5.5: Solve x^{2}y′′ − x(2 + x)y′ + (2 + x^{2})y = 0 p(x) = −\frac{x(2+x)}{x^{2}} = −\frac{2+x}{x}. Thus x_{0} = 0 is a singular value. q(x) = \frac{2+x^{2}}{x^{2}} also implies x_{0} = 0 is a singular value. xp(x) = −(2 + x) and x^{2}q(x) = 2 + x^{2}. Thus x_{0} = 0 is a regular singular value. Suppose y = ∑_{n=0}^{∞} a_{n}x^{n+r} is a solution. WLOG assume a_{0} ≠ 0 (otherwise one can reindex the summation). Then y′ = ∑_{n=0}^{∞}(n + r)a_{n}x^{n+r−1} and y′′ = ∑_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r−2} x^{2}y′′ − x(2 + x)y′ + (2 + x^{2})y = x^{2} ∑_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r−2} − (2x + x^{2}) ∑_{n=0}^{∞}(n + r)a_{n}x^{n+r−1} + (2 + x^{2}) ∑_{n=0}^{∞} a_{n}x^{n+r} = ∑_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r} − ∑_{n=0}^{∞} 2(n + r)a_{n}x^{n+r} − ∑_{n=0}^{∞}(n + r)a_{n}x^{n+r+1}

+ ∑_{n=0}^{∞} 2a_{n}x^{n+r} + ∑_{n=0}^{∞} a_{n}x^{n+r+2}

= ∑_{n=0}^{∞}[(n + r)(n + r − 1) − 2(n + r) + 2]a_{n}x^{n+r} − Σ_{n=1}^{∞}(n + r − 1)a_{n−1}x^{n+r} + Σ_{n=2}^{∞}a_{n−2}x^{n+r} = [r(r − 1) − 2r + 2]a_{0}x^{r} + [(1 + r)r − 2(1 + r) + 2]a_{1}x^{r+1} − ra_{0}x^{r+1}

+Σ_{n=2}^{∞}[(n + r)(n + r − 1) − 2(n + r) + 2]a_{n}x^{n+r} − Σ_{n=2}^{∞}(n + r − 1)a_{n−1}x^{n+r} + Σ_{n=2}^{∞}a_{n−2}x^{n+r} = [r(r − 1) − 2r + 2]a_{0}x^{r} + ([(1 + r)r − 2(1 + r) + 2]a_{1} − ra_{0})x^{r+1}

+Σ_{n=2}^{∞}([(n + r)(n + r − 1) − 2(n + r) + 2]a_{n} − (n + r − 1)a_{n−1} + a_{n−2})x^{n+r}

= [r^{2} − r − 2r + 2]a_{0}x^{r} + ([r + r^{2} − 2 − 2r + 2]a_{1} − ra_{0})x^{r+1}

+Σ_{n=2}^{∞}([(n + r)(n + r − 3) + 2]a_{n} − (n + r − 1)a_{n−1} + a_{n−2})x^{n+r}

= [r^{2} − 3r + 2]a_{0}x^{r} + ([r^{2} − r]a_{1} − ra_{0})x^{r+1} +Σ_{n=2}^{∞}([n^{2} + 2rn + r^{2} − 3n − 3r + 2]a_{n} − (n + r − 1)a_{n−1} + a_{n−2})x^{n+r} = 0 Set all coefficients = 0: Since a_{0} ≠ 0, r^{2} − 3r + 2 = (r − 2)(r − 1) = 0 implies r = 1, 2. r^{2} − 3r + 2 = 0 is the indicial equation [r^{2} − r]a_{1} = ra_{0} implies (r − 1)a_{1} = a_{0}. Thus if r = 1, a_{0} = 0, a contradiction. If r = 2, a_{1} = a_{0} [n^{2} +2rn+r^{2} −3n−3r+2]a_{n} −(n+r−1)a_{n−1} +a_{n−2} = [n^{2} +2rn−3n]a_{n} −(n+r−1)a_{n−1} +a_{n−2} = 0 a_{n} = \frac{(n+r−1)a_{n−1}−a_{n−2}}{n^{2}+2rn−3n} = \frac{(n+1)a_{n−1}−a_{n−2}}{n^{2}+4n−3n} = \frac{(n+1)a_{n−1}−a_{n−2}}{n^{2}+n} = \frac{(n+1)a_{n−1}−a_{n−2}}{n(n+1)} a_{2} = \frac{3a_{1}−a_{0}}{6} = \frac{3a_{0}−a_{0}}{6} = \frac{2a_{0}}{6} = \frac{a_{0}}{3} a_{3} = \frac{4a_{2}−a_{1}}{(3)(4)} = \frac{4( \frac{a_{0}}{3} )−a_{0}}{(3)(4)} = \frac{4a_{0}−3a_{0}}{(3)^{2}(4)} = \frac{a_{0}}{(3)^{2}(4)} a_{4} = \frac{5a_{3}−a_{2}}{(4)(5)} = ^{\frac{(5 \frac{a_{0}}{(3)^{2}(4)} )−( \frac{a_{0}}{3} )}{(4)(5)}} = \frac{5a_{0}−3(4)a_{0}}{3^{2}(4)^{2}(5)} = \frac{7a_{0}}{3^{2}(4)^{2}(5)} a_{5} = \frac{6a_{4}−a_{3}}{(5)(6)} = ^{\frac{6( \frac{7a_{0}}{3^{2}(4)^{2}(5)} )−( \frac{a_{0}}{(3)^{2}(4)} )}{(5)(6)}} = \frac{6(7a_{0})−(20a_{0})}{(3)^{2}(4)^{2}(5)^{2}(6)} = \frac{22a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)} a_{6} = \frac{7a_{5}−a_{4}}{(6)(7)} = ^{\frac{7( \frac{22a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)} )− \frac{7a_{0}}{3^{2}(4)^{2}(5)}}{(6)(7)}} = \frac{7(22a_{0})−30(7a_{0})}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)} = \frac{−56a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)} a_{7} = \frac{8a_{6}−a_{5}}{(7)(8)} = ^{\frac{8( \frac{−56a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)} )− \frac{22a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)}}{(7)(8)}} = \frac{8(−56a_{0})−42∗22a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)^{2}(8)} = \frac{−1372a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)^{2}(8)} y = x^{2}(a_{0}+a_{0}x+\frac{a_{0}}{3} x^{2}+\frac{a_{0}}{(3)^{2}(4)}x^{3}+\frac{7a_{0}}{3^{2}(4)^{2}(5)}x^{4}+\frac{22a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)}x^{5}+\frac{−56a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)}x^{6}+\frac{−1372a_{0}}{(3)^{2}(4)^{2}(5)^{2}(6)^{2}(7)^{2}(8)}x^{7}+...)

