(2x+4y)+(2x−2y)y′=0

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (2x + 4y) + (2x - 2y)y' = 0

This article will explore the solution of the first-order differential equation (2x + 4y) + (2x - 2y)y' = 0. We will use the method of exact differential equations to solve this problem.

Identifying an Exact Differential Equation

A differential equation of the form M(x, y) dx + N(x, y) dy = 0 is considered exact if ∂M/∂y = ∂N/∂x.

To check if our equation is exact, we need to rewrite it in the standard form:

(2x + 4y) dx + (2x - 2y) dy = 0

Here, M(x, y) = 2x + 4y and N(x, y) = 2x - 2y. Let's calculate the partial derivatives:

• ∂M/∂y = 4
• ∂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, which is a function μ(x, y) that, when multiplied by the original equation, will make it exact.

We can use the following formula to find the integrating factor:

μ(x, y) = exp[∫(∂N/∂x - ∂M/∂y)/N dx] = exp[∫(2 - 4)/(2x - 2y) dx] = exp[∫(-1)/(x - y) dx] = 1/(x - y)

Solving the Exact Differential Equation

Multiplying the original equation by the integrating factor, we get:

(2 + 4y/(x - y)) dx + (2 - 2y/(x - y)) dy = 0

Now, the equation is exact because:

• ∂(2 + 4y/(x - y))/∂y = 4/(x - y)
• ∂(2 - 2y/(x - y))/∂x = 4/(x - y)

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

• ∂F/∂x = 2 + 4y/(x - y)
• ∂F/∂y = 2 - 2y/(x - y)

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

F(x, y) = 2x + 4y ln|x - y| + g(y)

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

Differentiating this expression with respect to y, we get:

∂F/∂y = 4 ln|x - y| + 4y/(x - y) + g'(y)

Comparing this with the second equation, we have:

g'(y) = -2

Integrating this, we get:

g(y) = -2y + C

where C is an arbitrary constant.

Therefore, the general solution of the differential equation is:

F(x, y) = 2x + 4y ln|x - y| - 2y + C = 0

This implicit solution represents the family of curves that satisfy the given differential equation.