# Integrating over functions, not variables, in a first order differential equation?

Suppose I'm trying to solve the differential equation

$y'(x) = k\frac{y(x)}{x}$

where $k$ is a constant and $y'(x)$ is $\frac{\partial (y(x))}{\partial x}$. One could isolate the $y$s on one side and the $x$s on another to get

$\frac{1}{y(x)} \partial (y(x)) = k \frac{1}{x} \partial x$

Integrating on both sides:

$\int_{x_{0}}^{x} \frac{1}{y(x)} \partial (y(x)) = \int_{x_{0}}^{x} k \frac{1}{x} \partial x$

$\ln(y(x)/y(x_{0}))= k \ln(x/x_{0})$

... and a little algebra results in:

$y(x) = y(x_{0})(\frac{x}{x_{0}})^{k}$

This is pretty standard. But what if instead I had the equation

$\frac{\partial (y(x(t)))}{\partial (x(t))} = k(t) \frac{y(x(t))}{x(t)}$

where $k$ is now a function of time, and so is $x$? Assuming that $k$ and $x$ are analytic, is there any way to solve for $y$? I've tried using the naïve method above and integrating across the function $x$ rather than the variable $x$

$\int_{x_{0}(t)}^{x(t)} \frac{1}{y(x(t))} \partial (y(x(t))) = \int_{x_{0}(t)}^{x(t)} k(t) \frac{1}{x(t)} \partial (x(t))$

but my intuition tells me it's not so simple. Once $t$ gets involved, I don't think the initial conditions for $x$ and $y$ stay stable if you try integrating over a function; also, I'm not even sure if integrating over a function is something you can do at all.

Help?

(I know extremely little about differential equations, apologies if I'm missing something obvious)