# Thread: now i have problem that cant be proofed

1. ## now i have problem that cant be proofed

if Y=[X(1-X)]½

prove that: Y[second derivative]+[first derivative]²+1=0

i think there is something wroung in it

2. Hello, mohamedsafy!

It took me a few tries, but I got it . . .

$\text{If }\,y \:=\:\left(x-x^2\right)^{\frac{1}{2}}$ .[1]

prove that: . $y\left(\frac{d^2y}{dx^2}\right)+ \left(\frac{dy}{dx}\right)^2 +1 \:=\:0$
First, we find the derivatives . . .

$\frac{dy}{dx} \;=\;\tfrac{1}{2}(x - x^2)^{-\frac{1}{2}}(1-2x) \quad\Rightarrow\quad\frac{dy}{dx}\;=\;\frac{1-2x}{2\sqrt{x-x^2}}$ .[2]

$\frac{d^2y}{dx^2} \;=\;\frac{2\sqrt{x-x^2}(-2) - (1-2x)\cdot2\cdot\frac{1}{2}(x-x^2)^{-\frac{1}{2}}(1-2x)}{4(x-x^2)}$

. . which simplifies to: . $\frac{d^2y}{dx^2} \;=\;\frac{-1}{4(x-x^2)^{\frac{3}{2}}}$ .[3]

Then: .. . $y\;\;\cdot\;\;\left(\frac{d^2y}{dx^2}\right) \quad+ \quad\;\;\left(\frac{dy}{dx}\right)^2\;\; +\;\; 1$ . .
Substitute [1], [2] and [3]

. . $= \;\overbrace{\sqrt{x-x^2}}\cdot\overbrace{\frac{-1}{4(x-x^2)^{\frac{3}{2}}}} + \overbrace{\left(\frac{1-2x}{2\sqrt{x-x^2}}\right)^2\,} + 1$

. . $= \;\frac{-1}{4(x-x^2)} + \frac{(1-2x)^2}{4(x-x^2} + \frac{4(x-x^2)}{4(x-x^2)} \;=\;\frac{-1 + (1-2x)^2 + 4(x-x^2)}{4(x-x^2)}$

. . $= \;\frac{-1 + 1 - 4x + 4x^2 + 4x - 4x^2}{4(x-x^2)} \;=\;\frac{0}{4(x-x^2)} \;=\;0\quad\hdots$ There!

3. ## oh thxxxxx

thank u now i knew where was my error
thankx for all people in this forum u r really helping me too much to pass my exam tommorow i really apreciae it