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## UnkleRhaukus Group Title Verify a solution to the differential equation one year ago one year ago

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1. UnkleRhaukus Group Title

2. UnkleRhaukus Group Title

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

3. UnkleRhaukus Group Title


4. UnkleRhaukus Group Title


5. UnkleRhaukus Group Title

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

6. AccessDenied Group Title

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...

7. UnkleRhaukus Group Title

im not sure

8. wio Group Title

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

9. wio Group Title

Hmmm oh yeah, there is that $$x$$ in there...

10. UnkleRhaukus Group Title

i dunno

11. wio Group Title

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

12. AccessDenied Group Title

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

13. UnkleRhaukus Group Title

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

14. wio Group Title

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

15. AccessDenied Group Title

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

16. wio Group Title

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

17. wio Group Title

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

18. wio Group Title

@UnkleRhaukus did you look at the formal statement?

19. UnkleRhaukus Group Title

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

20. wio Group Title

Did you try writing it out to see what happens?

21. AccessDenied Group Title

$$\displaystyle \frac{\text{d}}{\text{d} x} \int_{a(x)}^{b(x)} y(u, \; x) \; \text{d}u = \int_{a(x)}^{b(x)} y_x (u, \; x) \; \text{d}u + y(b(x), \; x) \; b'(x) - y(a(x), \; x) \; a'(x)$$ $$a(x) = 0$$, $$b(x) = x$$, $$y(u, \; x) = f(u) \; \sin(k(x - u))$$ I think, anyways. I haven't used this theorem before... (:

22. wio Group Title

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

23. AccessDenied Group Title

Yeah, I see that now. Interesting. :)

24. wio Group Title

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

25. wio Group Title

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

26. wio Group Title

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

27. wio Group Title

@UnkleRhaukus @AccessDenied could you clear this up?

28. wio Group Title

How can a single variable function have a partial derivative?

29. wio Group Title

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

30. UnkleRhaukus Group Title

um

31. wio Group Title

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 $$x$$.

32. wio Group Title

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?

33. wio Group Title

@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?

34. UnkleRhaukus Group Title

im not sure

35. wio Group Title

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

36. wio Group Title

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

37. wio Group Title

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

38. wio Group Title

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

39. UnkleRhaukus Group Title

i dont understand

40. AccessDenied Group Title

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.

41. AccessDenied Group Title

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.