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

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

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

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

AccessDeniedBest ResponseYou've already chosen the best response.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...
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

AccessDeniedBest 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... (:
 one year ago

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

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

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

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

wioBest ResponseYou've already chosen the best response.2
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

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

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

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

wioBest ResponseYou've already chosen the best response.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 \(x\).
 one year ago

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

wioBest 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?
 one year ago

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

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

wioBest 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!
 one year ago

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

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

AccessDeniedBest ResponseYou've already chosen the best response.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.
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.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.
 one year ago
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