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

6 min read Jun 17, 2024

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

This article will guide you through solving the differential equation (x+y-1)dx + (2x+2y+1)dy = 0, which is a first-order, non-linear differential equation. We'll use the method of exact differential equations to find its solution.

Understanding Exact Differential Equations

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

∂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.

Checking for Exactness

In our equation, M(x,y) = x+y-1 and N(x,y) = 2x+2y+1. Let's calculate the partial derivatives:

∂M/∂y = 1

∂N/∂x = 2

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

Making the Equation Exact

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

Finding the Integrating Factor

Check for a function of x or y only: If either ∂(N/M)/∂x or ∂(M/N)/∂y is a function of either x or y only, then the integrating factor is given by:

μ(x) = exp(∫(∂(N/M)/∂x)dx) or μ(y) = exp(∫(∂(M/N)/∂y)dy)
Calculate (N/M): (N/M) = (2x+2y+1)/(x+y-1)
Find ∂(N/M)/∂x: ∂(N/M)/∂x = (2(x+y-1) - (2x+2y+1))/(x+y-1)^2 = -3/(x+y-1)^2
Calculate μ(x): μ(x) = exp(∫(-3/(x+y-1)^2)dx) = exp(3/(x+y-1))

Now, we multiply the original equation by the integrating factor μ(x) = exp(3/(x+y-1)):

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

This makes the equation exact.

Solving the Exact Equation

Let's denote the left side of the new equation as:

F(x,y) = exp(3/(x+y-1))[(x+y-1)dx + (2x+2y+1)dy]

Since the equation is now exact, we can find a solution by finding a function F(x,y) such that:

dF = Mdx + Ndy

To find F(x,y), we can integrate either M or N. Let's integrate M:

F(x,y) = ∫Mdx = ∫exp(3/(x+y-1))(x+y-1)dx

We can use the substitution u = x+y-1, du = dx, and simplify to get:

F(x,y) = ∫exp(3/u)udu = exp(3/u) + g(y)

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

Now, we can find g(y) by taking the partial derivative of F(x,y) with respect to y:

∂F/∂y = -3exp(3/(x+y-1))/(x+y-1)^2 + g'(y)

This should be equal to N:

-3exp(3/(x+y-1))/(x+y-1)^2 + g'(y) = exp(3/(x+y-1))(2x+2y+1)

Solving for g'(y) and integrating, we get:

g(y) = 2exp(3/(x+y-1))(x+y) + C

where C is an arbitrary constant.

Therefore, the general solution of the differential equation is:

F(x,y) = exp(3/(x+y-1)) + 2exp(3/(x+y-1))(x+y) + C = 0

Final Solution

The solution to the differential equation (x+y-1)dx + (2x+2y+1)dy = 0 is:

exp(3/(x+y-1)) + 2exp(3/(x+y-1))(x+y) = C