# Help solving a simple system of partial differential equations

I would like to solve the following system of partial differential equations: \begin{align} \frac{\partial f(x,y)}{\partial x}&= \frac{a f(x,y)}{x}, \\ \frac{\partial f(x,y)}{\partial y}&= \frac{b f(x,y)}{y}, \end{align} subject to an initial condition, $$f(x_0,y_0)=z_0$$. If I consider the variable $$y$$ to be fixed, divide both sides of the first equation by $$f(x,y)$$, and integrate both sides with respect to $$x$$, I obtain: $$\log(f(x,y))=a\log(x)+c_1(y),$$ for some function $$c_1(y)$$. Or, equivalently: $$f(x,y)=e^{c_1(y)}x^a.$$ I can do the same for the second equation to obtain: $$f(x,y)=e^{c_2(x)}y^b,$$ for some function $$c_2(x)$$. Using the initial condition, I can also infer that: $$e^{c_1(y_0)}=\frac{z_0}{x_0^a}, \; \; \text{and} \; \; e^{c_2(x_0)}=\frac{z_0}{y_0^b}.$$ But, I am not sure how to proceed from here. Intuitively, I think that the solution should be:$$f(x,y)=z_0\left(\frac{x}{x_0}\right)^{a}\left(\frac{y}{y_0}\right)^{b}.$$ Is what I have above correct? How should I proceed? Am I on the right track? Is there a better way to solve this system (possibly because it is of the form of an exact differential equation)? Any help would be appreciated!

If $$f$$ satisfies $$\frac{\partial f(x,y)}{\partial x}=\frac{af(x,y)}{x}$$, then $$f(x,y)=x^ag(y)$$ for some differentiable function $$g$$. Now since this $$f(x,y)=x^ag(y)$$ has to satisfy the equation $$\frac{\partial f(x,y)}{\partial y}=\frac{bf(x,y)}{y}$$ as well, so we get $$x^a\frac{dg}{dy}=\frac{bx^ag(y)}{y}$$, i.e., $$g(y)=cy^b$$ for some constant $$c$$. Putting this value of $$g$$ in $$f$$, we get $$f(x,y)=cx^ay^b$$. Now $$f(x_0,y_0)=z_0$$ implies $$c=\frac{z_0}{x_0^ay_0^b}$$. Finally putting the value of $$c$$ in the expression of $$f$$ we have, $$f(x,y)=z_0\big(\frac{x}{x_0}\big)^a\big(\frac{y}{y_0}\big)^b$$.