the DE?
 
  • #3
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I am confused. Which of these is the proposed general solution and which is the DE?
 
  • #4
gabbagabbahey
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DE stands for differential equation. The proposed general solution is [itex]x(t)=x_p(t)+Ce^{at}[/itex] What do you get when you substitute this into the differential equation?
 
  • #5
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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
gabbagabbahey
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Why not just differentiate [itex]x(t)[/itex]?
 
  • #7
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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
gabbagabbahey
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You already know that [itex]x'_p(t)=ax_p(t)+f(t)[/itex] since the particular solution must satisfy the differential equation, so why not substitute this into your above result?
 
  • #9
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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
gabbagabbahey
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No, just rearrange what you've got: [itex]x'(t)=a(x_p(t)+ce^{at})+f(t)[/itex] but what is [itex]a(x_p(t)+ce^{at})[/itex]? ;)
 
  • #11
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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
gabbagabbahey
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Well, you've just shown that [itex]x(t)=x_p(t)+Ce^{at}[/itex] satisfies the DE [itex]x'(t)=ax(t)+f(t)[/itex]....Doesn't that mean that x(t) IS the general solution?
 
  • #13
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I guess so! The problem had a hint suggesting I look at the difference x - x_p. What did this mean?
 
  • #14
gabbagabbahey
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Well, the general solution is the sum of the particular solution and the complimentary solution (i.e. [itex]x(t)=x_p(t)+x_c(t)[/itex]. So, [itex]x(t)-x_p(t)=x_c(t)[/itex]. The complimentary solution must satisfy the complimentary, homogeneous DE [itex]x'_c(t)=ax_c(t)[/itex]. Does [itex]x(t)-x_p(t)[/itex] satisfy the homogeneous DE?
 

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