Find general solution to this PDE

Question

$$x^2z{\partial z\over\partial x} + y^2z{\partial z \over \partial y} = x+y$$ My attempt: $${dx\over x^2z}={dy\over y^2z}={dz\over x+y}$$

Notice ${dx\over x^2z}={dy\over y^2z} \text{ we can multiply by } z \text{ and after integration} \Rightarrow {1\over x} = {1\over y} + C_1$
Here I stuck, 'cause I can't find new combination to integrate. I'm sure there are a lot of people quite good at finding these combination. So my question is also what are you looking at first. Is there common tricks to get this done?

Answer 1

The beginning of your calculus is correct. Next step : $${dx\over x^2}={dy\over y^2}=\frac{dx-dy}{x^2-y^2}\qquad\text{see note below}.$$ $${dx\over x^2z}={dy\over y^2z}=\frac{dx-dy}{(x+y)(x-y)z}={dz\over x+y}$$ $$\frac{dx-dy}{(x-y)}=zdz$$ $$ z^2-2\ln|x-y|=C_2$$ General solution : $$z^2-2\ln|x-y|=F\left(\frac{1}{x}-\frac{1}{y} \right)$$ $F$ is an arbitrary function, to be determined according to some boundary condition. $$z(x,y)=\pm\sqrt{2\ln|x-y|+F\left(\frac{1}{x}-\frac{1}{y} \right)}$$ NOTE :

Well-known property of the fractions :

$\frac{A}{B}=\frac{C}{D}=\frac{k_1A+k_2C}{k_1B+k_2D}$

In the above equation : $k_1=1$ and $k_2=-1$ , chosen in order to simplify further calculus.

Answer 2

Hint

$${dx\over x^2z}={dy\over y^2z}={dz\over x+y}$$ $${dy\over y^2z}={dz\over x+y}$$ $$(x+y){dy\over y^2}=zdz$$

But you can express from first equation $x=h(y)$ substitute and integrate simply. $${1\over x} = {1\over y} + C_1 \implies x=\frac {y}{yC_1+1}$$ Solve the following equation $$(\frac {y}{yC_1+1}+y){dy\over y^2}=zdz$$ $$(\frac {1}{yC_1+1}+1){dy\over y}=zdz$$ $$2\ln|y|-\ln |y+\frac 1 {C_1}|=\frac {z^2}2+C_2$$ $$\ln|y|-\ln |xC_1|=\frac {z^2}2+C_2$$ $$\ln|y-x|=\frac {z^2}2+C_2$$ $$C_2=2\ln|y-x|- {z^2}$$ $$H({1\over x} - {1\over y}) =2\ln|y-x|- {z^2}$$ $$z^2=2\ln|y-x|-H({1\over x} - {1\over y})$$ $$\boxed{z(x,y)=\pm\sqrt {2\ln|y-x|-H({1\over x} - {1\over y})}}$$