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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!

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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, $f(x,y)=z_0\big(\frac{x}{x_0}\big)^a\big(\frac{y}{y_0}\big)^b$.

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