5.5: Solve xy′′ + (x − 6)y′ − 3y = 0 p(x) = \frac{x−6}{x}. Thus x_{0} = 0 is a singular value. q(x) = \frac{−3}{x} also implies x_{0} = 0 is a singular value. xp(x) = x − 6 and x^{2}q(x) = 3x. Thus x_{0} = 0 is a regular singular value. Suppose y = ∑_{n=0}^{∞} a_{n}x^{n+r} is a solution. WLOG assume a_{0} ≠ 0 (otherwise one can reindex the summation). Then y′ = ∑_{n=0}^{∞}(n + r)a_{n}x^{n+r−1} and y′′ = ∑_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r−2} xy′′ + (x − 6)y′ − 3y = 0 x ∑_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r−2} + (x − 6) ∑_{n=0}^{∞}(n + r)a_{n}x^{n+r−1} − 3 ∑_{n=0}^{∞} a_{n}x^{n+r} = 0 ∑_{n=0}^{∞}(n+r)(n+r−1)a_{n}x^{n+r−1}+∑_{n=0}^{∞}(n+r)a_{n}x^{n+r} −∑_{n=0}^{∞} 6(n+r)a_{n}x^{n+r−1}−3 ∑_{n=0}^{∞} a_{n}x^{n+r} = 0 ∑_{n=0}^{∞}[(n + r)(n + r − 1) − 6(n + r)]a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0 ∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0 ∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + Σ_{n=1}^{∞}(n − 1 + r − 3)a_{n−1}x^{n+r−1} = 0 r(r − 7)a_{0}x^{r−1} + Σ_{n=1}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + Σ_{n=1}^{∞}(n + r − 4)a_{n−1}x^{n+r−1} = 0 r(r − 7)a_{0}x^{r−1} + Σ_{n=1}^{∞}[(n + r)(n + r − 7)a_{n} + (n + r − 4)a_{n−1}]x^{n+r−1} = 0 ∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0 Set all coefficients = 0: Since a_{0} ≠ 0, r(r −7) = 0 implies r = 0, 7. r(r −7) = 0 is the indicial equation (n + r)(n + r − 7)a_{n}(n + r − 4)a_{n−1} = [n^{2} + nr − 7n + rn + r(r − 7)]a_{n} + (n + r − 4)a_{n−1} = [n^{2} + 2nr − 7n]a_{n} + (n + r − 4)a_{n−1} = 0

impliesa_{n} = \frac{(n+r−4)a_{n−1}}{n^{2}+2nr−7n}

r = 7: a_{n} = \frac{(n+3)a_{n−1}}{n^{2}+7n} r = 0: a_{n} = \frac{(n−4)a_{n−1}}{n^{2}−7n} = \frac{(n−4)a_{n−1}}{n(n−7)}

a_{1} = \frac{−3a_{0}}{−6} = \frac{a_{0}}{2}

a_{2} = \frac{−2a_{1}}{2(−5)} = \frac{−a_{0}}{2(−5)} = \frac{a_{0}}{10}

a_{3} = \frac{−1a_{2}}{3(−4)} = \frac{a_{0}}{(3)(4)(10)} = \frac{a_{0}}{120}

a_{4} = 0, a_{5} = 0, ...

