Question 1 of 7: Power-Series Solution of $y''-4xy=0$

Question 1: Power-Series Solution of $y''-4xy=0$ (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The linear second-order ODE $y''-4xy=0$; the coefficient functions ($p(x)=0$, $q(x)=-4x$) are polynomials, hence analytic everywhere, so $x=0$ is an ordinary point.

Find. Two linearly independent power-series solutions $y_1(x)$, $y_2(x)$ about $x=0$.

Approach. Substitute $y=\sum_{k=0}^{\infty}a_kx^{k}$, shift indices so every term carries the same power of $x$, read off the recurrence relation between coefficients, then generate the two independent solution families from the two free constants $a_0$, $a_1$.

1. Substitute the series and align powers. With $y''=\sum_{k\ge2}k(k-1)a_kx^{k-2}$, the ODE reads $$\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2} - 4\sum_{k=0}^{\infty}a_kx^{k+1}=0.$$ Re-index the first sum with $m=k-2$ and the second with $m=k+1$: $$\sum_{m=0}^{\infty}(m+2)(m+1)a_{m+2}x^{m} - \sum_{m=1}^{\infty}4a_{m-1}x^{m}=0$$ (the first sum already starts cleanly at $m=0$ since its $m=-2,-1$ terms carry the factor $(m+2)(m+1)=0$).
2. Read off the recurrence. Matching the coefficient of $x^{0}$ gives $2a_2=0\Rightarrow a_2=0$. Matching $x^{m}$ for $m\ge1$ gives $(m+2)(m+1)a_{m+2}=4a_{m-1}$. Writing $k=m+2$, $$\boxed{a_k = \frac{4}{k(k-1)}\,a_{k-3}}, \qquad k\ge3.$$ Because the recurrence steps back by 3, the coefficients split into three independent families keyed by $a_0$, $a_1$, and $a_2=0$ — and since $a_2=0$, every coefficient with index $\equiv2\pmod3$ vanishes identically. Only $a_0$ and $a_1$ survive as free parameters, giving exactly two independent series.
3. Build $y_1$ (choose $a_0=1$, $a_1=0$). Applying the recurrence to indices $3,6,9,\dots$: $$a_3=\frac{4a_0}{3\cdot2}=\frac{2}{3},\quad a_6=\frac{4a_3}{6\cdot5}=\frac{4}{45},\quad a_9=\frac{4a_6}{9\cdot8}=\frac{2}{405},\ \dots$$ $$\boxed{y_1(x) = 1 + \frac{2}{3}x^{3} + \frac{4}{45}x^{6} + \frac{2}{405}x^{9} + \cdots}$$
4. Build $y_2$ (choose $a_0=0$, $a_1=1$). Applying the recurrence to indices $4,7,10,\dots$: $$a_4=\frac{4a_1}{4\cdot3}=\frac{1}{3},\quad a_7=\frac{4a_4}{7\cdot6}=\frac{2}{63},\quad a_{10}=\frac{4a_7}{10\cdot9}=\frac{4}{2835},\ \dots$$ $$\boxed{y_2(x) = x + \frac{1}{3}x^{4} + \frac{2}{63}x^{7} + \frac{4}{2835}x^{10} + \cdots}$$
5. Confirm linear independence. $y_1(0)=1,\ y_1'(0)=0$ and $y_2(0)=0,\ y_2'(0)=1$, so the Wronskian $W(0)=y_1(0)y_2'(0)-y_1'(0)y_2(0)=1\neq0$ — $y_1,y_2$ are linearly independent, and since $p,q$ are polynomials (entire), both series converge for all $x$.