(1+e^x/y)dx+e^x/y(1-x/y)dy=0

5 min read Jun 16, 2024

Solving the Differential Equation: (1 + e^x/y)dx + e^x/y(1 - x/y)dy = 0

This equation is a first-order, non-linear differential equation. To solve it, we can use the following steps:

1. Identifying the Type of Equation

The given equation is not exact, as the partial derivative of the coefficient of dx with respect to y is not equal to the partial derivative of the coefficient of dy with respect to x. However, it can be made exact by multiplying the entire equation by an integrating factor.

2. Finding the Integrating Factor

We can identify an integrating factor by observing that the equation can be written in the following form:

M(x,y) dx + N(x,y) dy = 0

where:

M(x,y) = 1 + e^x/y
N(x,y) = e^x/y(1 - x/y)

Let's check if the following condition holds:

(∂M/∂y - ∂N/∂x)/N = f(x)

where f(x) is a function of x only.

Calculating the partial derivatives:

∂M/∂y = -e^x/y^2
∂N/∂x = e^x/y - e^x/y^2

Substituting these into the condition:

(-e^x/y^2 - (e^x/y - e^x/y^2))/(e^x/y(1 - x/y)) = -1/(1 - x/y)

We see that the result is a function of x and y, not just x. Therefore, the equation is not exact. However, it can be made exact by multiplying the entire equation by the integrating factor:

μ(x,y) = y

3. Multiplying by the Integrating Factor

Multiplying the given equation by μ(x,y) = y we get:

y(1 + e^x/y)dx + ye^x/y(1 - x/y)dy = 0

Simplifying the equation:

(y + e^x)dx + e^x(1 - x/y)dy = 0

Now, we can verify that this equation is exact.

∂(y + e^x)/∂y = 1
∂(e^x(1 - x/y))/∂x = e^x(1 - x/y)

Since the partial derivatives are equal, the equation is exact.

4. Solving the Exact Equation

To solve the exact equation, we need to find a function Φ(x,y) such that:

∂Φ/∂x = y + e^x
∂Φ/∂y = e^x(1 - x/y)

Integrating the first equation with respect to x, we get:

Φ(x,y) = xy + e^x + h(y)

where h(y) is an arbitrary function of y.

Taking the partial derivative of this expression with respect to y, we get:

∂Φ/∂y = x + h'(y)

Comparing this to the second equation, we can see that:

h'(y) = e^x(1 - x/y) - x

Integrating both sides with respect to y, we get:

h(y) = e^x(y - x) - xy + C

where C is the constant of integration.

Substituting this value of h(y) back into the expression for Φ(x,y), we get the general solution:

Φ(x,y) = xy + e^x + e^x(y - x) - xy + C = 0

Simplifying the solution, we get:

e^x(2y - x) + C = 0

Therefore, the general solution of the given differential equation is:

e^x(2y - x) + C = 0

where C is an arbitrary constant.