How to prove $(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$

Question

$$E(x)=\int_0^{\frac{\pi}{2}} \sqrt{1-x^2 \sin^2 t}\, dt$$ Where $E(x)$ is complete elliptic integral of the second kind.

$u=\sin t$

$$E(x)=\int_0^{1} \frac{\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du$$

$$\frac{dE(x)}{dx}=-x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$$

$$\frac{d}{dx}(x\frac{dE(x)}{dx})=-2x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du-x^2\int_0^{1} \frac{xu^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$$

$$\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{-2xu^2(1-x^2 u^2)-x^3u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$$

$$(x^2-1)\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du \tag1$$

$$xE(x)=\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du \tag 2$$

According to Wikipedia, Equation 1 and 2 are equal but I could not prove it. Could you please help me to prove that?

$$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$$ http://en.wikipedia.org/wiki/Elliptic_integral

EDIT:

If $$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$$ is true, then

$$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})-xE(x)=0$$

$$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du=0$$ must be. And then

$$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)-x (1-x^2 u^2)^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$$

$$-x\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$$

If Wikipedia differential equation is true ,

$$\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$$

$$\int_0^{1} \frac{1-u^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{u^2(1-x^2u^2)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$$

$$\int_0^{1} \frac{\sqrt{1-u^2}}{(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$$ must be true too. Now I need to prove that last equation. Any idea how to proceed? Thanks a lot for advice.

Answer 1

Let's call on the help of E's older brother, $K(x)$. That is the complete elliptic integral of the first kind:

$$K(x)=\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-x^2 \sin^2 t}}$$ Together, they form the following system:

$$x\frac{dE}{dx}=E-K$$

$$x\frac{dK}{dx}=\frac{E}{1-x^2}-K$$ which is easy to prove.

Placing them into the diff-equation we get

$$(x^2-1)\frac{d}{dx}(x\frac{dE(x)}{dx})=(x^2-1)\left(\frac{dE}{dx}-\frac{dK}{dx}\right )$$ $$=\frac{x^2-1}{x}\left(E-K-\frac{E}{1-x^2}+K\right )$$

$$=\frac{x^2-1}{x}\frac{1-x^2-1}{1-x^2}E=xE $$
Edit: Here is the proof of

$$x\frac{dK}{dx}=\frac{E}{1-x^2}-K$$

$$\frac{dK}{dx}=\left [\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{(1-x^2 \sin^2 t)^3}}-\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-x^2 \sin^2 t}}\right ]\frac{1}{x}$$ But

$$\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{(1-x^2 \sin^2 t)^3}}=\frac{1}{1-x^2}\int_0^{\frac{\pi}{2}} \sqrt{1-x^2 \sin^2 t}\, dt$$ The last result follows from the obvious equality:

$$\frac{1-x^2}{\sqrt{(1-x^2 \sin^2 t)^3}}=$$

$$=\sqrt{1-x^2 \sin^2 t}-x^2\frac{d}{dt}\left (\frac{\sin t \cos t}{\sqrt{1-x^2 \sin^2 t}}\right )$$

Answer 2

The derivative of the complete elliptic integral of the second kind$E(x)$ is given by,

$$ E'(x) = {\frac {{\it E} \left( x \right) }{x}}-{\frac {{\it K} \left( x \right) }{x}} \,,$$ where $K(x)$ is the complete elliptic integral of the first kind. Multiplying the above equation by $x$ gives $$ x E'(x) = {\it E}(x) - {\it K(x) } $$

$$\Rightarrow (x E'(x))' = {\frac {{\it E} \left( x \right) }{x}}-{\frac {{\it E} \left( x \right) }{ \left( 1-{x}^{2} \right) x}} =\frac{x E(x)}{x^2-1} \,.$$

Multiplying both sides of the last equation by $ (x^2-1) $ yields the desired result.