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UnkleRhaukus

Verify a solution to the differential equation

  • one year ago
  • one year ago

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

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

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

    • one year ago
  5. UnkleRhaukus
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    now i guess i just have to apply the product rule in reverse ,

    • one year ago
  6. AccessDenied
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    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...

    • one year ago
  7. UnkleRhaukus
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    im not sure

    • one year ago
  8. wio
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    Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?

    • one year ago
  9. wio
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    Hmmm oh yeah, there is that \(x\) in there...

    • one year ago
  10. UnkleRhaukus
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    i dunno

    • one year ago
  11. wio
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    What inspired the partial derivative, was it just a guess?

    • one year ago
  12. AccessDenied
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    I believe that is an integration rule being used here: http://en.wikipedia.org/wiki/Leibniz_integral_rule

    • one year ago
  13. UnkleRhaukus
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    @wio im not sure, it worked on the previous question

    • one year ago
  14. wio
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    Ah, I see. I was a big confused as to the fact that we have a multivariable function

    • one year ago
  15. AccessDenied
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    Actually, the part titled "Formal Statement" on that page looks useful here...

    • one year ago
  16. wio
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    What would \(b(\theta)\) ad \(\theta\) be?

    • one year ago
  17. wio
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    \(b(\theta) = x \) and \(\theta = x\)?

    • one year ago
  18. wio
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    @UnkleRhaukus did you look at the formal statement?

    • one year ago
  19. UnkleRhaukus
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    yeah , but im not sure how to use that in my problem

    • one year ago
  20. wio
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    Did you try writing it out to see what happens?

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

    • one year ago
  22. wio
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    @AccessDenied actually it seems when you write it out, the added parts go to 0

    • one year ago
  23. AccessDenied
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    Yeah, I see that now. Interesting. :)

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

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

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

    • one year ago
  27. wio
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    @UnkleRhaukus @AccessDenied could you clear this up?

    • one year ago
  28. wio
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    How can a single variable function have a partial derivative?

    • one year ago
  29. wio
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    imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

    • one year ago
  30. UnkleRhaukus
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    um

    • one year ago
  31. wio
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    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 \(x\).

    • one year ago
  32. wio
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    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?

    • one year ago
  33. wio
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    @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?

    • one year ago
  34. UnkleRhaukus
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    im not sure

    • one year ago
  35. wio
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    If you do this: \[ \large \frac{dy}{dx} = \int^x_0 f(u)\cos k(x-u) du \]

    • one year ago
  36. wio
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    then try the thing that @AccessDenied mentioned.. I wonder

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

    • one year ago
  38. wio
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    @UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

    • one year ago
  39. UnkleRhaukus
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    i dont understand

    • one year ago
  40. AccessDenied
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    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.

    • one year ago
  41. AccessDenied
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    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.

    • one year ago
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