How do I solve $(x+y^2)y' = 1 \quad y(0)=-1$? Now I have:
$$(x+y^2)y' = 1 \quad y(0)=-1$$
So Cauchy's problem is out of discussion till we find general solution, after some experements I found out that this equation is not separable as it is, so I was thinking of some kind of substitution, like $x+y^2 = u$, but I do not know how to proceed from here, I have never worked with substitutions in DEs yet.
 A: $$(x+y^2)\frac{dy}{dx} = 1 \implies x+y^2 = \frac{dx}{dy}$$
This is a linear ODE with respect to $y$, multiplying suitable integrating factor gives
$$\frac{d}{dy}(xe^{-y}) = y^2e^{-y}$$
I am sure you can continue after this. You might not be able to solve explicit for $x$, but we will have an implicitly defined solution.
A: HINT: first, rewrite the equation:
$$\text{y}'\left(x\right)\cdot\left(x+\text{y}\left(x\right)\right)-1=0\tag1$$
Now, let $\text{A}\left(x,\text{y}\right)=-1$ and $\text{B}\left(x,\text{y}\right)=x+\text{y}^2$, than this is not an exact equation, because:
$$\left(\frac{\partial\text{A}\left(x,\text{y}\right)}{\partial\text{y}}=0\right)\ne\left(1=\frac{\partial\text{B}\left(x,\text{y}\right)}{\partial x}\right)\tag2$$
Now, find an integrating factor $\rho\left(\text{y}\right)$ such that:
$$\frac{\partial}{\partial\text{y}}\left(\text{A}\left(x,\text{y}\right)\cdot\rho\left(\text{y}\right)\right)=\frac{\partial}{\partial x}\left(\text{B}\left(x,\text{y}\right)\cdot\rho\left(\text{y}\right)\right)\space\Longleftrightarrow\space$$
$$-\rho'\left(\text{y}\right)=\rho\left(\text{y}\right)\space\Longleftrightarrow\space\ln\left|\rho\left(\text{y}\right)\right|=-\text{y}\tag3$$
Multiply both sides of $\left(1\right)$ by $\rho\left(\text{y}\right)$
