Question

find a three - term recurrence relation for solutions of the form ( y=sum_{n = 0}^{infty}c_{n}x^{n} ). then find the first three nonzero terms in each of two linearly independent solutions
( (x^{2}-2)y + 2xy + 2xy = 0 )
the three - term recurrence relation is ( c_{2}=0,c_{n + 2}=\frac{2c_{n-1}+n(n + 1)c_{n}}{2(n + 2)(n + 1)} ) for ( ngeq1 ).
enter the first three nonzero terms in each of two linearly independent solutions. the first term of ( y_{1} ) is given
( y_{1}(x)=1+square+cdots )
( y_{2}(x)=square+cdots )

Explanation:

Step1: Determine coefficients for \(y_1(x)\)

Given \(y_1(x)=\sum_{n = 0}^{\infty}c_nx^n\) with \(c_0 = 1\), \(c_1=0\) (since \(y_1\) has no \(x\) - term initially).
For \(n = 1\) in the recurrence relation \(c_{n + 2}=\frac{2c_{n-1}+n(n + 1)c_n}{2(n + 2)(n + 1)}\), we have:
\(c_{3}=\frac{2c_{0}+1\times(1 + 1)c_1}{2(1+ 2)(1 + 1)}\)
Substitute \(c_0 = 1\) and \(c_1 = 0\)
\(c_{3}=\frac{2\times1+2\times0}{2\times3\times2}=\frac{1}{6}\)

For \(n = 2\) (but \(c_2 = 0\)), \(c_{4}=\frac{2c_{1}+2\times(2 + 1)c_2}{2(2+ 2)(2 + 1)} = 0\)

For \(n = 3\), \(c_{5}=\frac{2c_{2}+3\times(3 + 1)c_3}{2(3+ 2)(3 + 1)}\)
Substitute \(c_2 = 0\) and \(c_3=\frac{1}{6}\)
\(c_{5}=\frac{0+3\times4\times\frac{1}{6}}{2\times5\times4}=\frac{1}{20}\)

Step2: Determine coefficients for \(y_2(x)\)

Let \(y_2(x)=\sum_{n = 0}^{\infty}d_nx^n\) with \(d_0 = 0\), \(d_1 = 1\)
For \(n = 1\) in the recurrence relation \(d_{n + 2}=\frac{2d_{n-1}+n(n + 1)d_n}{2(n + 2)(n + 1)}\)
\(d_{3}=\frac{2d_{0}+1\times(1 + 1)d_1}{2(1+ 2)(1 + 1)}\)
Substitute \(d_0 = 0\) and \(d_1 = 1\)
\(d_{3}=\frac{0 + 2\times1}{2\times3\times2}=\frac{1}{6}\)

For \(n = 2\) ( \(d_2 = 0\)), \(d_{4}=\frac{2d_{1}+2\times(2 + 1)d_2}{2(2+ 2)(2 + 1)}=\frac{2\times1+6\times0}{2\times4\times3}=\frac{1}{12}\)

Answer:

\(y_1(x)=1+\frac{1}{6}x^{3}+\frac{1}{20}x^{5}+\cdots\)
\(y_2(x)=x+\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\cdots\)