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Diff eq. question

  1. Oct 4, 2008 #1
    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. jcsd
  3. Oct 4, 2008 #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?
     
  4. Oct 6, 2008 #3
    I am confused. Which of these is the proposed general solution and which is the DE?
     
  5. Oct 6, 2008 #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?
     
  6. Oct 6, 2008 #5
    Do I take the proposed general solution and solve for x_p(t) and then differentiate it? Then set it equal ot the DE?
     
  7. Oct 6, 2008 #6

    gabbagabbahey

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    Why not just differentiate [itex]x(t)[/itex]?
     
  8. Oct 6, 2008 #7
    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?
     
  9. Oct 6, 2008 #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?
     
  10. Oct 6, 2008 #9
    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?
     
  11. Oct 6, 2008 #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]? ;)
     
  12. Oct 6, 2008 #11
    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?!
     
  13. Oct 6, 2008 #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?
     
  14. Oct 6, 2008 #13
    I guess so! The problem had a hint suggesting I look at the difference x - x_p. What did this mean?
     
  15. Oct 6, 2008 #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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