Question

Find the coefficients \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in the series solution \(y=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the initial value problem. $$ \left(1-x+x^{2}\right) y^{\prime \prime}-(1-4 x) y^{\prime}+2 y=0, \quad y(1)=2, \quad y^{\prime}(1)=-1 $$

Short answer

Question: Find the coefficients for the power series solution of the given second-order linear differential equation up to at least N = 7, and write down the expression for y(x): $$(1 -x + x^2) y'' - (1 - 4x)y' + 2y = 0$$ with the initial conditions $$y(1) = 2, \ y'(1) = -1$$ Answer: The power series solution for the given initial value problem is approximately given by: $$y(x) = 2x + x^2 - \frac{1}{3} x^3 - \frac{1}{4} x^4 + \frac{2}{15} x^5 - \frac{1}{30} x^6 + \frac{1}{35} x^7 \cdots$$

Step-by-step solution

Write down the given differential equation and initial conditions.

The differential equation is given by: $$(1 -x + x^2) y'' - (1 - 4x)y' + 2y = 0$$ The initial conditions are $$y(1) = 2 \ \text{ and } \ y'(1) = -1$$

Write down the form of the power series solution.

The power series solution is given by $$y(x) = \sum_{n=0}^{\infty} a_n x^n$$

Differentiate the power series solution.

First, we find the first and second derivatives of y with respect to x: $$y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}$$ $$y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}$$

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

Now, we substitute y, y', and y'' into the differential equation and perform the multiplications: $$(1 - x + x^2) \left(\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}\right) - (1 - 4x)\left(\sum_{n=1}^{\infty} na_n x^{n-1}\right) + 2\left(\sum_{n=0}^{\infty} a_n x^n\right) = 0$$

Collect terms with the same power of x and set coefficients equal to zero.

By comparing the coefficients of the same powers of x, we get the following recurrence relationships: $$2a_0 = 0, (2-1)a_1 = 0$$ For n ≥ 2, $$(n^2 - n)a_n - 4(n-1)a_{n-1} + (n-1)(n-2)a_{n-2} = 0$$

Use the initial conditions to find the first two coefficients.

Since y(1) = 2 and y'(1) = -1, $$2 + \sum_{n=1}^{\infty} a_n = 2 \ \text{ and } \ \sum_{n=1}^{\infty} na_n = -1$$ From these conditions, we find out that \(a_0 = 0\), \(a_1 = 2\), and \(2a_2 - 3a_1 = -1 \implies a_2 = 1\).

Calculate the remaining coefficients using the recurrence relationships.

Starting from n = 3, apply the recurrence relationships obtained in Step 5: $$a_3 = - \frac{1}{3}$$ $$a_4 = - \frac{1}{4}$$ $$a_5 = \frac{2}{15}$$ $$a_6 = - \frac{1}{30}$$ $$a_7 = \frac{1}{35}$$ And so on, according to the power index of x.

Write down the power series solution with the calculated coefficients.

Based on the coefficients calculated up to the Nth term, the power series solution for the given initial value problem is: $$y(x) = 2x + x^2 - \frac{1}{3} x^3 - \frac{1}{4} x^4 + \frac{2}{15} x^5 - \frac{1}{30} x^6 + \frac{1}{35} x^7 \cdots$$