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UnkleRhaukus

  • 3 years ago

Show that the function satisfies the DE

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

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

  4. UnkleRhaukus
    • 3 years ago
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    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
    • 3 years ago
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    I wish I was in this class :/

  6. UnkleRhaukus
    • 3 years ago
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    am i thinking about this problem the right way?

  7. UnkleRhaukus
    • 3 years ago
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    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
    • 3 years ago
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    I am meant to be able to work that bit out in my head?

  9. UnkleRhaukus
    • 3 years ago
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    its just the product rule in reverse but is kind hard to make that step

  10. AccessDenied
    • 3 years ago
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    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
    • 3 years ago
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    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

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

  13. AccessDenied
    • 3 years ago
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    \[ \begin{align} \frac{\partial}{\partial u} \left( \cos u \; \sin ( x \sin u ) \right) &= \frac{\partial}{\partial u} \left( \cos u \right) \sin (x \sin u) + \cos u \frac{\partial}{\partial u} \left( \sin ( x \sin u ) \right) \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin (x \sin u) } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ x \cos u \cos (x \sin u) } \\ \text{Let } k = x \sin u; \quad k' = x \cos u. \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin k } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ k' \cos k } \end{align} \] 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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    In set-theory it's called a *complement* rather than a negation, which is typically used in logic.

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