Solve the PDE ∂z/∂x + ∂z/∂y = z2
∂z∂x+∂z∂y=z2The auxillairy equation are, dx=dy=dzz2Integrate the first two fractions. We have that; x=y+ax−y=aIntegrate the last two fractions y=−1z+by+1z=b.Hence the solution is f(a,b)=f(x−y,y+1z)=0.\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=z^2\\ \text{The auxillairy equation are, }\\ dx=dy=\frac{dz}{z^2}\\ \text{Integrate the first two fractions. We have that; }\\ x=y+a\\ x-y=a\\ \text{Integrate the last two fractions }\\ y=-\frac{1}{z}+b\\ y+\frac{1}{z}=b.\\ \text{Hence the solution is } f(a,b)=f(x-y,y+\frac{1}{z})=0.∂x∂z+∂y∂z=z2The auxillairy equation are, dx=dy=z2dzIntegrate the first two fractions. We have that; x=y+ax−y=aIntegrate the last two fractions y=−z1+by+z1=b.Hence the solution is f(a,b)=f(x−y,y+z1)=0.
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