题目
Problem
(a) Show that
d d x ( 1 2 x 4 − x 2 + 2 sin − 1 ( 1 2 x ) ) = 4 − x 2 . \frac{\mathrm{d}}{\mathrm{d}x}
\bigg(
\frac{1}{2}x\sqrt{4 - x^2}
+ 2\sin^{-1}\bigg(\frac{1}{2}x\bigg)
\bigg)
=
\sqrt{4 - x^2}. d x d ( 2 1 x 4 − x 2 + 2 sin − 1 ( 2 1 x ) ) = 4 − x 2 . [3]
(b) Find the solution of the differential equation
2 d y d x + y 2 + x = 2 2 − x 2\frac{\mathrm{d}y}{\mathrm{d}x}
+
\frac{y}{2 + x}
=
2\sqrt{2 - x} 2 d x d y + 2 + x y = 2 2 − x for which y = 1 2 y = \frac{1}{2} y = 2 1 x = 1 x = 1 x = 1
[8]
题目中文翻译
(a) 证明
d d x ( 1 2 x 4 − x 2 + 2 sin − 1 ( 1 2 x ) ) = 4 − x 2 . \frac{\mathrm{d}}{\mathrm{d}x}
\bigg(
\frac{1}{2}x\sqrt{4 - x^2}
+ 2\sin^{-1}\bigg(\frac{1}{2}x\bigg)
\bigg)
=
\sqrt{4 - x^2}. d x d ( 2 1 x 4 − x 2 + 2 sin − 1 ( 2 1 x ) ) = 4 − x 2 . (3)
(b) 求微分方程
2 d y d x + y 2 + x = 2 2 − x 2\frac{\mathrm{d}y}{\mathrm{d}x}
+
\frac{y}{2 + x}
=
2\sqrt{2 - x} 2 d x d y + 2 + x y = 2 2 − x 的解，且当 x = 1 x = 1 x = 1 y = 1 2 y = \frac{1}{2} y = 2 1
(8)
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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前一项用乘积法则，反正弦项用链式法则。把含分母 4 − x 2 \sqrt{4-x^2} 4 − x 2
答题过程
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For ∣ x ∣ < 2 |x|<2 ∣ x ∣ < 2
d d x [ 1 2 x 4 − x 2 + 2 arcsin ( x 2 ) ] = 1 2 4 − x 2 − x 2 2 4 − x 2 + 2 4 − x 2 = 1 2 4 − x 2 + 4 − x 2 2 4 − x 2 = 4 − x 2 . \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac12x\sqrt{4-x^2}
+2\arcsin\bigg(\frac x2\bigg)\bigg]
=&\,\frac12\sqrt{4-x^2}-\frac{x^2}{2\sqrt{4-x^2}}
+\frac{2}{\sqrt{4-x^2}}\\
=&\,\frac12\sqrt{4-x^2}+\frac{4-x^2}{2\sqrt{4-x^2}}
=\sqrt{4-x^2}.
\end{align*} d x d [ 2 1 x 4 − x 2 + 2 arcsin ( 2 x ) ] = = 2 1 4 − x 2 − 2 4 − x 2 x 2 + 4 − x 2 2 2 1 4 − x 2 + 2 4 − x 2 4 − x 2 = 4 − x 2 .
(b)
解法一
思路
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先把 y ′ y' y ′ 1 1 1 2 + x \sqrt{2+x} 2 + x 2 − x \sqrt{2-x} 2 − x
答题过程
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On the interval − 2 < x < 2 -2<x<2 − 2 < x < 2
y ′ + y 2 ( 2 + x ) = 2 − x . y'+\frac{y}{2(2+x)}=\sqrt{2-x}. y ′ + 2 ( 2 + x ) y = 2 − x . Its integrating factor is
e ∫ 1 2 ( 2 + x ) d x = e 1 2 ln ( 2 + x ) = 2 + x . e^{\int\frac1{2(2+x)}\,\mathrm{d}x}
=e^{\frac12\ln(2+x)}=\sqrt{2+x}. e ∫ 2 ( 2 + x ) 1 d x = e 2 1 l n ( 2 + x ) = 2 + x . Hence
d d x ( y 2 + x ) = 4 − x 2 . \frac{\mathrm{d}}{\mathrm{d}x}\big(y\sqrt{2+x}\big)=\sqrt{4-x^2}. d x d ( y 2 + x ) = 4 − x 2 . By (a),
y 2 + x = 1 2 x 4 − x 2 + 2 arcsin ( x 2 ) + C . y\sqrt{2+x}=\frac12 x\sqrt{4-x^2}+2\arcsin\bigg(\frac x2\bigg)+C. y 2 + x = 2 1 x 4 − x 2 + 2 arcsin ( 2 x ) + C . At x = 1 x=1 x = 1 y = 1 2 y=\frac12 y = 2 1 arcsin ( 1 2 ) = π 6 \arcsin(\frac12)=\frac\pi6 arcsin ( 2 1 ) = 6 π
3 2 = 3 2 + π 3 + C . \frac{\sqrt3}{2}=\frac{\sqrt3}{2}+\frac\pi3+C. 2 3 = 2 3 + 3 π + C . Thus C = − π / 3 C=-\pi/3 C = − π /3
y = 1 2 x 4 − x 2 + 2 arcsin ( x / 2 ) − π / 3 2 + x . y=\frac{\frac12 x\sqrt{4-x^2}+2\arcsin(x/2)-\pi/3}{\sqrt{2+x}}. y = 2 + x 2 1 x 4 − x 2 + 2 arcsin ( x /2 ) − π /3 . 