题目
Problem
(a) Show that
d d x ( 1 2 x x 2 − 9 − 9 2 cosh − 1 x 3 ) = x 2 − 9 . \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x\sqrt{x^2-9}-\frac{9}{2}\cosh^{-1}\frac{x}{3}\right)=\sqrt{x^2-9}. d x d ( 2 1 x x 2 − 9 − 2 9 cosh − 1 3 x ) = x 2 − 9 . [3]
(b) Find the solution of the differential equation
x d y d x − y = x 2 x 2 − 9 , x\frac{\mathrm{d}y}{\mathrm{d}x}-y=x^2\sqrt{x^2-9}, x d x d y − y = x 2 x 2 − 9 , given that y = 1 y=1 y = 1 x = 3 x=3 x = 3 y = f ( x ) y=f(x) y = f ( x )
[9]
题目中文翻译
(a) 证明
d d x ( 1 2 x x 2 − 9 − 9 2 cosh − 1 x 3 ) = x 2 − 9 \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x\sqrt{x^2-9}-\frac{9}{2}\cosh^{-1}\frac{x}{3}\right)=\sqrt{x^2-9} d x d ( 2 1 x x 2 − 9 − 2 9 cosh − 1 3 x ) = x 2 − 9 (3)
(b) 求微分方程
x d y d x − y = x 2 x 2 − 9 x\frac{\mathrm{d}y}{\mathrm{d}x}-y=x^2\sqrt{x^2-9} x d x d y − y = x 2 x 2 − 9 的解，已知当 x = 3 x=3 x = 3 y = 1 y=1 y = 1 y = f ( x ) y=f(x) y = f ( x )
(9)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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第一项用乘积法则，第二项用反双曲余弦的链式法则。将两项中含根式分母的部分合并。
答题过程
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For x > 3 x>3 x > 3 arcosh ( x / 3 ) \operatorname{arcosh}(x/3) arcosh ( x /3 ) 1 / x 2 − 9 1/\sqrt{x^2-9} 1/ x 2 − 9
d d x [ x 2 x 2 − 9 − 9 2 arcosh x 3 ] = 1 2 x 2 − 9 + x 2 2 x 2 − 9 − 9 2 x 2 − 9 = 1 2 x 2 − 9 + x 2 − 9 2 x 2 − 9 = x 2 − 9 . \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\bigg[
\frac x2\sqrt{x^2-9}-\frac92\operatorname{arcosh}\frac x3\bigg]
=&\,\frac12\sqrt{x^2-9}+\frac{x^2}{2\sqrt{x^2-9}}\\
&\,\hspace{2pt}-\frac{9}{2\sqrt{x^2-9}}\\
=&\,\frac12\sqrt{x^2-9}+\frac{x^2-9}{2\sqrt{x^2-9}}\\
=&\,\sqrt{x^2-9}.
\end{align*} d x d [ 2 x x 2 − 9 − 2 9 arcosh 3 x ] = = = 2 1 x 2 − 9 + 2 x 2 − 9 x 2 − 2 x 2 − 9 9 2 1 x 2 − 9 + 2 x 2 − 9 x 2 − 9 x 2 − 9 .
(b)
解法一
思路
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在 x ≥ 3 x\ge3 x ≥ 3 x 2 x^2 x 2 ( y / x ) ′ (y/x)' ( y / x ) ′ x = 3 x=3 x = 3
答题过程
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For x > 3 x>3 x > 3
y ′ − 1 x y = x x 2 − 9 . y'-\frac1x y=x\sqrt{x^2-9}. y ′ − x 1 y = x x 2 − 9 . An integrating factor is e − ∫ ( 1 / x ) d x = 1 / x e^{-\int(1/x)\,\mathrm{d}x}=1/x e − ∫ ( 1/ x ) d x = 1/ x
d d x ( y x ) = x 2 − 9 . \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac yx\bigg)=\sqrt{x^2-9}. d x d ( x y ) = x 2 − 9 . By (a),
y x = x 2 x 2 − 9 − 9 2 arcosh x 3 + C . \frac yx=\frac x2\sqrt{x^2-9}-\frac92\operatorname{arcosh}\frac x3+C. x y = 2 x x 2 − 9 − 2 9 arcosh 3 x + C . Both non-constant terms on the right tend to zero at x = 3 x=3 x = 3 y ( 3 ) = 1 y(3)=1 y ( 3 ) = 1 C = 1 3 C=\frac13 C = 3 1 x ≥ 3 x\ge3 x ≥ 3
y = x [ x 2 x 2 − 9 − 9 2 arcosh x 3 + 1 3 ] . \begin{align*}
y=&\,x\bigg[\frac x2\sqrt{x^2-9}\\
&\,\hspace{2pt}-\frac92\operatorname{arcosh}\frac x3+\frac13\bigg].
\end{align*} y = x [ 2 x x 2 − 9 − 2 9 arcosh 3 x + 3 1 ] . 