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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{equation*} \frac{\mathrm d^2 x}{\mathrm dy^2} +k^2y=f(x),\qquad k\in\mathbb R \end{equation*}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % nth derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % nth partial derivative \begin{align*} y&=\frac1k\int\limits_0^x f(u)\sin k(xu) \dd u\\ \de yx&=\frac1k\int\limits_0^x \pa {}{x}\Big[f(u)\sin k(xu)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_x(u)\sin k(xu)+kf(u)\cos k(xu) \dd u\\ \\ \\ \den yx2&=\frac1k\int\limits_0^x \pa {}x\Big[f_x(u)\sin k(xu)+kf(u)\cos k(xu)\Big] \dd u\\ &=\frac1k\int\limits_0^x f_{xx}(u)\sin k(xu)+kf_x(u)\cos k(xu)\\&\qquad\qquad+kf_x(u)\cos k(xu)kf(u)\sin k(xu)\dd u\\ &=\frac1k\int\limits_0^x\Big( f_{xx}(u)kf(u)\Big)\sin k(xu)+2kf_x(u)\cos k(xu)\dd u\\ \end{align*}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % nth derivative \newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % nth partial derivative {\small\begin{align*} \den xy2 +k^2y =f(x)\\ \frac1k\int\limits_0^x\Big( f_{xx}(u)kf(u)\Big)\sin k(xu)+2kf_x(u)\cos k(xu)\dd u\\+k\int\limits_0^x f(u)\sin k(xu) \dd u=f(x)\\ \int\limits_0^x\Big( \frac{f_{xx}(u)}kf(u)\Big)\sin k(xu)+2f_x(u)\cos k(xu)\\+ kf(u)\sin k(xu) \dd u=f(x) \\ \int\limits_0^x\left( \frac{f_{xx}(u)}k+(k1)f(u)\right)\sin k(xu)+2f_x(u)\cos k(xu) \dd u=f(x) \\ \end{align*}}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0now i guess i just have to apply the product rule in reverse ,

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, 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...

wio
 one year ago
Best ResponseYou've already chosen the best response.2Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?

wio
 one year ago
Best ResponseYou've already chosen the best response.2Hmmm oh yeah, there is that \(x\) in there...

wio
 one year ago
Best ResponseYou've already chosen the best response.2What inspired the partial derivative, was it just a guess?

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.0I believe that is an integration rule being used here: http://en.wikipedia.org/wiki/Leibniz_integral_rule

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0@wio im not sure, it worked on the previous question

wio
 one year ago
Best ResponseYou've already chosen the best response.2Ah, I see. I was a big confused as to the fact that we have a multivariable function

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.0Actually, the part titled "Formal Statement" on that page looks useful here...

wio
 one year ago
Best ResponseYou've already chosen the best response.2What would \(b(\theta)\) ad \(\theta\) be?

wio
 one year ago
Best ResponseYou've already chosen the best response.2\(b(\theta) = x \) and \(\theta = x\)?

wio
 one year ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus did you look at the formal statement?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0yeah , but im not sure how to use that in my problem

wio
 one year ago
Best ResponseYou've already chosen the best response.2Did you try writing it out to see what happens?

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.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... (:

wio
 one year ago
Best ResponseYou've already chosen the best response.2@AccessDenied actually it seems when you write it out, the added parts go to 0

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I see that now. Interesting. :)

wio
 one year ago
Best ResponseYou've already chosen the best response.2Ummm does \[ \sin k(xu) = \sin(kx)\cos(ku)  \sin(ku)\cos(kx) \]make things any easier? grasping at straws here.

wio
 one year ago
Best ResponseYou've already chosen the best response.2You know one thing I've been wondering about.... Since \(f(u)\) is obviously a single variable function, does it have a partial derivative?

wio
 one year ago
Best ResponseYou've already chosen the best response.2Either it's constant with respect to \(x\) or \(\frac{d}{dx}f(u) = f'(u) u'(x)\) based on chain rule.

wio
 one year ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus @AccessDenied could you clear this up?

wio
 one year ago
Best ResponseYou've already chosen the best response.2How can a single variable function have a partial derivative?

wio
 one year ago
Best ResponseYou've already chosen the best response.2imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

wio
 one year ago
Best ResponseYou've already chosen the best response.2I 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 \(x\).

wio
 one year ago
Best ResponseYou've already chosen the best response.2And 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?

wio
 one year ago
Best ResponseYou've already chosen the best response.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?

wio
 one year ago
Best ResponseYou've already chosen the best response.2If you do this: \[ \large \frac{dy}{dx} = \int^x_0 f(u)\cos k(xu) du \]

wio
 one year ago
Best ResponseYou've already chosen the best response.2then try the thing that @AccessDenied mentioned.. I wonder

wio
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \frac{d^2y}{dx^2} = k\int^x_0 f(u)\sin k(xu) du +f(x) \] Oh pellet, it works!

wio
 one year ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, 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
 one year ago
Best ResponseYou've already chosen the best response.0From 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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