(1+lnx+y/x)dx=(1-lnx)dy

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation: (1 + ln(x) + y/x)dx = (1 - ln(x))dy

This article will delve into the solution of the given differential equation. We will use techniques of exact differential equations to solve for the general solution.

Identifying the Form and Conditions

The given differential equation is in the form of:

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

Where:

• M(x, y) = 1 + ln(x) + y/x
• N(x, y) = 1 - ln(x)

For a differential equation to be exact, the following condition must hold:

∂M/∂y = ∂N/∂x

Let's check if our equation satisfies this condition:

• ∂M/∂y = 1/x
• ∂N/∂x = -1/x

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

Transforming to an Exact Equation

We can make the equation exact by finding an integrating factor. An integrating factor, μ(x, y), is a function that we multiply the entire equation by to make it exact. There are two common types of integrating factors:

1. μ(x) depends only on x
2. μ(y) depends only on y

In our case, we can use an integrating factor that depends only on x. This is because the difference between ∂M/∂y and ∂N/∂x is a function of x only:

∂M/∂y - ∂N/∂x = 1/x + 1/x = 2/x

To find the integrating factor μ(x), we use the formula:

μ(x) = exp(∫ (∂M/∂y - ∂N/∂x)/N dx)

In this case:

μ(x) = exp(∫ (2/x)/(1 - ln(x)) dx) = exp(-2 ln(1 - ln(x))) = (1 - ln(x))^-2

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

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

Now, we can verify that the new equation is exact:

• ∂M/∂y = 1/x(1 - ln(x))^2
• ∂N/∂x = 1/x(1 - ln(x))^2

Since ∂M/∂y = ∂N/∂x, the equation is now exact.

Solving the Exact Equation

Now that we have an exact equation, we can solve for the general solution using the following steps:

1. Find a function F(x, y) such that:

   • ∂F/∂x = M(x, y)
   • ∂F/∂y = N(x, y)

2. The general solution is given by:

   • F(x, y) = C (where C is an arbitrary constant)

Let's find F(x, y):

• ∂F/∂x = (1 + ln(x) + y/x) / (1 - ln(x))^2

  • Integrating with respect to x, we get:
    • F(x, y) = -1/(1 - ln(x)) + y/(1 - ln(x)) + g(y) (where g(y) is an arbitrary function of y)

• ∂F/∂y = 1/(1 - ln(x)) + g'(y) = (1 - ln(x)) / (1 - ln(x))^2

  • Comparing both sides, we find:
    • g'(y) = 0
    • g(y) = C_1 (where C_1 is an arbitrary constant)

Therefore, the general solution of the given differential equation is:

-1/(1 - ln(x)) + y/(1 - ln(x)) + C_1 = C

Or, we can simplify it as:

y = C(1 - ln(x)) + 1 - C_1(1 - ln(x))

This is the general solution of the given differential equation.