(1-xy)ydx-x(1+xy)dy=0

3 min read Jun 16, 2024

Solving the Differential Equation (1-xy)ydx - x(1+xy)dy = 0

This article explores the solution to the given differential equation:

(1-xy)ydx - x(1+xy)dy = 0

This equation is a first-order homogeneous differential equation. Let's break down the steps to solve it:

First, we rearrange the equation to make it easier to work with:

(1-xy)ydx = x(1+xy)dy

Then, we divide both sides by xy(1+xy):

(1-xy)/(xy(1+xy))dx = (1+xy)/(xy(1+xy))dy

This simplifies to:

(1/x - 1/y)dx = (1/y + 1/x)dy

To solve the equation, we introduce a substitution:

v = y/x

This means:

y = vx

We also need to find dy:

dy = vdx + xdv

Now we substitute y = vx and dy = vdx + xdv into our simplified equation:

(1/x - 1/(vx))dx = (1/(vx) + 1/x)(vdx + xdv)

Simplifying further:

(1/x - 1/(vx))dx = (v/x + 1/x)dx + (1/v + 1)dv

Now we can separate the variables x and v:

(1/x - 1/(vx) - v/x - 1/x)dx = (1/v + 1)dv

This simplifies to:

(-1/v)dx = (1/v + 1)dv

We integrate both sides of the equation:

∫(-1/v)dx = ∫(1/v + 1)dv

This gives us:

-ln|v| = ln|v| + v + C

We can now solve for v:

2ln|v| = -v - C

ln|v|^2 = -v - C

|v|^2 = e^(-v-C)

v^2 = Ce^(-v)

Finally, we substitute back v = y/x:

(y/x)^2 = Ce^(-y/x)

Therefore, the solution to the differential equation (1-xy)ydx - x(1+xy)dy = 0 is:

y^2 = Cx^2e^(-y/x)

This is the implicit solution to the differential equation.

Note: The constant 'C' is an arbitrary constant of integration.