(1+e^x/y)dx+e^x/y(1-xy)dy=0 By Exact Differential Equation

7 min read Jun 16, 2024

(1+e^x/y)dx+e^x/y(1-xy)dy=0 By Exact Differential Equation

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

This article will demonstrate how to solve the given differential equation using the method of exact differential equations.

1. Identifying the Equation as Exact

The given differential equation is:

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

We need to check if this equation is exact. A differential equation of the form:

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

is exact if and only 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.

2. Finding an Integrating Factor

To make the equation exact, we need to find an integrating factor. This is a function, μ(x,y), that when multiplied to the equation will make it exact.

We can find the integrating factor by using the following formulas:

If (∂N/∂x - ∂M/∂y)/M is a function of y only, then μ(y) = exp(∫(∂N/∂x - ∂M/∂y)/M dy)
If (∂M/∂y - ∂N/∂x)/N is a function of x only, then μ(x) = exp(∫(∂M/∂y - ∂N/∂x)/N dx)

Let's try the first formula:

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

This expression is a function of both x and y, so this formula doesn't help.

Now, let's try the second formula:

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

This expression is a function of x and y only, so we can use it to find the integrating factor.

μ(x) = exp(∫(-1/(1-xy)) dx) = exp(ln(1-xy)) = 1-xy

3. Multiplying by the Integrating Factor

Now, we multiply the original equation by the integrating factor, μ(x) = (1-xy):

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

This simplifies to:

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

4. Checking for Exactness

Now, let's check if this new equation is exact:

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

Calculating the partial derivatives:

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

Simplifying, we see that ∂M/∂y = ∂N/∂x. Therefore, the equation is now exact.

5. Solving the Exact Equation

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

∂F/∂x = M(x,y)
∂F/∂y = N(x,y)

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

F(x,y) = x - (x^2*y)/2 + e^x(1-x)/y + g(y)

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

Now, we differentiate F(x,y) with respect to y and set it equal to N(x,y):

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

Solving for g'(y), we get:

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

Integrating with respect to y, we get:

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

where C is an arbitrary constant.

Substituting g(y) back into F(x,y), we get the general solution of the differential equation:

F(x,y) = x - (x^2*y)/2 + e^x(1-x)/y - e^x(1-xy)/y + x^2y/2 - e^x(1-x)/y + C = 0

Simplifying the equation, the final solution is:

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

This implicit equation represents the solution to the given differential equation.