## UnkleRhaukus 2 years ago Show that the function satisfies the DE

1. UnkleRhaukus

2. UnkleRhaukus

$\begin{equation*} xJ''+J'+xJ=0 \end{equation*}$ \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

\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

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

I wish I was in this class :/

6. UnkleRhaukus

7. UnkleRhaukus

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

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

9. UnkleRhaukus

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

10. AccessDenied

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

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

12. UnkleRhaukus

\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

\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

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

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

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

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