1. ## Calculus problem

1) $y^3 -y = 2x$
show,
$9(x^2 - \frac{1}{27})\frac{d^2 y}{dx^2} + 9x\frac{dy}{dx}=y$

i tried this few times but i couldn't get the answer

2)
$\int_0^\pi \frac{x}{1-sin(x)cos(\alpha)}dx= \frac{\pi(\pi -\alpha)}{sin(\alpha)}$

with substitution and using $\int_0^a f(x) =\int_0^a f(a-x)$

2. first one quite straightforward so i think you should show us some progress.

as for the second, just put $x\longrightarrow\pi-x$ and your integral becomes $\int_{0}^{\pi }{\frac{\pi -x}{1-\sin (x)\cos (\alpha) }},$ thus it remains to compute $\int_{0}^{\pi }{\frac{dx}{1-\sin (x)\cos (\alpha) }}$ which shouldn't be that tough.

3. hi there

I had a look at question 1

Using implicit differenitation I found the following

$\frac{dy}{dx} = \frac{2}{3y^2-1}$

$\frac{d^2y}{dx^2} = \frac{-24}{(3y^2-1)^3}$

and $x = \frac{1}{2}(y^3-y)$

Try to sub these values into the LHS side of the DE and simplyfy, hopefully you will get y.