A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

I need to find the solution for these differential equations: 1) [(D^2)+D]y = x^2 + 3x + e^3x 2) y'' + 4y = 8cos2x -4x I know their complementary functions which are 1) yc = c1+c2e^-x 2) yc = c1cos2x +c2sin2x But I don't know how to get the particular functions to find the solution. Please help!!!

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    familiar with the method of undetermined coefficients ?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep. but i'm quite confused on how to use that

  3. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    for the first one, try to search a solution like this: \[A{e^{3x}} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    can you use it on below two differential equations separately ? [(D^2)+D]y = x^2 + 3x [(D^2)+D]y = e^3x

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this right? [(D^2)+D]y = e^3x y= c1+c2e^-x + (1/12)e^3x

  6. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yes!

  7. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes only find a particular solution for each of the equations : [(D^2)+D]y = x^2 + 3x [(D^2)+D]y = e^3x

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but i don't know how to solve for this [(D^2)+D]y = x^2 + 3x

  9. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    you have to consider a solution like this one: \[{B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

  10. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    we are applying the superposition principle of solutions

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i still can't understand ;;

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    you have to replace \(y'\) and \(y''\) with the first and second derivative of \({B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\)

  13. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    then you have to use the principle of identity of polynomials

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this right? B2=-1 B3=1/2 B4= 1/3 B5=0

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you may use wolfram to double check http://www.wolframalpha.com/input/?i=solve+y%27%27%2By%27+%3D+x%5E2+%2B+3x+%2B+e%5E%283x%29 it seems you have got it correctly! good job !

  16. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    correct!

  17. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the first one can be made less boring by multiplying by \(e^x\) and seeing the LHS as \( (e^x y^{\prime})^{\prime} \), n'est-ce pas?

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how about the B1?

  19. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \(B_1\) is added to \(c_1\) so we get the new constant: \(k_1=c_1+B_1\)

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks to all!!! can you still help me for the second one y'' + 4y = 8cos2x -4x??

  21. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    for the second one, you have to try a solution like this one: \[{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right) + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

  22. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    or, if you prefer, you can split the problem, as @ganeshie8 has indicated before, and then try these solutions separately: \[{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right)\] \[{B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

  23. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    please wait, I have made a typo, the right solution is: \[x\left\{ {{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right)} \right\} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is this the answer to the second one? y=c1cos2x+c2sin2x+2xsinx-x

  25. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    do you mean: y=c1cos2x+c2sin2x+2xsin(2x)-x

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah yes!

  27. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ok! correct!

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks a lot!!!

  29. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    :)

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can i still ask you a question?

  31. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yes!

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks. This is still a problem for differential equations. I have trouble solving these type of problems. Find the constant EMF E if C=(1/20)farad, R=20 ohms, q=0 when t=0, and q=2 coulomb when t=1 sec

  33. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    is it a series RC circuit?

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohm's law

  35. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    we can write the subsequent ODE: \[ - \frac{Q}{C} + RI = {E_0}\cos \left( {\omega t} \right)\]

  36. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    or, more precisely, we have this: \[\frac{{dV}}{{dt}} + \frac{V}{{RC}} = 0\] where \(V \) is the electric voltage

  37. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \(V= V(t)\) is the unknown function, of course

  38. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    furthermore we have this equation: \[I = - C\frac{{dV}}{{dt}}\]

  39. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    and, of course: \[Q\left( t \right) = CV\left( t \right)\]

  40. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    those equations can solve your problem

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks

  42. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    :)

  43. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.