Question

Show that the function satisfies the DE

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

I wish I was in this class :/

Answer 5

am i thinking about this problem the right way?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

\[ \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...

Answer 13

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\(-\)

Answer 14

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!

Answer 15

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}

Answer 16

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