$\displaystyle y''+(\frac{x}{1+x})y'+(\frac{1}{1+x})y=0$
for x>-1
i dont know how to solve it
i was told to find a solution by the form of y1=e^x
and getting a solution
how?
If you multily by $\displaystyle 1 + x$ then
$\displaystyle (1+x)y'' + xy' + y = 0$. Now $\displaystyle \dfrac{d}{dx}(xy'-y) = x y'' $ and $\displaystyle \dfrac{d}{dx} (xy) = x y' + y$.
So, your original equation can be written as
$\displaystyle \dfrac{d}{dx} \left(y' + xy' - y + xy \right) = 0$.
Your turn!