Answer to Question #277437 in Differential Equations for Kent

Question

Find the solution of Non-Exact D.E

Given:

(x+4y^3)dy-ydx=0

Answer 1

Let us find the general solution of the differential equation $(x + 4y^3)dy - ydx =0$ which is equivalent to $- yx' +x + 4y^3=0.$ 

Note that $y=0$ is a solution. Let us divide both part of the last differential equation by $-y^2.$Then we get $\frac{1}{y}x' -\frac{1}{y^2}x =4y.$ 

It follows that $(\frac{1}{y}x)' = 4y,$ and hence $\frac{1}{y}x=2y^2+C.$ 

We conclude that the general solution is $x=2y^3+Cy$ or $y=0.$