IAL 2021 Jan Q5 | 國際數學站

Given that

\begin{align*} (2 - x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 3y \end{align*}

(a) show that

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{1}{(2 - x^2)}\left(2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\left(1 - 5\frac{\mathrm{d}y}{\mathrm{d}x}\right) - 5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3\frac{\mathrm{d}y}{\mathrm{d}x}\right) \end{align*}

(5)

Given also that $y = 3$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{4}$ at $x = 0$

(b) obtain a series solution for $y$ in ascending powers of $x$ with simplified coefficients, up to and including the term in $x^3$

(4)