The given second-order linear differential equation is \(y'' + 3x^2 y' + 5xy = 0\).
To find the recurrence relationship and the first four non-zero terms of each fundamental solution, we can use the Frobenius method, assuming that the solutions can be written in the form of a power series \(y = \sum_{n=0}^\infty a_nx^{n+r}\), where \(r\) is a root of the indicial equation.
First, we find the indicial equation by substituting \(y = x^r \sum_{n=0}^\infty a_nx^n\) into the differential equation and equating coefficients of \(x^{n+r}\) to zero:
\[
r(r-1) + 3rx + 5x = 0
\]
Solving for \(r\), we get the roots \(r = 0\) and \(r = -2\).
For the root \(r = 0\):
The recurrence relation is \(a_{n+2} = -\frac{5a_n}{(n+2)(n+1)}\).
The first four non-zero terms are:
\(a_0, a_2, a_4, a_6\).
For the root \(r = -2\):
The recurrence relation is \(a_{n+2} = -\frac{3a_n}{(n+2)(n+1)}\).
The first four non-zero terms are:
\(a_{-2}, a_0, a_2, a_4\).
Please note that in the case of negative indices, the term \(a_{-2}\) represents the coefficient corresponding to the term \(x^{-2}\). The recurrence relations provided above allow you to generate the terms of each fundamental solution.