(x+y-1)dx+(2x+2y-3)dy=0

5 min read Jun 17, 2024

Solving the Differential Equation (x+y-1)dx + (2x+2y-3)dy = 0

This article will guide you through solving the differential equation (x+y-1)dx + (2x+2y-3)dy = 0. We will utilize the method of exact differential equations to find the general solution.

Understanding Exact Differential Equations

A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is considered exact if the following condition holds:

∂M/∂y = ∂N/∂x

Where ∂M/∂y represents the partial derivative of M with respect to y, and ∂N/∂x represents the partial derivative of N with respect to x.

Applying the Method

Identify M and N: In our equation, M(x, y) = x + y - 1 and N(x, y) = 2x + 2y - 3.
Check for Exactness:
- ∂M/∂y = 1
- ∂N/∂x = 2

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

Making the Equation Exact

We can use an integrating factor to transform the equation into an exact one. Let's find an integrating factor μ(x, y) that satisfies:

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

In our case, ∂M/∂y - ∂N/∂x = -1. Let's try an integrating factor that depends only on x, μ(x). This simplifies the equation:

μ(x) * (-1) = d(μ(x) * (2x + 2y - 3))/dx - d(μ(x) * (x + y - 1))/dy

μ(x) = -d(μ(x) * (2x + 2y - 3))/dx + d(μ(x) * (x + y - 1))/dy

Simplifying further, we get:

μ(x) = -2μ'(x)x - 3μ'(x) + μ'(x)x + μ'(x)

μ(x) = -μ'(x)x - 2μ'(x)

Now, we have a separable differential equation:

μ'(x)/(μ(x) + 2μ'(x)) = -1/x

Solving this equation, we get:

μ(x) = 1/x

Solving the Exact Equation

Now, we multiply the original differential equation by our integrating factor μ(x) = 1/x:

(1/x)(x + y - 1)dx + (1/x)(2x + 2y - 3)dy = 0

(1 + y/x - 1/x)dx + (2 + 2y/x - 3/x)dy = 0

Now, this equation is exact:

∂(1 + y/x - 1/x)/∂y = 1/x
∂(2 + 2y/x - 3/x)/∂x = -2/x^2 + 3/x^2 = 1/x

To find the solution, we need a function F(x, y) such that:

∂F/∂x = 1 + y/x - 1/x
∂F/∂y = 2 + 2y/x - 3/x

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

F(x, y) = x + yln(x) - ln(x) + g(y)

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

Differentiating this expression with respect to y, we get:

∂F/∂y = ln(x) + g'(y)

Comparing this to the second equation for ∂F/∂y, we see that:

g'(y) = 2 + 2y/x - 3/x - ln(x)

Since g'(y) should only depend on y, we notice that g'(y) = 2. Integrating, we get:

g(y) = 2y + C

Finally, substituting g(y) back into F(x, y), we get the general solution:

F(x, y) = x + yln(x) - ln(x) + 2y + C = 0

x + yln(x) + 2y - ln(x) + C = 0

This represents the general solution of the given differential equation.