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. $$ (2+x) y^{\prime \prime}+(1+x) y^{\prime}+3 y=0, \quad y(0)=4, \quad y^{\prime}(0)=3 $$

Short answer

Based on the provided solution, find the series solution to the IVP. The series solution to the given IVP is: $$ y = 4 - 6x + x^2 + \cdots + a_N x^N + \cdots $$ where \(a_i\)s, for \(i > 3\), can be found by the recurrence relation: $$ (n+2)(n+1)a_{n+2}+na_{n+1}+(n-1)a_n = 0, \quad \text{for }n \geq 2 $$

Step-by-step solution

Rewrite the IVP using power series

Rewrite the given IVP in terms of a power series: $$ (2+x) \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}+(1+x) \sum_{n=1}^{\infty} n a_n x^{n-1}+3 \sum_{n=0}^{\infty} a_n x^{n} = 0 $$

Determine relationships between coefficients

Match the coefficients of each term with the same power of x, and rewrite the equation as: $$ [2a_2 + 3a_0] + [2 \cdot 3a_3 + 2a_2 + 3a_1 + a_0]x + \sum_{n=2}^{\infty} [(n+2)(n+1)a_{n+2}+na_{n+1}+(n-1)a_n]x^n = 0 $$ For the series to converge and satisfy the given IVP, all coefficients of each power of x must be equal to zero. Thus, we have: $$ 2a_2 + 3a_0 = 0 \\ 2 \cdot 3a_3 + 2a_2 + 3a_1 + a_0 =0\\ (n+2)(n+1)a_{n+2}+na_{n+1}+(n-1)a_n = 0, \quad \text{for }n \geq 2 $$

Find the coefficients using the initial conditions

Using the initial conditions: $$ y(0) = \sum_{n=0}^{\infty} a_n \cdot 0^{n} = a_0 = 4 \\ y'(0) = \sum_{n=1}^{\infty} n a_n \cdot 0^{n-1} = a_1 = 3 $$ Now we can solve for the coefficients \(a_2, a_3, ..., a_N\). From the relationships we derived in step 2: $$ 2a_2 + 3(4) = 0 \Rightarrow a_2 = -6 \\ 2 \cdot 3a_3 + 2(-6) + 3(3) + 4 = 0 \Rightarrow a_3 = 1 \\ (n+2)(n+1)a_{n+2}+na_{n+1}+(n-1)a_n = 0, \quad \text{for }n \geq 2 $$ We can continue using the recurrence relation for \(n\geq 2\) to find as many values of \(a_n\) as needed.

Write the final series solution

Combining the coefficients we found and the power series, the solution to the IVP is: $$ y = 4 - 6x + x^2 + \cdots + a_N x^N + \cdots $$ where \(a_i\)s, for \(i > 3\), can be found by the recurrence relation we established in step 2.