(1-y^2)(1+log X)dx+2xy Dy=0

Jasonbradley

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Jun 16, 2024

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

This article explores the solution to the given differential equation:

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

This equation is a first-order, non-linear differential equation because it involves derivatives of the first order and the terms are not linear in the dependent variable y. We will use the method of separation of variables to solve this equation.

Separating Variables

1. Rearrange the equation:

   We begin by isolating the terms involving x and y on their respective sides of the equation:

   2xy dy = -(1-y^2)(1+log x) dx 
   

2. Divide by appropriate terms:

   To separate the variables, we divide both sides by (1-y^2) and 2xy:

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

Now we have the variables x and y separated on different sides of the equation.

Integration

1. Integrate both sides:

   Integrate both sides of the equation with respect to their respective variables:

   ∫ dy/(1-y^2) = - ∫ (1+log x)/(2x) dx
   

2. Solve the integrals:

   The left-hand side can be solved using partial fraction decomposition. The right-hand side can be solved using substitution.

   • Left-hand side:

     ∫ dy/(1-y^2) = ∫ (1/2) * [1/(1-y) + 1/(1+y)] dy = (1/2) * [ln(1+y) - ln(1-y)] + C1
     

   • Right-hand side:

     Let u = 1+log x. Then du = 1/x dx.

     - ∫ (1+log x)/(2x) dx = - (1/2) ∫ u/x * (x du) = - (1/2) ∫ u du = - (1/4)u^2 + C2 
     

3. Combine the results:

   Combining the results of the integrations and replacing u with 1+log x, we get:

   (1/2) * [ln(1+y) - ln(1-y)] = - (1/4)(1+log x)^2 + C 
   

   where C = C2 - C1 is the constant of integration.

Solution

Finally, we can express the solution explicitly for y:

ln(1+y) - ln(1-y) = - (1/2)(1+log x)^2 + 2C

Using the property of logarithms, ln(a) - ln(b) = ln(a/b), we get:

ln[(1+y)/(1-y)] = - (1/2)(1+log x)^2 + 2C

Exponentiating both sides:

(1+y)/(1-y) = exp[ - (1/2)(1+log x)^2 + 2C]

Simplifying and solving for y:

y = [exp[ - (1/2)(1+log x)^2 + 2C] - 1] / [exp[ - (1/2)(1+log x)^2 + 2C] + 1]

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

y = [exp[ - (1/2)(1+log x)^2 + 2C] - 1] / [exp[ - (1/2)(1+log x)^2 + 2C] + 1]

This solution involves an arbitrary constant C, which can be determined by applying an initial condition if provided.