(1-x^2)y''+2xy'=0 Y1=1

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (1-x^2)y'' + 2xy' = 0 with Initial Condition y(1) = 1

This article will guide you through solving the second-order linear differential equation (1-x^2)y'' + 2xy' = 0, with the initial condition y(1) = 1.

Recognizing the Equation Type

The given equation is a Legendre's Equation, a special type of differential equation commonly found in physics and engineering. It arises when applying the method of separation of variables to solve Laplace's equation in spherical coordinates.

The Frobenius Method

To solve this equation, we'll use the Frobenius method. This method involves finding a series solution of the form:

y(x) = ∑_(n=0)^∞ a_n x^(n+r)

where r is a constant and the coefficients a_n are to be determined.

Steps:

1. Find the indicial equation: Substitute the series solution into the differential equation, and equate the coefficients of the lowest powers of x to zero. This will give you a quadratic equation known as the indicial equation.
2. Solve the indicial equation: The roots of the indicial equation will give you the possible values for r.
3. Determine the recurrence relation: Substitute the series solution back into the differential equation, and equate the coefficients of all powers of x to zero. This will give you a recurrence relation for the coefficients a_n.
4. Find the solution: Use the recurrence relation to find the first few terms of the series solution.

Note: In this case, the indicial equation will have two distinct roots, which means that two linearly independent solutions can be obtained using the Frobenius method.

Applying the Method

1. Finding the Indicial Equation:

After substituting the series solution into the differential equation and simplifying, we get:

∑(n=0)^∞ [(n+r)(n+r-1)a_n x^(n+r-2) - (n+r)(n+r-1)a_n x^(n+r)] + 2∑(n=0)^∞ (n+r)a_n x^(n+r-1) = 0

By equating the coefficients of the lowest power of x (x^(r-2)) to zero, we get the indicial equation:

r(r-1) = 0

2. Solving the Indicial Equation:

The roots of the indicial equation are:

r = 0, 1

3. Determining the Recurrence Relation:

Equating the coefficients of x^(n+r-1) to zero, we get:

[(n+r+1)(n+r)a_(n+1) - (n+r)(n+r-1)a_n] + 2(n+r)a_n = 0

Simplifying, we obtain the recurrence relation:

a_(n+1) = (n+r-1)/(n+r+1) * a_n

4. Finding the Solutions:

For r = 0:

a_1 = -1 * a_0, a_2 = 0 * a_1 = 0, a_3 = 1/3 * a_2 = 0, ...

Therefore, the first solution is:

y_1(x) = a_0 (1 - x)

For r = 1:

a_1 = 0 * a_0 = 0, a_2 = 1/3 * a_1 = 0, a_3 = 2/5 * a_2 = 0, ...

Therefore, the second solution is:

y_2(x) = a_0 x

The General Solution:

The general solution is a linear combination of the two linearly independent solutions:

y(x) = C_1 (1 - x) + C_2 x

Applying the Initial Condition:

Using the initial condition y(1) = 1, we get:

1 = C_1 (1 - 1) + C_2 * 1

Solving for C_2, we find C_2 = 1.

Therefore, the particular solution to the differential equation (1-x^2)y'' + 2xy' = 0 with initial condition y(1) = 1 is:

y(x) = x