Answer to Question #244286 in Differential Equations for Jac

Question

Find the differential equations of the following equations by integrating factors by inspection. Show complete solution.

ydx + (x + x^3 y^2)dy = 0

Answer 1

ydx + (x + x³ y²)dy = 0

Comparing with Mdx+Ndy=0 we get

M = y and N = x + x³y²

So $\frac{\delta{M}}{\delta{y}}=1 , \frac{\delta{N}}{\delta{x}} = 1+3x²y²$

$\frac{\delta{M}}{\delta{y}}≠\frac{\delta{N}}{\delta{x}}$

So this is not an exact differential equation.

Given equation can be written as

y.1.dx + x(1+ x² y²)dy = 0

i.e of the form yf(xy)dx+xg(xy)dy=0

So integrating factor is $\frac{1}{Mx-Ny}=\frac{1}{xy-xy-x³y³}=-\frac{1}{x³y³}$

Multiplying both sides by integrating factor we get

$\frac{y}{-x³y³}dx + \frac{x+x³y²}{-x³y³}dy=0$

=> $-\frac{1}{x³y²}dx + (-\frac{1}{x²y³}-\frac{1}{y})dy=0$

This is an exact differential equation.

So the general solution is

$-\int{\frac{1}{x³y²}dx} - \int{\frac{1}{y}}dy= C$ ,where $C$ is integration constant

=> $\frac{1}{2x²y²}-ln|y| = C$