Chapter 8.7, Problem 26E

To obtain two linearly independent solutions to

$x^2{y}^{″}+\left(x+x^2\right)y^{\prime }+y=0$, $x>0$, $\left(46\right)$

complete the following steps.

a. Verify that $\left(46\right)$ has a regular singular point at $x=0$ and that the associated indicial equation has complex roots $\pm i$.

b. As discussed in Section $8.5$, we can express

$\begin{array}{rll}{x}^{\alpha +i\beta }& =& x^{\alpha }{x}^{i\beta }\\ & =& x^{\alpha }\cos \left(\beta \ln x\right)+i{x}^{\alpha }\sin \left(\beta \ln x\right)\end{array}$.

Deduce from this formula that $\frac{d}{dx}{x}^{\alpha +i\beta }=\left(\alpha +i\beta \right)x^{\alpha -1+i\beta }$.

c. Set $y\left(x\right)={\displaystyle {\sum }_{n=0}^{\infty }{a}_n{x}^{n+i}}$, where the coefficients now are complex constants, and substitute this series into equation $\left(46\right)$ using the result of part (b).

d. Setting the coefficients of like powers equal to zero, derive the recurrence relation

$a_n=-\frac{n-1+i}{{\left(n+i\right)}^2+1}{a}_{n-1}$, for $n\ge 1$.

e. Taking $a_0=1$, compute the coefficients $a_1$ and $a_2$ and thereby obtain the first few terms of a complex solution to $\left(46\right)$.

f. By computing the real and imaginary parts of the solution obtained in part (e), derive the following linearly independent real solutions to $\left(46\right)$:

$\begin{array}{rll}{y}_1\left(x\right)& =& \left[\cos \left(\ln x\right)\right]\left\{1-\frac{2}{5}x+\frac{1}{10}{x}^2+\mathrm{...}\right\}\\ +\left[\sin \left(\ln x\right)\right]\left\{\frac{1}{5}x-\frac{1}{20}{x}^2+\mathrm{...}\right\}\end{array}$,

$\begin{array}{rll}{y}_2\left(x\right)& =& \left[\cos \left(\ln x\right)\right]\left\{-\frac{1}{5}x+\frac{1}{20}{x}^2+\mathrm{...}\right\}\\ +\left[\sin \left(\ln x\right)\right]\left\{1-\frac{2}{5}x+\frac{1}{10}{x}^2+\mathrm{...}\right\}\end{array}$.