Question

Verify a solution to the differential equation

Answer 1

\[\begin{equation*}
\frac{\mathrm d^2 x}{\mathrm dy^2} +k^2y=f(x),\qquad k\in\mathbb R
\end{equation*}\]

Answer 2

\[\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*}\]

Answer 3

\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal
\newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative
\newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative
\newcommand\den[3]{\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th derivative
\newcommand\pan[3]{\frac{\partial^#3 #1}{\partial#2^#3}} % n-th partial derivative
{\small\begin{align*}
\den xy2 +k^2y =f(x)\\
\frac1k\int\limits_0^x\Big( f_{xx}(u)-kf(u)\Big)\sin k(x-u)+2kf_x(u)\cos k(x-u)\dd u\\+k\int\limits_0^x f(u)\sin k(x-u) \dd u=f(x)\\
\int\limits_0^x\Big( \frac{f_{xx}(u)}k-f(u)\Big)\sin k(x-u)+2f_x(u)\cos k(x-u)\\+ kf(u)\sin k(x-u) \dd u=f(x)
\\
\int\limits_0^x\left( \frac{f_{xx}(u)}k+(k-1)f(u)\right)\sin k(x-u)+2f_x(u)\cos k(x-u) \dd u=f(x)
\\
\end{align*}}\]

Answer 4

now i guess i just have to apply the product rule in reverse ,

Answer 5

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...

Answer 6

im not sure

Answer 7

Why are you pulling that partial derivative in? Isn't it implied by fundamental theorem that the function inside the integral is the derivative?

Answer 8

Hmmm oh yeah, there is that \(x\) in there...

Answer 9

i dunno

Answer 10

What inspired the partial derivative, was it just a guess?

Answer 11

I believe that is an integration rule being used here:
http://en.wikipedia.org/wiki/Leibniz_integral_rule

Answer 12

@wio im not sure, it worked on the previous question

Answer 13

Ah, I see. I was a big confused as to the fact that we have a multivariable function

Answer 14

Actually, the part titled "Formal Statement" on that page looks useful here...

Answer 15

What would \(b(\theta)\) ad \(\theta\) be?

Answer 16

\(b(\theta) = x \) and \(\theta = x\)?

Answer 17

@UnkleRhaukus did you look at the formal statement?

Answer 18

yeah , but im not sure how to use that in my problem

Answer 19

Did you try writing it out to see what happens?

Answer 20

\( \displaystyle \frac{\text{d}}{\text{d} x} \int_{a(x)}^{b(x)} y(u, \; x) \; \text{d}u = \int_{a(x)}^{b(x)} y_x (u, \; x) \; \text{d}u + y(b(x), \; x) \; b'(x) - y(a(x), \; x) \; a'(x) \)
\(a(x) = 0\), \(b(x) = x\), \(y(u, \; x) = f(u) \; \sin(k(x - u)) \)

I think, anyways. I haven't used this theorem before... (:

Answer 21

@AccessDenied actually it seems when you write it out, the added parts go to 0

Answer 22

Yeah, I see that now. Interesting. :)

Answer 23

Ummm does \[
\sin k(x-u) = \sin(kx)\cos(ku) - \sin(ku)\cos(kx)
\]make things any easier? grasping at straws here.

Answer 24

You know one thing I've been wondering about....
Since \(f(u)\) is obviously a single variable function, does it have a partial derivative?

Answer 25

Either it's constant with respect to \(x\) or \(\frac{d}{dx}f(u) = f'(u) u'(x)\) based on chain rule.

Answer 26

@UnkleRhaukus @AccessDenied could you clear this up?

Answer 27

How can a single variable function have a partial derivative?

Answer 28

imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

Answer 29

um

Answer 30

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 \(x\).

Answer 31

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?

Answer 32

@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?

Answer 33

im not sure

Answer 34

If you do this: \[ \large
\frac{dy}{dx} = \int^x_0 f(u)\cos k(x-u) du
\]

Answer 35

then try the thing that @AccessDenied mentioned.. I wonder

Answer 36

\[\large
\frac{d^2y}{dx^2} = -k\int^x_0 f(u)\sin k(x-u) du +f(x)
\]
Oh pellet, it works!

Answer 37

@UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

Answer 38

i dont understand

Answer 39

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.

Answer 40

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.