题目
Problem
(a) Show that an appropriate integrating factor for
x 2 − 1 d y d x + y = x 2 − x x 2 − 1 \sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1} x 2 − 1 d x d y + y = x 2 − x x 2 − 1 is x + x 2 − 1 x+\sqrt{x^2-1} x + x 2 − 1
[4]
(b) Hence find the solution of the differential equation
x 2 − 1 d y d x + y = x 2 − x x 2 − 1 \sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1} x 2 − 1 d x d y + y = x 2 − x x 2 − 1 for which y = 1 y=1 y = 1 x = 5 4 x=\frac{5}{4} x = 4 5 y = f ( x ) y=f(x) y = f ( x )
[7]
题目中文翻译
(a) 证明微分方程
x 2 − 1 d y d x + y = x 2 − x x 2 − 1 \sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1} x 2 − 1 d x d y + y = x 2 − x x 2 − 1 的一个适当积分因子为 x + x 2 − 1 x+\sqrt{x^2-1} x + x 2 − 1
(4)
(b) 由此求微分方程
x 2 − 1 d y d x + y = x 2 − x x 2 − 1 \sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1} x 2 − 1 d x d y + y = x 2 − x x 2 − 1 的解，满足当 x = 5 4 x=\frac{5}{4} x = 4 5 y = 1 y=1 y = 1 y = f ( x ) y=f(x) y = f ( x )
(7)
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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在含初始点的 x > 1 x>1 x > 1 x 2 − 1 \sqrt{x^2-1} x 2 − 1 y y y
答题过程
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On the interval x > 1 x>1 x > 1 x 2 − 1 \sqrt{x^2-1} x 2 − 1
y ′ + y x 2 − 1 = x 2 x 2 − 1 − x . y'+\frac{y}{\sqrt{x^2-1}}=\frac{x^2}{\sqrt{x^2-1}}-x. y ′ + x 2 − 1 y = x 2 − 1 x 2 − x . The coefficient of y y y ( x 2 − 1 ) − 1 / 2 (x^2-1)^{-1/2} ( x 2 − 1 ) − 1/2 x > 1 x>1 x > 1
μ = exp ( ∫ d x x 2 − 1 ) = e arcosh x = e ln ( x + x 2 − 1 ) = x + x 2 − 1 . \begin{align*}
\mu=&\,\exp\bigg(\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}\bigg)\\
=&\,e^{\operatorname{arcosh}x}\\
=&\,e^{\ln(x+\sqrt{x^2-1})}\\
=&\,x+\sqrt{x^2-1}.
\end{align*} μ = = = = exp ( ∫ x 2 − 1 d x ) e arcosh x e l n ( x + x 2 − 1 ) x + x 2 − 1 .
解法二
思路
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也可直接求题给函数的导数，验证它的对数导数正好是标准形式中 y y y
答题过程
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Let μ = x + x 2 − 1 \mu=x+\sqrt{x^2-1} μ = x + x 2 − 1 x > 1 x>1 x > 1
μ ′ = 1 + x x 2 − 1 = μ x 2 − 1 . \mu'=1+\frac{x}{\sqrt{x^2-1}}
=\frac{\mu}{\sqrt{x^2-1}}. μ ′ = 1 + x 2 − 1 x = x 2 − 1 μ . Thus μ ′ / μ = 1 / x 2 − 1 \mu'/\mu=1/\sqrt{x^2-1} μ ′ / μ = 1/ x 2 − 1 μ \mu μ
(b)
解法一
思路
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乘上积分因子后，右边含有 x − x 2 − 1 x-\sqrt{x^2-1} x − x 2 − 1 x + x 2 − 1 x+\sqrt{x^2-1} x + x 2 − 1 1 1 1
答题过程
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Multiplying the standard form by μ = x + x 2 − 1 \mu=x+\sqrt{x^2-1} μ = x + x 2 − 1
d d x ( μ y ) = x ( x 2 − ( x 2 − 1 ) ) x 2 − 1 = x x 2 − 1 . \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}(\mu y)
=&\,\frac{x\big(x^2-(x^2-1)\big)}{\sqrt{x^2-1}}\\
=&\,\frac{x}{\sqrt{x^2-1}}.
\end{align*} d x d ( μ y ) = = x 2 − 1 x ( x 2 − ( x 2 − 1 ) ) x 2 − 1 x . With u = x 2 − 1 u=x^2-1 u = x 2 − 1 d u = 2 x d x \mathrm{d}u=2x\,\mathrm{d}x d u = 2 x d x 1 2 ∫ u − 1 / 2 d u = u + C \frac12\int u^{-1/2}\,\mathrm{d}u=\sqrt u+C 2 1 ∫ u − 1/2 d u = u + C μ y = x 2 − 1 + C \mu y=\sqrt{x^2-1}+C μ y = x 2 − 1 + C x = 5 4 x=\frac54 x = 4 5 3 4 \frac34 4 3 μ = 2 \mu=2 μ = 2 2 = 3 4 + C 2=\frac34+C 2 = 4 3 + C
y = x 2 − 1 + 5 4 x + x 2 − 1 . y=\frac{\sqrt{x^2-1}+\frac54}{x+\sqrt{x^2-1}}. y = x + x 2 − 1 x 2 − 1 + 4 5 . 