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

Jasonbradley

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Jun 16, 2024

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

This article will explore the solution of the second-order linear differential equation:

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

Identifying the Type of Differential Equation

First, we need to identify the type of differential equation we are dealing with. This equation is:

• Second-order: It involves the second derivative of y.
• Linear: The dependent variable y and its derivatives appear only to the first power and are not multiplied together.
• Homogeneous: The right-hand side of the equation is zero.

Finding the Solution

Solving this differential equation involves finding two linearly independent solutions, which form a fundamental set of solutions. The general solution will then be a linear combination of these two solutions.

1. Finding the Roots of the Auxiliary Equation:

• We can rewrite the given equation as:

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

• Now, we assume a solution of the form y = e^(rx). Substituting this into the equation and simplifying, we get the auxiliary equation:

  (1-2x-x^2)r^2 + 2(1+x)r - 2 = 0

• This is a quadratic equation in r. However, it has variable coefficients, making it a little tricky to solve directly.

2. Using the Frobenius Method:

The Frobenius method is a powerful technique for solving linear differential equations with variable coefficients. This method involves finding solutions of the form:

• y = x^r * Σ_(n=0)^∞ a_n * x^n

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

• By substituting this into the original equation and solving for the coefficients, we can obtain two linearly independent solutions.

3. Finding the Two Linearly Independent Solutions:

Applying the Frobenius method, we find that the two linearly independent solutions are:

• y_1(x) = 1/(1+x)
• y_2(x) = (1+x)

4. General Solution:

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

• y(x) = c_1 * y_1(x) + c_2 * y_2(x)

where c_1 and c_2 are arbitrary constants.

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

y(x) = c_1/(1+x) + c_2(1+x)

Important Note:

Solving this equation using the Frobenius method involves some complex algebraic manipulations and finding recurrence relations for the coefficients. It is a more advanced technique that requires a good understanding of series solutions of differential equations.