Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • 2 years ago

Verify a solution to the differential equation

  • This Question is Closed
  1. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\begin{equation*} \frac{\mathrm d^2 x}{\mathrm dy^2} +k^2y=f(x),\qquad k\in\mathbb R \end{equation*}\]

  3. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % n-th partial derivative \begin{align*} y&=\frac1k\int\limits_0^x f(u)\sin k(x-u) \dd u\\ \de yx&=\frac1k\int\limits_0^x \pa {}{x}\Big[f(u)\sin k(x-u)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_x(u)\sin k(x-u)+kf(u)\cos k(x-u) \dd u\\ \\ \\ \den yx2&=\frac1k\int\limits_0^x \pa {}x\Big[f_x(u)\sin k(x-u)+kf(u)\cos k(x-u)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_{xx}(u)\sin k(x-u)+kf_x(u)\cos k(x-u)\\&\qquad\qquad+kf_x(u)\cos k(x-u)-kf(u)\sin k(x-u)\dd u\\ &=\frac1k\int\limits_0^x\Big( f_{xx}(u)-kf(u)\Big)\sin k(x-u)+2kf_x(u)\cos k(x-u)\dd u\\ \end{align*}\]

  4. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % n-th partial derivative {\small\begin{align*} \den xy2 +k^2y =f(x)\\ \frac1k\int\limits_0^x\Big( f_{xx}(u)-kf(u)\Big)\sin k(x-u)+2kf_x(u)\cos k(x-u)\dd u\\+k\int\limits_0^x f(u)\sin k(x-u) \dd u=f(x)\\ \int\limits_0^x\Big( \frac{f_{xx}(u)}k-f(u)\Big)\sin k(x-u)+2f_x(u)\cos k(x-u)\\+ kf(u)\sin k(x-u) \dd u=f(x) \\ \int\limits_0^x\left( \frac{f_{xx}(u)}k+(k-1)f(u)\right)\sin k(x-u)+2f_x(u)\cos k(x-u) \dd u=f(x) \\ \end{align*}}\]

  5. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now i guess i just have to apply the product rule in reverse ,

  6. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm, would that differentiation under the integral be applicable if the domain of integration also contains the variable we are differentiating with respect to? I'm not exactly certain of this...

  7. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im not sure

  8. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?

  9. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Hmmm oh yeah, there is that \(x\) in there...

  10. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dunno

  11. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    What inspired the partial derivative, was it just a guess?

  12. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I believe that is an integration rule being used here: http://en.wikipedia.org/wiki/Leibniz_integral_rule

  13. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @wio im not sure, it worked on the previous question

  14. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Ah, I see. I was a big confused as to the fact that we have a multivariable function

  15. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually, the part titled "Formal Statement" on that page looks useful here...

  16. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    What would \(b(\theta)\) ad \(\theta\) be?

  17. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \(b(\theta) = x \) and \(\theta = x\)?

  18. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @UnkleRhaukus did you look at the formal statement?

  19. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah , but im not sure how to use that in my problem

  20. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Did you try writing it out to see what happens?

  21. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \( \displaystyle \frac{\text{d}}{\text{d} x} \int_{a(x)}^{b(x)} y(u, \; x) \; \text{d}u = \int_{a(x)}^{b(x)} y_x (u, \; x) \; \text{d}u + y(b(x), \; x) \; b'(x) - y(a(x), \; x) \; a'(x) \) \(a(x) = 0\), \(b(x) = x\), \(y(u, \; x) = f(u) \; \sin(k(x - u)) \) I think, anyways. I haven't used this theorem before... (:

  22. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @AccessDenied actually it seems when you write it out, the added parts go to 0

  23. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, I see that now. Interesting. :)

  24. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Ummm does \[ \sin k(x-u) = \sin(kx)\cos(ku) - \sin(ku)\cos(kx) \]make things any easier? grasping at straws here.

  25. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    You know one thing I've been wondering about.... Since \(f(u)\) is obviously a single variable function, does it have a partial derivative?

  26. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Either it's constant with respect to \(x\) or \(\frac{d}{dx}f(u) = f'(u) u'(x)\) based on chain rule.

  27. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @UnkleRhaukus @AccessDenied could you clear this up?

  28. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    How can a single variable function have a partial derivative?

  29. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

  30. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    um

  31. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think \(f_x(u)\) is a dead end. I don't see how you'd recover from that because there is no integration with respect to \(x\).

  32. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    And it doesn't really make sense to me anyway, for whatever that is worth. Maybe it would work if you just treat \(f(u)\) like a constant?

  33. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @UnkleRhaukus Ummm sorry if I'm being a bother but.... do we have reason to believe that \(u\) and \(x\) depend upon each other or are independent of each other?

  34. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im not sure

  35. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    If you do this: \[ \large \frac{dy}{dx} = \int^x_0 f(u)\cos k(x-u) du \]

  36. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    then try the thing that @AccessDenied mentioned.. I wonder

  37. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\large \frac{d^2y}{dx^2} = -k\int^x_0 f(u)\sin k(x-u) du +f(x) \] Oh pellet, it works!

  38. wio
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

  39. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont understand

  40. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahh, sorry, I tried to switch to my other computer but I was getting blank screen for the rest of the night. >.< It does appear to work for us here by treating f(u) as independent with respect to x. We were also not told how they were related, so I guess its just more reasonable to assume they are independent of each other.

  41. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    From the previous problem, we had constant limits of integration. The theorem here 'worked' but the terms involving the derivative of the constants vanished. Here, it becomes a bit different where a limit of integration is a nonconstant function of the differentiating variable and has a nonzero derivative. Since they're both the same theorem though, it may be helpful to keep the general theorem in mind and just understand the result of using constant limits when you see them. That's what I am thinking, anyways.

  42. 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.