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UnkleRhaukus

  • one year ago

Verify a solution to the differential equation

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

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

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

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

  6. AccessDenied
    • one year ago
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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...

  7. UnkleRhaukus
    • one year ago
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    im not sure

  8. wio
    • one year ago
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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?

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

  10. UnkleRhaukus
    • one year ago
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    i dunno

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

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

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

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

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

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

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

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

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

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

  21. AccessDenied
    • one year ago
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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... (:

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

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

  24. wio
    • one year ago
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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.

  25. wio
    • one year ago
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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?

  26. wio
    • one year ago
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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.

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

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

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

  30. UnkleRhaukus
    • one year ago
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    um

  31. wio
    • one year ago
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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 \(x\).

  32. wio
    • one year ago
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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?

  33. wio
    • one year ago
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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?

  34. UnkleRhaukus
    • one year ago
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    im not sure

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

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

  37. wio
    • one year ago
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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!

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

  39. UnkleRhaukus
    • one year ago
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    i dont understand

  40. AccessDenied
    • one year ago
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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.

  41. AccessDenied
    • one year ago
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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.

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