y'' + 2x (y')^2= 0
show the general solution, then find an explicit formula for a solution y(x), which also satisfies intial conditions y'(0) =1 and y (0) = 0
where to start?
Let $\displaystyle u=y^{\prime}\implies u^{\prime}=y^{\prime\prime}$.
The equation becomes $\displaystyle u^{\prime}+2xu^2=0\implies u^{\prime}=-2xu^2$. Apply separation of variables to solve for u, where $\displaystyle y^{\prime}\!\left(0\right)=1\implies u\!\left(0\right)=1$.
Then back substitute and solve the resulting differential equation for y.
Can you take it from here?
You mean du/u^2
Good. so $\displaystyle u= \frac{1}{x- c}$intergrate u ^-2 du = intergrate -2x dx
-u ^-1 = -x^+c
And now, since $\displaystyle u= \frac{dy}{dx}$, you still need to solve
$\displaystyle \frac{dy}{dx}= \frac{1}{x- c}$
That should be an easy integral and will introduce a second "constant of integration".