Compute partial derivative w.r.t initial condition Could anyone help me on the following problem? Thank you very much!
Define $X: \mathcal{R} \times \mathcal{R} \rightarrow \mathcal{R}$ by:
$$\frac{dX}{dt}(t,x_0)=X(t,x_0)+\sin(X^2(t,x_0)),$$
$X(0,x_0)=x_0.$
Question: Find $\frac{\partial X}{\partial x_0}(t,0).$
 A: In the following I interpreted your total derivative with respect to $t$ as a partial derivative. Willie explained in a comment that it might be intended to signify that $x_0$ is a parameter. There's no substantial difference between those two interpretations, so I'll leave the answer in my original notation, which considers $X$ as a function of two variables.
It's not clear to me what the subscript on $x_0$ is for, since $x$ doesn't occur anywhere without that subscript; I'll just write $x$ instead to simplify things.
I'll assume that we can exchange the partial derivatives with respect to $t$ and $x$; I think this follows from the Cauchy–Kowalevski theorem, but as Willie points out in his comment we don't need the full force of that theorem in this case.
Differentiating the differential equation with respect to $x$ yields
$$
\begin{eqnarray}
\def\part#1#2#3{\frac{\partial^2#1}{\partial#2\partial#3}}\def\par#1#2{\frac{\partial#1}{\partial#2}}\par{}t\par Xx
&=&
\par{}x\par Xt
\\
&=&
\par{}x\left(X+\sin X^2\right)
\\
&=&
\par Xx+2X\par Xx\cos X^2\;.
\end{eqnarray}
$$
Since the solution for $x=0$ is $X(t,0)=0$, substituting $x=0$ yields
$$\par{}t\par Xx(t,0)=\par Xx(t,0)\;.$$
For the initial value
$$\par Xx(0,0)=\left.\par{}xX(0,x)\right|_{x=0}=1\;,$$
the solution to this differential equation is
$$\par Xx(t,0)=\mathrm e^t\;.$$
