Solving differential equation using a factor

Question

I want to solve $$(x^2-y)dx+xdy=0$$ Since in this form it is not an exact dif eq I multiply by a $\mu(x,y)$ factor so that it reduces to one. $$\mu (x^2-y)dx+\mu xdy=0$$ With the conditions $P_y' = Q_y'$ and $$P=\mu (x^2-y)$$ $$Q=\mu x$$ $$\mu _y'(x^2-y)-\mu=x\mu_x'+\mu$$ From here I don't know how to find $\mu$ since one is derivated with respect to $x$ and one to $y$. Could you help me? I took this easier example so that I can understand this method.

Answer 1

$$P=x^2-y\\Q=x\\\frac{\partial P}{\partial y}=-1\\\frac{\partial Q}{\partial x}=1$$

Check if the operations beyond satisfy a Polynomial that either depends only on x or y.

$$G(x)=\frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{Q}\implies G(x)=-\frac2x$$

$$G(y)=\frac{\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}}{P}\implies G(y)=\frac{2}{x^2-y}$$ The Integrating factor:

$$\mu(x)=e^{-\int\frac2xdx}=x^{-2}$$ Now The ODE is exact: $$(1-x^{-2}y)dx+x^{-1}dy=0$$ Check: $$\frac{\partial P}{\partial y}=-x^{-2}\\\frac{\partial Q}{\partial x}=-x^{-2}$$

You can solve it from here I guess.

Answer 2

$$(x^2-y)dx+xdy=0$$ $$Pdx+Qdy=0$$ You can take an integrating factor that depends only on x $$s=\partial_y P-\partial_x Q=-2$$ $$\int \frac {d \mu}{\mu}=\int \frac {\partial_y P-\partial_x Q} Q dx$$ $$\ln (\mu)=-2\int \frac {dx}x \implies \mu(x)=\frac 1 {x^2}$$ Then the equation becomes exact $$(1-\frac y {x^2})dx+\frac 1xdy=0 \implies \partial_y P=\partial_x Q$$