Is it possible to solve this ODE by separation of variables?

Question

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?

Answer 1

Separable equations are exact, $$ \frac{dy}{dx}=f(x)g(y)\implies g(y)dy-f(x)dx=0 $$ This one is not since $$ \frac{\partial}{\partial y} y=1\ne \frac{\partial}{\partial x}( 3x-14y^4)=3 $$

Answer 2

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.

Answer 3

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, $$\boxed{x(y)=\frac K {y^3}=2y^4+\frac {C} {y^3}}$$