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

6 min read Jun 16, 2024

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

This article will guide you through the steps of solving the given differential equation:

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

This equation is a first-order differential equation and can be solved using the method of exact differential equations.

Step 1: Check for Exactness

A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is said to be exact if:

∂M/∂y = ∂N/∂x

In our case:

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

Let's calculate the partial derivatives:

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

Since ∂M/∂y ≠ ∂N/∂x, the given differential equation is not exact.

Step 2: Finding an Integrating Factor

To make the equation exact, we need to find an integrating factor, which is a function μ(x, y) that satisfies:

μ(x, y) * (∂M/∂y - ∂N/∂x) = ∂(μN)/∂x - ∂(μM)/∂y

We can try finding a function μ(x, y) that depends only on x or y.

Let's assume μ = μ(x). Then, the above condition becomes:

μ(x) * (-e^x/y^2 - e^x/y + xe^x/y^2) = ∂(μ(x) * e^x/y(1 - xy))/∂x - ∂(μ(x) * (1 + e^x/y))/∂y

Simplifying:

μ(x) * (-e^x/y + xe^x/y^2) = μ'(x) * e^x/y(1 - xy) + μ(x) * e^x/y - μ(x) * e^x/y^2

Rearranging:

μ'(x) * e^x/y(1 - xy) = -μ(x) * xe^x/y^2

This simplifies to:

μ'(x)/μ(x) = -x/(1 - xy)

Integrating both sides with respect to x:

ln|μ(x)| = -∫(x/(1 - xy)) dx = ln|1 - xy| + C

Therefore, the integrating factor is:

μ(x) = 1 - xy

Step 3: Multiplying by the Integrating Factor

Multiplying the original differential equation by the integrating factor:

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

This simplifies to:

(1 - xy + e^x/y - xe^x)dx + e^x(1 - xy)^2/y dy = 0

Now, the equation is exact because:

∂[(1 - xy + e^x/y - xe^x)]/∂y = -x - e^x/y^2 + xe^x/y^2 = -x - e^x/y^2 ∂[e^x(1 - xy)^2/y]/∂x = e^x(1 - xy)^2/y - 2xe^x(1 - xy)/y = -x - e^x/y^2

Step 4: Finding the Solution

Since the equation is now exact, there exists a function F(x, y) such that:

∂F/∂x = M(x, y) = (1 - xy + e^x/y - xe^x) ∂F/∂y = N(x, y) = e^x(1 - xy)^2/y

Integrating the first equation with respect to x:

F(x, y) = x - x^2y/2 + e^x/y - xe^x + g(y)

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

Differentiating F(x, y) with respect to y and equating it to N(x, y):

∂F/∂y = -x^2/2 - e^x/y^2 + g'(y) = e^x(1 - xy)^2/y

Solving for g'(y):

g'(y) = e^x(1 - xy)^2/y + x^2/2 + e^x/y^2

Integrating both sides with respect to y:

g(y) = -e^x(1 - xy)/y + x^2y/4 + C

Substituting g(y) back into F(x, y):

F(x, y) = x - x^2y/2 + e^x/y - xe^x - e^x(1 - xy)/y + x^2y/4 + C

Finally, the general solution of the differential equation is given by:

F(x, y) = C

x - x^2y/4 + e^x/y - xe^x - e^x(1 - xy)/y = C

where C is an arbitrary constant.

This solution represents a family of curves that are solutions to the given differential equation.