UnkleRhaukus
  • UnkleRhaukus
Verify a solution to the differential equation
Differential Equations
jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
\[\begin{equation*} \frac{\mathrm d^2 x}{\mathrm dy^2} +k^2y=f(x),\qquad k\in\mathbb R \end{equation*}\]
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % n-th partial derivative \begin{align*} y&=\frac1k\int\limits_0^x f(u)\sin k(x-u) \dd u\\ \de yx&=\frac1k\int\limits_0^x \pa {}{x}\Big[f(u)\sin k(x-u)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_x(u)\sin k(x-u)+kf(u)\cos k(x-u) \dd u\\ \\ \\ \den yx2&=\frac1k\int\limits_0^x \pa {}x\Big[f_x(u)\sin k(x-u)+kf(u)\cos k(x-u)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_{xx}(u)\sin k(x-u)+kf_x(u)\cos k(x-u)\\&\qquad\qquad+kf_x(u)\cos k(x-u)-kf(u)\sin k(x-u)\dd u\\ &=\frac1k\int\limits_0^x\Big( f_{xx}(u)-kf(u)\Big)\sin k(x-u)+2kf_x(u)\cos k(x-u)\dd u\\ \end{align*}\]

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UnkleRhaukus
  • UnkleRhaukus
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % n-th partial derivative {\small\begin{align*} \den xy2 +k^2y =f(x)\\ \frac1k\int\limits_0^x\Big( f_{xx}(u)-kf(u)\Big)\sin k(x-u)+2kf_x(u)\cos k(x-u)\dd u\\+k\int\limits_0^x f(u)\sin k(x-u) \dd u=f(x)\\ \int\limits_0^x\Big( \frac{f_{xx}(u)}k-f(u)\Big)\sin k(x-u)+2f_x(u)\cos k(x-u)\\+ kf(u)\sin k(x-u) \dd u=f(x) \\ \int\limits_0^x\left( \frac{f_{xx}(u)}k+(k-1)f(u)\right)\sin k(x-u)+2f_x(u)\cos k(x-u) \dd u=f(x) \\ \end{align*}}\]
UnkleRhaukus
  • UnkleRhaukus
now i guess i just have to apply the product rule in reverse ,
AccessDenied
  • AccessDenied
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...
UnkleRhaukus
  • UnkleRhaukus
im not sure
anonymous
  • anonymous
Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?
anonymous
  • anonymous
Hmmm oh yeah, there is that \(x\) in there...
UnkleRhaukus
  • UnkleRhaukus
i dunno
anonymous
  • anonymous
What inspired the partial derivative, was it just a guess?
AccessDenied
  • AccessDenied
I believe that is an integration rule being used here: http://en.wikipedia.org/wiki/Leibniz_integral_rule
UnkleRhaukus
  • UnkleRhaukus
@wio im not sure, it worked on the previous question
anonymous
  • anonymous
Ah, I see. I was a big confused as to the fact that we have a multivariable function
AccessDenied
  • AccessDenied
Actually, the part titled "Formal Statement" on that page looks useful here...
anonymous
  • anonymous
What would \(b(\theta)\) ad \(\theta\) be?
anonymous
  • anonymous
\(b(\theta) = x \) and \(\theta = x\)?
anonymous
  • anonymous
@UnkleRhaukus did you look at the formal statement?
UnkleRhaukus
  • UnkleRhaukus
yeah , but im not sure how to use that in my problem
anonymous
  • anonymous
Did you try writing it out to see what happens?
AccessDenied
  • AccessDenied
\( \displaystyle \frac{\text{d}}{\text{d} x} \int_{a(x)}^{b(x)} y(u, \; x) \; \text{d}u = \int_{a(x)}^{b(x)} y_x (u, \; x) \; \text{d}u + y(b(x), \; x) \; b'(x) - y(a(x), \; x) \; a'(x) \) \(a(x) = 0\), \(b(x) = x\), \(y(u, \; x) = f(u) \; \sin(k(x - u)) \) I think, anyways. I haven't used this theorem before... (:
anonymous
  • anonymous
@AccessDenied actually it seems when you write it out, the added parts go to 0
AccessDenied
  • AccessDenied
Yeah, I see that now. Interesting. :)
anonymous
  • anonymous
Ummm does \[ \sin k(x-u) = \sin(kx)\cos(ku) - \sin(ku)\cos(kx) \]make things any easier? grasping at straws here.
anonymous
  • anonymous
You know one thing I've been wondering about.... Since \(f(u)\) is obviously a single variable function, does it have a partial derivative?
anonymous
  • anonymous
Either it's constant with respect to \(x\) or \(\frac{d}{dx}f(u) = f'(u) u'(x)\) based on chain rule.
anonymous
  • anonymous
@UnkleRhaukus @AccessDenied could you clear this up?
anonymous
  • anonymous
How can a single variable function have a partial derivative?
anonymous
  • anonymous
imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?
UnkleRhaukus
  • UnkleRhaukus
um
anonymous
  • anonymous
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 \(x\).
anonymous
  • anonymous
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?
anonymous
  • anonymous
@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?
UnkleRhaukus
  • UnkleRhaukus
im not sure
anonymous
  • anonymous
If you do this: \[ \large \frac{dy}{dx} = \int^x_0 f(u)\cos k(x-u) du \]
anonymous
  • anonymous
then try the thing that @AccessDenied mentioned.. I wonder
anonymous
  • anonymous
\[\large \frac{d^2y}{dx^2} = -k\int^x_0 f(u)\sin k(x-u) du +f(x) \] Oh pellet, it works!
anonymous
  • anonymous
@UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.
UnkleRhaukus
  • UnkleRhaukus
i dont understand
AccessDenied
  • AccessDenied
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.
AccessDenied
  • AccessDenied
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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