Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • 2 years ago

Show that the function satisfies the DE

  • This Question is Closed
  1. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  2. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\begin{equation*} xJ''+J'+xJ=0 \end{equation*}\] \[\begin{align*} J(x)&=\frac2\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ J'(x)&=\frac2\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{-2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)+\,\mathrm du\\ \\ J''(x)&=\frac{-2}\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{-2}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du\\ \\ \end{align*} \]

  3. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ \begin{align*} &xJ''+J'+xJ\\ &=\frac{-2x}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du+\frac{-2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du+\frac{2x}\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}-x\sin^2(u)\cos\left(x\sin (u)\right)-\sin(u)\sin\left(x\sin (u)\right)+x\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x(1-\sin^2(u))\cos(x\sin(u))-\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x\cos^2(u)\cos(x\sin(u))-\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ \end{align*} \]

  4. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think i need to find [?]\[\begin{equation*}\frac{\partial }{\partial u}\Big[\quad?\quad\Big]=x\cos^2(u)\cos(x\sin(u))-\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]

  5. malevolence19
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I wish I was in this class :/

  6. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    am i thinking about this problem the right way?

  7. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm wolfram tells me \[\begin{equation*}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]=x\cos^2(u)\cos(x\sin(u))-\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]

  8. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am meant to be able to work that bit out in my head?

  9. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    its just the product rule in reverse but is kind hard to make that step

  10. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The work appears correct to me here. I can't say I'd be able to get that antiderivative easily though, in my head... lol

  11. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the wolfram step-by-step solution starts by assuming we can tell the factors in the product rule, after we have worked out these , the problem is simple, but i have trouble seeing the factors

    1 Attachment
  12. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\begin{align*}&=\frac{2}\pi\int\limits_0^{\pi/2}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]\,\mathrm du\\ &=\frac{2}\pi\cos(u)\sin\left(x\sin(u)\right)\Big|_0^{\pi/2}\,\mathrm du\\ &=\frac{2}\pi\left[\cos(\tfrac\pi2)\sin\left(x\sin(\tfrac\pi2)\right)-\cos(0)\sin\left(x\sin(0)\right)\right]\,\mathrm du\\ \\ &=0\\\end{align*}\]

  13. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[ \begin{align} \frac{\partial}{\partial u} \left( \cos u \; \sin ( x \sin u ) \right) &= \frac{\partial}{\partial u} \left( \cos u \right) \sin (x \sin u) + \cos u \frac{\partial}{\partial u} \left( \sin ( x \sin u ) \right) \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin (x \sin u) } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ x \cos u \cos (x \sin u) } \\ \text{Let } k = x \sin u; \quad k' = x \cos u. \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin k } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ k' \cos k } \end{align} \] I think it looks a little clearer when you replace the weird composition of functins and find the derivative of the inner function / replace that...

  14. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that is a good method letting k= x sin u , because it is common to both terms thanks. P.S. \(\neg\) is not the same as\(-\)

  15. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ooh, I see what that is now. I kept thinking \neg was a negative sign and the bent end was just a weird latex thing. i remember seeing it in something else now.. lol :P You're welcome, and thanks!

  16. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    negation is used in logic and sets, for example if the universal set is U={1,2,3,4,5,6,7,8} and a subset is S={1,2} the negation of S ¬S={3,4,5,6,7,8}

  17. oldrin.bataku
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In set-theory it's called a *complement* rather than a negation, which is typically used in logic.

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.