How do you solve $x\dfrac{\partial z}{\partial x}+2y\dfrac{\partial z}{\partial y}=x^2y+z$?

Question

The given: $x\dfrac{\partial z}{\partial x}+2y\dfrac{\partial z}{\partial y}=x^2y+z$
Transform the PDE into an ODE: $ \frac{dy}{x} = \frac{dy}{2y} = \frac{dz}{x^2y+z}$.
From the first two equations obtain $ \frac{dy}{x} = \frac{dy}{2y} \rightarrow φ_1= \frac{y}{x^2}$
But I have no idea to find the other. Please enlighten me. Thank you.

Answer 1

Hint:

Auxiliary equation:

$$\frac{dx}{x} = \frac{dy}{2y} = \frac{dz}{x^2y+z}$$

Solving $\frac{dx}{x} = \frac{dy}{2y}$

We have $x^2=cy\tag{1}$ for some constant $c$.

Solve $$\frac{dy}{2y} = \frac{dz}{x^2y+z}$$

$$\implies \frac{dy}{2y} = \frac{dz}{cy^2+z}\quad [\text{using (1)}] $$

$$\implies \frac{dz}{dy} = \frac{cy^2+z}{2y}$$

$$\implies \frac{dz}{dy} -\frac{z}{2y}=\frac{c}{2}y$$

Which is easy to solve, linear first order ode.

Can you finish from here?

Answer 2

$$\dfrac{dx}{x} = \dfrac{dy}{2y} = \dfrac{dz}{x^2y+z}$$ For the second constant of integration: $$\dfrac{ydx^2}{2x^2y} = \dfrac{x^2dy}{2x^2y} = \dfrac{dz}{x^2y+z}$$ $$\dfrac{dx^2y}{4x^2y} = \dfrac{dz}{x^2y+z}$$ $$\dfrac{u+z}{4u} = \dfrac{dz}{du}$$ Where $u=x^2y$. This is a first order DE. Substitute $z=tu$: $$1+t=4t'u+4t$$ $$4t'u=1-3t$$ This DE is separable.