Differential Equation $y'-P(x)y=Q(y)$ Solve: $$(xy^4+y).dx=xdy$$
I tried but it ended into $$y'-P(x)y=Q(y)$$
Had it been $Q(x)$ , I would have been able to solve.[Lenier D.E.].
But how to solve this one?
 A: You can also write the ODE as: $$\frac{dy}{dx}=y^4+\frac{1}{x}y,~~x\neq 0$$ so it is a Bernoulli First order ODE. Try to substitute $y^{-3}=w$ and solve the following OE instead: $$w'-\frac{1}{3}w=-\frac{1}{3}$$ instead.
A: We have $(xy^4+y)dx-xdy=0$. We consider the topic of exact equations, and within it, the method of integrating factor. (Integrating factor for first order linear differential equations is comparably simpler, but our example is not amenable to this.) 
We consider this equation as the form $M(x,y)dx+N(x,y)dy=0$. If $M_y=N_x$, the equation is exact, and we may proceed directly in solving. Here, this is not the case, $M_y=4xy^3+1\ne-1=N_x$. We have to research what the integrating factor is for our problem. 
Luckily we satisfy the condition ${M_y-N_x\over -M}$ is a function of $y$ only. 
That is ${M_y-N_x\over -M}={4xy^3+1-1\over -x}=-4y^3$. 
Then, we our given that the integrating factor is 
$$\mu(y)=e^{-\int4y^3 dy}=e^{-y^4}.$$
The idea behind all this is that for exact equations, there is some $\phi(x,y)$ such that 
$$d\phi={\partial\phi\over\partial x}dx+{\partial\phi\over\partial y}dy=\mu Mdx+\mu Ndy.$$
I'll hand it off for you and others to discuss, in case I'm mistaken. 
