题目
Problem
(a) Use the substitution u = 1 + x 2 u=1+x^2 u = 1 + x 2
∫ x 1 + x 2 d x . \int\frac{x}{\sqrt{1+x^2}}\,\mathrm{d}x. ∫ 1 + x 2 x d x . [2]
(b) Find the solution of the differential equation
x d y d x − y = x 2 sinh − 1 x , x\frac{\mathrm{d}y}{\mathrm{d}x}-y=x^2\sinh^{-1}x, x d x d y − y = x 2 sinh − 1 x , given that y = 1 y=1 y = 1 x = 1 x=1 x = 1 y = f ( x ) y=f(x) y = f ( x )
[10]
题目中文翻译
(a) 使用代换 u = 1 + x 2 u=1+x^2 u = 1 + x 2
∫ x 1 + x 2 d x \int\frac{x}{\sqrt{1+x^2}}\,\mathrm{d}x ∫ 1 + x 2 x d x (2)
(b) 求微分方程
x d y d x − y = x 2 sinh − 1 x x\frac{\mathrm{d}y}{\mathrm{d}x}-y=x^2\sinh^{-1}x x d x d y − y = x 2 sinh − 1 x 的解，已知当 x = 1 x=1 x = 1 y = 1 y=1 y = 1 y = f ( x ) y=f(x) y = f ( x )
(10)
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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代换 u = 1 + x 2 u=1+x^2 u = 1 + x 2 x d x x\,\mathrm{d}x x d x 1 2 d u \frac12\mathrm{d}u 2 1 d u
答题过程
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With u = 1 + x 2 u=1+x^2 u = 1 + x 2 d u = 2 x d x \mathrm{d}u=2x\,\mathrm{d}x d u = 2 x d x
∫ x 1 + x 2 d x = 1 2 ∫ u − 1 / 2 d u = 1 + x 2 + C . \int\frac{x}{\sqrt{1+x^2}}\,\mathrm{d}x
=\frac12\int u^{-1/2}\,\mathrm{d}u=\boxed{\sqrt{1+x^2}+C}. ∫ 1 + x 2 x d x = 2 1 ∫ u − 1/2 d u = 1 + x 2 + C .
(b)
解法一
思路
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在含初始点的 x > 0 x>0 x > 0 ( y / x ) ′ = arsinh x (y/x)'=\operatorname{arsinh}x ( y / x ) ′ = arsinh x
答题过程
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On the interval x > 0 x>0 x > 0
y ′ − 1 x y = x arsinh x . y'-\frac1x y=x\operatorname{arsinh}x. y ′ − x 1 y = x arsinh x . 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 ) = arsinh x . \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac yx\bigg)=\operatorname{arsinh}x. d x d ( x y ) = arsinh x . For ∫ arsinh x d x \int\operatorname{arsinh}x\,\mathrm{d}x ∫ arsinh x d x u = arsinh x u=\operatorname{arsinh}x u = arsinh x d v = d x \mathrm{d}v=\mathrm{d}x d v = d x d u = 1 1 + x 2 d x \mathrm{d}u=\frac1{\sqrt{1+x^2}}\,\mathrm{d}x d u = 1 + x 2 1 d x v = x v=x v = x
y x = x arsinh x − ∫ x 1 + x 2 d x = x arsinh x − 1 + x 2 + C . \begin{align*}
\frac yx=&\,x\operatorname{arsinh}x-\int\frac{x}{\sqrt{1+x^2}}\,\mathrm{d}x\\
=&\,x\operatorname{arsinh}x-\sqrt{1+x^2}+C.
\end{align*} x y = = x arsinh x − ∫ 1 + x 2 x d x x arsinh x − 1 + x 2 + C . At x = 1 x=1 x = 1 1 = arsinh 1 − 2 + C 1=\operatorname{arsinh}1-\sqrt2+C 1 = arsinh 1 − 2 + C arsinh 1 = ln ( 1 + 2 ) \operatorname{arsinh}1=\ln(1+\sqrt2) arsinh 1 = ln ( 1 + 2 )
y = x [ x arsinh x − 1 + x 2 + 1 + 2 − ln ( 1 + 2 ) ] . \begin{align*}
y=&\,x\Big[x\operatorname{arsinh}x-\sqrt{1+x^2}\\
&\,\hspace{2pt}+1+\sqrt2-\ln(1+\sqrt2)\Big].
\end{align*} y = x [ x arsinh x − 1 + x 2 + 1 + 2 − ln ( 1 + 2 ) ] . 