Question

Find the general/particular solution of the following Differential Equations

(Non-Exact D.E)

(x + 4y³)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 $y=0.$