Is it possible to solve this ODE by separation of variables? Consider the equation $ydx +3xdy =14y^4dy$. Is there a clever trick which would allow to solve this equation by the method of separation of variables?
 A: No it's not seperable in y
$$ydx +3xdy =14y^4dy$$
$$y +3xy' =14y^4y'$$
$$yx'+3x=14y^4$$
it's a classical equation thats easy to solve
First solve the homogeneous equation
$$yx'+3x=0 \implies \ln|x|=-3\ln|y|+K \implies x=\frac K {y^3}$$
Now plug K in the non homogeneous equation and consider K as $K(y)$ Solve the first order ( separable equation) to find K
$$y(K'y^{-3}-3Ky^{-4})+3Ky^ {-3}=14y^4 $$
Then you get a seperable equation in $K(y)$
$$K'=14y^6$$
Integrate
$$\int dK=14\int y^6dy$$
$$\implies K( y)=2y^7+C$$
Therefore,
$$\boxed{x(y)=\frac K {y^3}=2y^4+\frac {C} {y^3}}$$
A: This equation $$ydx +3xdy =14y^4dy$$ is not separable.
Note that if you try to factor $dy$, you will get an expression involving both $x$ and $y.$
$$ydx=(14y^4-3x)dy$$
We can solve the equation if we solve for $x$ as a function of $y.$
Note that  $$dx/dy=(14y^3-3x/y)$$ Which is linear.
A: Separable equations are exact, 
$$
\frac{dy}{dx}=f(x)g(y)\implies g(y)dy-f(x)dx=0
$$
This one is not since
$$
\frac{\partial}{\partial y} y=1\ne \frac{\partial}{\partial x}(
3x-14y^4)=3
$$
