Finding derivatives w.r.t. initial conditions So, I've got this problem I can't solve.
I have the differential equation:
$$y'(x, y_0)=y^2(x, y_0)+y(x, y_0)*sin(x)$$
where $y_0$ is the value of $y$ at $x=0$.
I want to find:
$$\frac{\partial y}{\partial y_0}|_{y_0=0}$$
I think I am on the right track. I have differentiated w.r.t. $y_0$ and I have gotten:
$$u'(x)=(2*y(x,0)+sin(x))*u(x)$$
where:
$$u(x)=\frac{\partial y}{\partial y_0}|_{y_0=0}$$
which would be great (and solvable) if not for the $y(x,0)$ part, which I do not know how to find. Anybody have any ideas?
 A: $y(x,0)$ is the particular solution to the differential equation with initial value $y_0 = 0$. In other words, it's a function of one variable, say $f(x) = y(x,0)$ satisfying
$$ f'(x) = f^2(x) + f(x) \sin x \\ f(0) = 0 $$
This function is just $f(x) = 0$.
So substituting $y(x,0)=0$, you get
$$ u'(x) = u(x) \sin x $$
Most likely you can take it from there.
A: This is a Bernoulli equation. Consider
$$
(y^{-1})'=-1-\sin(x)(y^{-1})
$$
which is linear and allows to apply the solution formula for first order linear equations. You are left with some integrals that can not be expressed in elementary functions.

Solve the linear DE with integrating factor
$$
(e^{1-\cos x}y(x)^{-1})'=-e^{1-\cos x}\\
e^{1-\cos x}y(x)^{-1}=y_0^{-1}-\int_0^xe^{1-\cos s}ds
$$
to get the solution and its derivative to the initial value
$$
y(x;y_0)=\frac{y_0e^{1-\cos x}}{1-y_0\int_0^xe^{1-\cos s}ds}
\\
\frac{\partial y(x;y_0)}{\partial y_0}=\frac{e^{1-\cos x}}{1-y_0\int_0^xe^{1-\cos s}ds}+\frac{y_0e^{1-\cos x}\int_0^xe^{1-\cos s}ds}{\left(1-y_0\int_0^xe^{1-\cos s}ds\right)^2}
$$
so that finally
$$
\left.\frac{\partial y(x;y_0)}{\partial y_0}\right|_{y_0=0}=e^{1-\cos x}
$$
