Question 1 of 7: Power-Series Solution about an Ordinary Point
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, 04-BS-5 Advanced Mathematics — three-hour closed-book paper; one approved Casio or Sharp calculator and one 8.5″×11″ two-sided aid sheet are permitted. Seven questions are printed and any five constitute a complete paper, all questions being of equal value (20 marks each). All seven questions are solved below, because the set is a study resource rather than an examination attempt.
Reference texts for 04-BS-5 Advanced Mathematics. E. Kreyszig, Advanced Engineering Mathematics, 10th ed. — Ch. 5 (series solutions of ODEs), Ch. 11 (Fourier series and Fourier transforms), Ch. 19 (numerics: root finding, interpolation, integration), Ch. 20 (numeric linear algebra); R. L. Burden and J. D. Faires, Numerical Analysis, 9th ed. — Ch. 2 (bisection, Newton, fixed-point iteration), Ch. 3 (divided-difference interpolation), Ch. 4 (Romberg integration); G. Strang, Introduction to Linear Algebra, 6th ed. — Ch. 6 (eigenvalues, Cayley–Hamilton).
Reading of the paper. Where a printed symbol is unclear, the reading adopted is stated explicitly and justified from the paper's own internal evidence (for example Question 7's matrix, which is recovered exactly from the characteristic equation the question prints).
Question 1: Power-Series Solution about an Ordinary Point (20 marks)
Given. A second-order linear homogeneous equation with polynomial coefficients, $p(x)=x^{2}+4$, $q(x)=-4x$ and $r(x)=-2$, to be solved about the expansion centre $x_{0}=0$.
Given data
Quantity
Value
Leading coefficient $p(x)$
$x^{2}+4$, with $p(0)=4 \ne 0$
First-derivative coefficient $q(x)$
$-4x$
Zeroth-order coefficient $r(x)$
$-2$
Expansion centre
$x_{0}=0$ (ordinary point)
Singular points of the equation
$x=\pm 2i$
Find. Two linearly independent power-series solutions $y_{1}(x)$ and $y_{2}(x)$ valid in a neighbourhood of $x=0$, together with the recurrence that generates their coefficients and the radius of convergence.
The two independent series solutions of $(x^{2}+4)y'' - 4xy' - 2y = 0$ built about $x=0$: the even solution $y_{1}$ (normalised by $a_{0}=1$) and the odd solution $y_{2}$ (normalised by $a_{1}=1$). Both are analytic inside $|x| \lt 2$.
Approach. Substitute an undetermined power series about $x=0$, shift the index on the term that carries $x^{n-2}$ so that every sum runs over the same power, equate the coefficient of $x^{n}$ to zero to obtain a two-term recurrence, then read off the two solutions generated by the independent choices $(a_{0},a_{1})=(1,0)$ and $(0,1)$.
Confirm that $x=0$ really is an ordinary point. Dividing through by $p(x)$ puts the equation in standard form $$y'' - \frac{4x}{x^{2}+4}\,y' - \frac{2}{x^{2}+4}\,y = 0 .$$
Both coefficient functions are quotients of polynomials whose denominator does not vanish at the origin, so both are analytic there and $x=0$ is an ordinary point. The nearest zeros of $x^{2}+4$ are $x=\pm 2i$, which fixes the radius of convergence in advance at $R=2$.
Insert an undetermined series and its derivatives. Take $y=\sum_{n=0}^{\infty}a_{n}x^{n}$, so that $y'=\sum_{n=1}^{\infty}n\,a_{n}x^{n-1}$ and $y''=\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}$. Substituting and expanding the product with $x^{2}+4$ gives
$$\sum n(n-1)a_{n}x^{n} \;+\; 4\sum n(n-1)a_{n}x^{n-2} \;-\; 4\sum n\,a_{n}x^{n} \;-\; 2\sum a_{n}x^{n} = 0 .$$
Only the second sum carries a shifted power.
Shift the index and collect the coefficient of $x^{n}$. Replacing $n$ by $n+2$ in the second sum turns $4n(n-1)a_{n}x^{n-2}$ into $4(n+2)(n+1)a_{n+2}x^{n}$, so every sum now runs from $n=0$ over the same power and
$$\sum_{n=0}^{\infty}\Big[\,4(n+2)(n+1)a_{n+2} + \big(n(n-1)-4n-2\big)a_{n}\Big]x^{n} = 0 .$$
A power series vanishes identically only if every coefficient vanishes.
Read off the recurrence. Setting the bracket to zero and simplifying $n(n-1)-4n-2 = n^{2}-5n-2$ gives
$$\boxed{\,a_{n+2} = \frac{-n^{2}+5n+2}{4(n+1)(n+2)}\;a_{n}\,},\qquad n = 0,1,2,\dots$$
The recurrence links coefficients two apart, so the even-indexed coefficients are all fixed by $a_{0}$ and the odd-indexed ones by $a_{1}$ — the two arbitrary constants a second-order equation must have. The numerator never vanishes for an integer $n$, so neither branch terminates: there is no polynomial solution.
Generate the even solution ($a_{0}=1$, $a_{1}=0$). The numerator takes the values $2, 8, 6, -4$ at $n=0,2,4,6$ while the denominator takes $8, 48, 120, 224$, so
$$a_{2}=\tfrac{2}{8}a_{0}=\tfrac14,\qquad a_{4}=\tfrac{8}{48}a_{2}=\tfrac{1}{24},\qquad a_{6}=\tfrac{6}{120}a_{4}=\tfrac{1}{480},\qquad a_{8}=\tfrac{-4}{224}a_{6}=-\tfrac{1}{26880}.$$
Collecting these even powers,
$$y_{1}(x) = 1 + \frac{x^{2}}{4} + \frac{x^{4}}{24} + \frac{x^{6}}{480} - \frac{x^{8}}{26880} + \cdots$$
which is an even function, as the recurrence forces.
Generate the odd solution ($a_{0}=0$, $a_{1}=1$). Now the numerator takes $6, 8, 2, -12$ at $n=1,3,5,7$ against denominators $24, 80, 168, 288$:
$$a_{3}=\tfrac{6}{24}a_{1}=\tfrac14,\qquad a_{5}=\tfrac{8}{80}a_{3}=\tfrac{1}{40},\qquad a_{7}=\tfrac{2}{168}a_{5}=\tfrac{1}{3360},\qquad a_{9}=\tfrac{-12}{288}a_{7}=-\tfrac{1}{80640}.$$
Hence
$$y_{2}(x) = x + \frac{x^{3}}{4} + \frac{x^{5}}{40} + \frac{x^{7}}{3360} - \frac{x^{9}}{80640} + \cdots$$
Confirm independence and state the general solution. At the centre the two solutions satisfy $y_{1}(0)=1$, $y_{1}'(0)=0$, $y_{2}(0)=0$, $y_{2}'(0)=1$, so their Wronskian there is
$$W(y_{1},y_{2})(0) = y_{1}(0)y_{2}'(0) - y_{1}'(0)y_{2}(0) = 1\cdot 1 - 0\cdot 0 = 1 \ne 0 ,$$
which is enough to prove linear independence on the whole interval of convergence. The general solution is therefore $y = C_{1}y_{1}(x) + C_{2}y_{2}(x)$, convergent for $|x| \lt 2$.
