Solve the differential equation below with initial conditions. Compute the first 6 coefficient. Find the general pattern.

(1-x)y"-y'+xy=0 y(0)=1, y'(0)=1

To solve the given differential equation (1-x)y'' - y' + xy = 0 with initial conditions y(0) = 1 and y'(0) = 1, we can use the power series method.

1. Assume a power series solution of the form y(x) = ∑(n=0 to ∞) aₙxⁿ, where aₙ are the coefficients to be determined.

2. Differentiate y(x) with respect to x to obtain y'(x) = ∑(n=0 to ∞) naₙxⁿ⁻¹ and y''(x) = ∑(n=0 to ∞) n(n-1)aₙxⁿ⁻².

3. Substitute the power series solution and its derivatives into the differential equation.

(1-x)∑(n=0 to ∞) n(n-1)aₙxⁿ⁻² - ∑(n=0 to ∞) naₙxⁿ⁻¹ + x∑(n=0 to ∞) aₙxⁿ = 0

4. Group terms with the same power of x together and rewrite the equation as follows:

∑(n=0 to ∞) n(n-1)aₙxⁿ⁻² - ∑(n=0 to ∞) naₙxⁿ⁻¹ + ∑(n=0 to ∞) aₙxⁿ - ∑(n=0 to ∞) aₙxⁿ⁺¹ = 0

5. Observe that the powers of x in each series are consecutive integers, so we can shift the indices of the series to combine terms with the same powers of x.

∑(n=0 to ∞) (n+2)(n+1)aₙ₊₂xⁿ⁻² - ∑(n=0 to ∞) (n+1)aₙ₊₁xⁿ⁻¹ + ∑(n=0 to ∞) aₙxⁿ - ∑(n=0 to ∞) aₙxⁿ⁺¹ = 0

6. We can start combining the terms from n = 0 to ∞ by equating the coefficients of each power of x to zero.

For the term with x⁰:
(2)(1)a₂ - (1)a₁ + a₀ = 0 ... (1)

For the term with x¹:
(3)(2)a₃ - (2)a₂ + a₁ - a₀ = 0 ... (2)

For the term with x²:
(4)(3)a₄ - (3)a₃ + a₂ - a₁ = 0 ... (3)


For the term with xⁿ:
(n+2)(n+1)aₙ₊₂ - (n+1)aₙ₊₁ + aₙ - aₙ₋₁ = 0 ... (n)

7. Apply the initial conditions: y(0) = 1 and y'(0) = 1.

From (1), we have:
(2)(1)a₂ - (1)a₁ + a₀ = 0
2a₂ - a₁ + a₀ = 0 ... (4)

From (2), we have:
(3)(2)a₃ - (2)a₂ + a₁ - a₀ = 0
6a₃ - 2a₂ + a₁ - a₀ = 0 ... (5)

Substituting y(x) = ∑(n=0 to ∞) aₙxⁿ and its derivative into the initial conditions:

y(0) = 1 gives a₀ = 1 ... (6)

y'(0) = 1 gives a₁ = 1 ... (7)

8. Solve the equations (4), (5), (6), and (7) simultaneously to determine the values of a₂, a₃, and so on.

By solving the system of equations, we can find the values of a₂, a₃, and the rest of the coefficients.

Continuing this process, we can find the coefficients a₄, a₅, a₆, and so on, by solving the system of equations generated from the differential equation and substituting the initial conditions.

The general pattern for the coefficients will emerge as the equation is solved iteratively.