题目

Problem

Given that $y = \mathrm{e}^x \sin x$

(a) show that

\begin{align*} \frac{\mathrm{d}^6y}{\mathrm{d}x^6} =&\, k\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

where $k$ is a constant to be determined.

(4)

(b) Hence determine the first $5$ non-zero terms in the Maclaurin series expansion for $y$ , giving each coefficient in simplest form.

(3)

解答

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连续求导会出现循环。先求到四阶导，发现 $y^{(4)}=-4y$ ，再继续求两次就能得到 $y^{(6)}=-4y''$ 。

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Given

\begin{align*} y=\mathrm{e}^x\sin x \end{align*}

differentiate:

\begin{align*} y' =&\, \mathrm{e}^x\sin x+\mathrm{e}^x\cos x\\[2mm] y'' =&\, 2\mathrm{e}^x\cos x\\[2mm] y''' =&\, 2\mathrm{e}^x\cos x-2\mathrm{e}^x\sin x\\[2mm] y^{(4)} =&\, -4\mathrm{e}^x\sin x\\[2mm] =&\, -4y \end{align*}

Therefore

\begin{align*} y^{(5)}=&\,-4y'\\[2mm] y^{(6)}=&\,-4y'' \end{align*}

Hence

\begin{align*} \frac{\mathrm{d}^6y}{\mathrm{d}x^6} =&\, -4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

\begin{align*} \boxed{k=-4} \end{align*}

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Maclaurin series 要在 $x=0$ 代入。第一项 $y(0)=0$ ，所以要继续算到第六阶，才能得到前 $5$ 个非零项。

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At $x=0$ ,

\begin{align*} y(0)=&\,0\\ y'(0)=&\,1\\ y''(0)=&\,2\\ y'''(0)=&\,2\\ y^{(4)}(0)=&\,0\\ y^{(5)}(0)=&\,-4\\ y^{(6)}(0)=&\,-8 \end{align*}

Therefore

\begin{align*} y =&\, 0+1x+\frac{2}{2!}x^2+\frac{2}{3!}x^3 +0x^4+\frac{-4}{5!}x^5+\frac{-8}{6!}x^6+\cdots\\[4mm] =&\, \boxed{x+x^2+\frac13x^3-\frac1{30}x^5-\frac1{90}x^6+\cdots} \end{align*}