IAL 2021 June Q4 | 國際數學站

Given that

\begin{align*} y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3y = 0 \end{align*}

(a) show that

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{28}{y^2}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 - \frac{24}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

(5)

Given also that $y = 8$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ at $x = 0$

(b) find a series solution for $y$ in ascending powers of $x$ , up to and including the term in $x^3$ , simplifying the coefficients where possible.

(4)