Answer to Question #277443 in Differential Equations for Dianne

Question

Find the solution of Non-exact D.E

Given:

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

Answer 1

Let us find the solution of differential equation $(x+4y^3)dy-ydx=0.$ It follows that $y=0$ is a solution. If $y\ne 0,$ then the last equation is equivalent to $x+4y^3-yx'=0,$ and hence to $yx'-x=4y^3.$ Let us divide both parts by $y^2.$ Then we get the differential equation $\frac{1}yx'-\frac{1}{y^2}x=4y,$ which is equivalent to $(\frac{1}yx)'=4y.$ It follows that $\frac{1}yx=2y^2+C.$ We conclude that the general solution of the differential equation is

$x=2y^3+Cy,\ y=0.$