Answer:
To find the power series solution of the initial value problem, we assume a power series solution of the form:
\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]
where \( a_n \) are coefficients to be determined, and the first two derivatives of \( y \) with respect to \( x \) can be expressed as follows:
\[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n \]
\[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \]
Now, we'll substitute these expressions into the differential equation:
\[ (2 + x^2) y'' - x y' + 4y = 0 \]
\[ (2 + x^2) \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - x \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n + 4 \sum_{n=0}^{\infty} a_n x^n = 0 \]
Next, we'll multiply out the terms and collect the coefficients of \( x^n \) to get a recurrence relation for the coefficients \( a_n \). Equating the coefficients to zero will give us the recurrence relation:
\[ \sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + 4a_n] x^n - \sum_{n=0}^{\infty} [(n+1)a_{n+1}] x^{n+1} = 0 \]
We can shift the index of the second summation and set the coefficients of \( x^n \) to zero:
\[ (n+2)(n+1)a_{n+2} + 4a_n - (n+1)a_{n+1} = 0 \]
Now, we can solve for the coefficients \( a_{n+2} \) in terms of \( a_n \):
\[ a_{n+2} = \frac{(n+1)a_{n+1} - 4a_n}{(n+2)(n+1)} \]
We already have the initial conditions given:
\[ y(0) = a_0 = -1 \]
\[ y'(0) = a_1 = 3 \]
Now, we can use the recurrence relation to find the first few coefficients:
For \( n = 0 \):
\[ a_2 = \frac{(0+1)a_1 - 4a_0}{(0+2)(0+1)} = \frac{(1)(3) - 4(-1)}{2} = \frac{7}{2} \]
For \( n = 1 \):
\[ a_3 = \frac{(1+1)a_2 - 4a_1}{(1+2)(1+1)} = \frac{(2)\left(\frac{7}{2}\right) - 4(3)}{6} = \frac{1}{2} \]
For \( n = 2 \):
\[ a_4 = \frac{(2+1)a_3 - 4a_2}{(2+2)(2+1)} = \frac{(3)\left(\frac{1}{2}\right) - 4\left(\frac{7}{2}\right)}{12} = -\frac{5}{8} \]
For \( n = 3 \):
\[ a_5 = \frac{(3+1)a_4 - 4a_3}{(3+2)(3+1)} = \frac{(4)\left(-\frac{5}{8}\right) - 4\left(\frac{1}{2}\right)}{20} = -\frac{17}{40} \]
For \( n = 4 \):
\[ a_6 = \frac{(4+1)a_5 - 4a_4}{(4+2)(4+1)} = \frac{(5)\left(-\frac{17}{40}\right) - 4\left(-\frac{5}{8}\right)}{30} = \frac{49}{192} \]
So, the first five nonzero terms in the power series solution of the initial value problem are:
\[ y(x) = -1 + 3x + \frac{7}{2}x^2 + \frac{1}{2}x^3 - \frac{5}{8}x^4 - \frac{17}{40}x^5 + \mathcal{O}(x^6) \]
Explanation: