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. $$ \left(1+x+3 x^{2}\right) y^{\prime \prime}+(2+15 x) y^{\prime}+12 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

Short answer

Question: Find the coefficients \(a_0, a_1, \ldots, a_N\) for \(N \geq 7\) of the power series solution to the given second-order homogeneous linear differential equation with initial conditions: \((1+x+3x^2)y''(x)+(2+15x)y'(x)+12y(x) = 0\), \(y(0)=0\), and \(y'(0)=1\). Answer: The coefficients are as follows: $$ a_0 = 0 \\ a_1 = 1 \\ a_2 = 0 \\ a_3 = 0 \\ a_n = -\frac{12}{n(n-1)(1+3(n-2))+15(n-1)}a_n, \, \text{for} \, n \geq 4. $$

Step-by-step solution

Expand the power series and its derivatives

First, we need to calculate the derivatives of the power series: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n \\ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \\ y''(x) = \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2} $$

Apply the given differential equation

Now we need to substitute the power series and its derivatives into the given differential equation: $$ (1+x+3x^2)y''(x)+(2+15x)y'(x)+12y(x) = 0 $$ Substitute the series expressions for \(y\), \(y'\), and \(y''\): $$ (1+x+3x^2) \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2}+(2+15x) \sum_{n=1}^{\infty} n a_n x^{n-1}+12 \sum_{n=0}^{\infty} a_n x^n = 0 $$

Equate coefficients of equal powers of x

To find the recurrence relation between the coefficients \(a_0, a_1, \ldots, a_N\), we need to equate coefficients of equal powers of x: $$ \sum_{n=2}^{\infty} (1+x+3x^2)n(n-1)a_n x^{n} + \sum_{n=1}^{\infty}(2+15x)na_n x^{n} + \sum_{n=0}^{\infty} 12a_n x^n = 0 $$ Combine the summations: $$ (2a_2) + (2a_3 + 3a_2)x + \sum_{n=4}^{\infty} [n(n-1)a_n(1+3(n-2))+15(n-1)a_n]x^n + \sum_{n=2}^{\infty} 12a_n x^n = 0 $$ Now, equate the coefficients of x^n: For n=0, $$ 2a_2=0 \Rightarrow a_2=0 $$ For n=1, $$ 2a_3+3a_2=0 \Rightarrow 2a_3=0 \Rightarrow a_3=0 $$ For n≥4: $$ [n(n-1)(1+3(n-2))+15(n-1)]a_n + 12a_n = 0 $$ Rearrange: $$ a_n = -\frac{12}{n(n-1)(1+3(n-2))+15(n-1)}a_n $$

Apply the initial conditions

Finally, we will use the initial conditions to find the coefficients \(a_0, a_1, \ldots, a_N\). $$ y(0)=0 \Rightarrow a_0=0 \\ y'(0)=1 \Rightarrow a_1=1 $$ Now that we have initial conditions and the recursion formula for \(a_n\), we can calculate coefficients \(a_2, a_3, \ldots, a_N\).