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UnkleRhaukus
 4 years ago
Verify a solution to the differential equation
UnkleRhaukus
 4 years ago
Verify a solution to the differential equation

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

AccessDenied
 4 years 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmmm oh yeah, there is that \(x\) in there...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What inspired the partial derivative, was it just a guess?

AccessDenied
 4 years 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
 4 years ago
Best ResponseYou've already chosen the best response.0@wio im not sure, it worked on the previous question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, I see. I was a big confused as to the fact that we have a multivariable function

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What would \(b(\theta)\) ad \(\theta\) be?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(b(\theta) = x \) and \(\theta = x\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus did you look at the formal statement?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did you try writing it out to see what happens?

AccessDenied
 4 years 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... (:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@AccessDenied actually it seems when you write it out, the added parts go to 0

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ummm does \[ \sin k(xu) = \sin(kx)\cos(ku)  \sin(ku)\cos(kx) \]make things any easier? grasping at straws here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You know one thing I've been wondering about.... Since \(f(u)\) is obviously a single variable function, does it have a partial derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Either it's constant with respect to \(x\) or \(\frac{d}{dx}f(u) = f'(u) u'(x)\) based on chain rule.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @AccessDenied could you clear this up?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How can a single variable function have a partial derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0And 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
 4 years ago
Best ResponseYou've already chosen the best response.0@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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you do this: \[ \large \frac{dy}{dx} = \int^x_0 f(u)\cos k(xu) du \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then try the thing that @AccessDenied mentioned.. I wonder

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large \frac{d^2y}{dx^2} = k\int^x_0 f(u)\sin k(xu) du +f(x) \] Oh pellet, it works!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

AccessDenied
 4 years 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
 4 years 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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