Thus y = a_{0} + \frac{a_{0}}{2} x + \frac{a_{0}}{10} x^{2} + \frac{a_{0}}{120}x^{3}

y′ = \frac{a_{0}}{2} + \frac{a_{0}}{5} x + \frac{a_{0}}{40} x^{2} and y′′ = \frac{a_{0}}{5} + \frac{a_{0}}{20} x

xy′′ + (x − 6)y′ − 3y = 0

x[\frac{a_{0}}{5} + \frac{a_{0}}{20} x] + (x − 6)[\frac{a_{0}}{2} + \frac{a_{0}}{5} x + \frac{a_{0}}{40} x^{2}] − 3[a_{0} + \frac{a_{0}}{2} x + \frac{a_{0}}{10} x^{2} + \frac{a_{0}}{120}x^{3}]

= \frac{a_{0}}{5} x + \frac{a_{0}}{20} x^{2} + \frac{a_{0}}{2} x + \frac{a_{0}}{5} x^{2} + \frac{a_{0}}{40} x^{3} − 3a_{0} − \frac{6a_{0}}{5} x − \frac{3a_{0}}{20} x^{2} − 3a_{0} − \frac{3a_{0}}{2} x − \frac{3a_{0}}{10} x^{2} − \frac{a_{0}}{40} x^{3}

= \frac{a_{0}}{20} x^{2} + \frac{a_{0}}{5} x^{2} − \frac{3a_{0}}{20} x^{2} − \frac{3a_{0}}{10} x^{2} + \frac{a_{0}}{5} x + \frac{a_{0}}{2} x − \frac{6a_{0}}{5} x − \frac{3a_{0}}{2} x − 3a_{0} − 3a_{0}

= \frac{a_{0}}{5} x^{2} − \frac{2a_{0}}{20} x^{2} − \frac{3a_{0}}{10} x^{2} − \frac{5a_{0}}{5} x − \frac{2a_{0}}{2} x − 3a_{0} − 3a_{0}

= \frac{2a_{0}}{10} x^{2} − \frac{1a_{0}}{10} x^{2} − \frac{3a_{0}}{10} x^{2} − a_{0}x − a_{0}x − 3a_{0} − 3a_{0} xy′′ + (x − 6)y′ − 3y = 0

xΣ_{n=2}^{∞}(n)(n − 1)a_{n}x^{n−2} + (x − 6)Σ_{n=1}^{∞}(n)a_{n}x^{n−1} − 3 ∑_{n=0}^{∞} a_{n}x^{n} = 0

Σ_{n=2}^{∞}(n)(n − 1)a_{n}x^{n−1} + Σ_{n=1}^{∞}(n)a_{n}x^{n} − Σ_{n=1}^{∞}6na_{n}x^{n−1} − 3 ∑_{n=0}^{∞} a_{n}x^{n} = 0

Σ_{n=1}^{∞}(n + 1)(n)a_{n+1}x^{n} + Σ_{n=2}^{∞}(n)a_{n}x^{n} − ∑_{n=0}^{∞} 6(n + 1)a_{n+1}x^{n} − 3 ∑_{n=0}^{∞} a_{n}x^{n} = 0

2a_{2}x−6a_{1}−12a_{2}x−3a_{0}+Σ_{n=2}^{∞}(n+1)(n)a_{n+1}x^{n}+Σ_{n=2}^{∞}(n)a_{n}x^{n}−Σ_{n=2}^{∞}6(n+1)a_{n+1}x^{n}−3Σ_{n=2}^{∞}a_{n}x^{n} = 0

−10a_{2}x − 6a_{1} − 3a_{0} + Σ_{n=2}^{∞}[(n + 1)(n)a_{n+1} + (n)a_{n} − 6(n + 1)a_{n+1} − 3a_{n}]x^{n} = 0

−10a_{2}x − 6a_{1} − 3a_{0} + Σ_{n=2}^{∞}[(n + 1)(n − 6)a_{n+1} + (n − 3)a_{n}]x^{n} = 0

(x − 6)m − 3mx − 3b = 0

xm − 6m − 3mx − 3b = 0

xm − 3mx − 6m − 3b = 0

−2mx − 6m − 3b = 0

∑_{n=0}^{∞}[(n + r)(n + r − 1) − 6(n + r)]a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0

∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0

∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + Σ_{n=1}^{∞}(n − 1 + r − 3)a_{n−1}x^{n+r−1} = 0

r(r − 7)a_{0}x^{r−1} + Σ_{n=1}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + Σ_{n=1}^{∞}(n + r − 4)a_{n−1}x^{n+r−1} = 0

r(r − 7)a_{0}x^{r−1} + Σ_{n=1}^{∞}[(n + r)(n + r − 7)a_{n} + (n + r − 4)a_{n−1}]x^{n+r−1} = 0

∑_{n=0}^{∞}(n + r)(n + r − 7)a_{n}x^{n+r−1} + ∑_{n=0}^{∞}(n + r − 3)a_{n}x^{n+r} = 0