题目

Problem

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)

解答

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先把原方程除以 $y$ ，解出 $y''$ ，再对 $y''$ 求导得到 $y'''$ 。最后把 $y''$ 代回去，目标式就只含 $y$ 和 $y'$ 。

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From

\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*}

divide by $y$ :

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

Differentiate with respect to $x$ :

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,4\frac{\mathrm{d}}{\mathrm{d}x} \left[ y^{-1}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right]\\[4mm] =&\,4\left[ -y^{-2}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 +2y^{-1}\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right]\\[4mm] =&\,-\frac{4}{y^2} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 +\frac{8}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Substitute

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

into the expression for $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}$ :

\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,-\frac{4}{y^2} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3\\[4mm] &\,\hspace{2pt}+\frac{8}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \left[ \frac{4}{y} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2-3 \right]\\[4mm] =&\,-\frac{4}{y^2} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 +\frac{32}{y^2} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3\\[4mm] &\,\hspace{2pt}-\frac{24}{y}\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,\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*}

This is the required result.

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Taylor series 到 $x^3$ 需要 $y(0)$ 、 $y'(0)$ 、 $y''(0)$ 、 $y'''(0)$ 。这里 $y''(0)$ 用原方程求， $y'''(0)$ 用 (a) 的结果求。

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At $x=0$ , $y=8$ and $y'=1$ . From the original equation,

\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*}

\begin{align*} 8\left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} -4+24=&\,0\\[4mm] 8\left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} =&\,-20\\[4mm] \left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} =&\,-\frac52. \end{align*}

Using the result from part (a),

\begin{align*} \left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=0} =&\,\frac{28}{8^2}(1)^3-\frac{24}{8}(1)\\[4mm] =&\,\frac{7}{16}-3\\[4mm] =&\,-\frac{41}{16}. \end{align*}

Therefore,

\begin{align*} y =&\,8+x+\frac{-\frac52}{2!}x^2 +\frac{-\frac{41}{16}}{3!}x^3\\[4mm] =&\,8+x-\frac54x^2-\frac{41}{96}x^3. \end{align*}