Solution. Here \(p(x) = 4x^2\), \(q(x) = -4x^2\) and \(r(x) = 1-2x\), and \(x = 0\) is a singular point. Since \[\lim _{x \to 0} x \frac {q(x)}{p(x)} = \lim _{x \to 0} x\cdot (-1) = 0 ,\qquad \lim _{x \to 0} x^2 \frac {r(x)}{p(x)} = \lim _{x \to 0} \frac {1-2x}{4} = \frac {1}{4} ,\] both limits exist, and \(x = 0\) is a regular singular point. Try \(y = \sum ^{\infty }_{k=0} a_kx^{k+r}\), with \[y' = \sum ^{\infty }_{k = 0} (k+r)a_k x^{k+r -1},\qquad y'' = \sum ^{\infty }_{k = 0} (k+r) (k+r -1)a_k x^{k+r -2} .\] Substituting, \[\sum ^{\infty }_{k = 0}4(k+ r)(k+r - 1) a_k x^{k+r} - \sum ^{\infty }_{k = 0} 4 (k+r)a_k x^{k + r + 1} + \sum ^{\infty }_{k = 0} a_k x^{k+ r} - \sum ^{\infty }_{k = 0}2a_k x^{k + r + 1} = 0 ,\] that is, \[\sum ^{\infty }_{k = 0}\big [ 4(k+r)(k + r -1) + 1\big ] a_k x^{k + r} - \sum ^{\infty }_{k = 0}\big [4(k+r) + 2\big ] a_k x^{k+r + 1} = 0 .\] Shifting the index in the second sum so that both contain \(x^{k+r}\), \[\big [4r(r-1)+1\big ]a_0x^r + \sum ^{\infty }_{k = 1}\Big (\big [4(k+r) (k+r-1) + 1\big ] a_k - \big [4(k+r - 1 ) + 2\big ] a_{k -1}\Big )x^{k +r} = 0 .\] The lowest power gives the indicial equation \(4r^2 - 4r + 1 = (2r - 1)^2 = 0\), with the double root \(r = \frac {1}{2}\). The other coefficients give, with \(r = \frac 12\), \[a_k = \frac {4\left (k - \frac 12\right ) + 2}{4\left (k+ \frac {1}{2}\right )\left ( k - \frac {1}{2}\right ) + 1}\, a_{k-1} = \frac {4k}{4k^2}\,a_{k-1} = \frac {a_{k-1}}{k} ,\qquad k = 1,2,3,\dots ,\] so \(a_k = \frac {a_0}{k!}\) and \[y_1(x) = a_0x^{1/2}\left (1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots \right ) = a_0 x^{1/2}e^x .\] Take \(a_0 = 1\). By case 2 of Theorem 5.7.1, the second solution involves \(\ln x\). It can be found by reduction of order: dividing by \(4x^2\), the equation is \(y'' - y' + \cdots = 0\), and the reduction of order formula \(y_2 = y_1\int \frac {e^{-\int (-1)\,dx}}{y_1^2}\,dx\) gives \[y_2 = \sqrt x\,e^x\int \frac {e^x}{xe^{2x}}\,dx = \sqrt x\,e^x\int \frac {e^{-x}}{x} \,dx = \sqrt x\,e^x\left (\ln x + \sum _{n=1}^\infty \frac {(-1)^nx^n}{n\cdot n!}\right ),\] on integrating \(\frac {e^{-x}}{x} = \frac 1x - 1 + \frac {x}{2!} - \cdots \) term by term. This has the form of case 2, \(y_2 = (\ln x)y_1 + x^{1/2}\sum _{k\geq 1}b_kx^k\). The general solution for \(x>0\) is \(y = c_1y_1 + c_2y_2\).

Solution. Here \(x\frac {q}{p} = \frac 12\) and \(x^2\frac {r}{p} = \frac {x}{2}\), so \(x = 0\) is a regular singular point. Substituting \(y = \sum a_kx^{k+r}\) and collecting the lowest power \(x^{r-1}\) gives \(\big [2r(r-1) + r\big ]a_0 = r(2r - 1)a_0 = 0\), so \(r = 0\) or \(r = \frac 12\), which do not differ by an integer: case 1. The general recurrence is \[a_k = -\frac {a_{k-1}}{(k + r)(2k + 2r - 1)} .\] For \(r = \frac 12\) this is \(a_k = -\frac {a_{k-1}}{k(2k + 1)}\), giving \[y_1 = x^{1/2}\left (1 - \frac {x}{3} + \frac {x^2}{30} - \frac {x^3}{630} + \cdots \right ) ,\] and for \(r = 0\) it is \(a_k = -\frac {a_{k-1}}{k(2k - 1)}\), giving \(y_2 = 1 - x + \frac {x^2}{6} - \frac {x^3}{90} + \cdots \).