(1 − X2)y'' + 2xy' = 0 Y1 = 1

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (1 − x²)y'' + 2xy' = 0 with Initial Condition y(1) = 1

This article will delve into the solution of the second-order linear differential equation (1 − x²)y'' + 2xy' = 0 with the initial condition y(1) = 1. This type of equation is known as a Legendre's differential equation, which arises in various areas of physics and mathematics, particularly in solving problems related to potential theory and spherical harmonics.

Understanding the Problem

We have a differential equation where the dependent variable is y and the independent variable is x. The equation involves the second derivative of y (y'') and the first derivative of y (y'). The initial condition y(1) = 1 provides us with a specific value of y at x = 1.

Solving the Differential Equation

The provided equation is a homogeneous differential equation, meaning it can be solved using the method of Frobenius. This method involves finding a series solution of the form:

y(x) = a₀ + a₁x + a₂x² + ...

where a₀, a₁, a₂, ... are constants to be determined.

Steps for Solving using Frobenius Method:

1. Substitute the series solution into the differential equation: This will give us an equation with terms involving x and its powers.

2. Equate coefficients of like powers of x to zero: Since the equation must hold for all values of x, each coefficient must be zero.

3. Solve the resulting system of equations for the coefficients: This will give us the values of a₀, a₁, a₂, ... and thus the series solution for y(x).

4. Apply the initial condition: We use the initial condition y(1) = 1 to determine the value of the constant of integration.

The series solution for this differential equation is found to be:

y(x) = a₀P₀(x) + a₁P₁(x) + a₂P₂(x) + ...

where P₀(x), P₁(x), P₂(x), ... are Legendre polynomials. These are a set of orthogonal polynomials that form a complete basis for the space of continuous functions on the interval [-1, 1].

Using the initial condition y(1) = 1, we can determine the coefficients:

y(1) = a₀P₀(1) + a₁P₁(1) + a₂P₂(1) + ... = 1

Since P₀(1) = 1 and all other Legendre polynomials are zero at x = 1, we have a₀ = 1.

Therefore, the solution to the differential equation with the given initial condition is:

y(x) = P₀(x) = 1

Conclusion

The differential equation (1 − x²)y'' + 2xy' = 0 with the initial condition y(1) = 1 has a solution given by the Legendre polynomial P₀(x) = 1. This solution can be obtained using the Frobenius method and applying the initial condition to determine the constants in the series solution. This solution highlights the important role of Legendre polynomials in solving this type of differential equation and its applications in various scientific fields.