Verify a solution to the differential equation

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Verify a solution to the differential equation

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

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