题目
Problem
(a) Show that an appropriate integrating factor for
x 2 + 16 d y d x + y = x x 2 + 16 \sqrt{x^2+16}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x\sqrt{x^2+16} x 2 + 16 d x d y + y = x x 2 + 16 is
1 4 x + 1 4 x 2 + 16 . \frac{1}{4}x+\frac{1}{4}\sqrt{x^2+16}. 4 1 x + 4 1 x 2 + 16 . [4]
(b) Hence find the solution of the differential equation
x 2 + 16 d y d x + y = x x 2 + 16 \sqrt{x^2+16}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x\sqrt{x^2+16} x 2 + 16 d x d y + y = x x 2 + 16 for which y = 6 y=6 y = 6 x = 3 x=3 x = 3
[6]
题目中文翻译
(a) 证明微分方程
x 2 + 16 d y d x + y = x x 2 + 16 \sqrt{x^2+16}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x\sqrt{x^2+16} x 2 + 16 d x d y + y = x x 2 + 16 的一个适当积分因子为
1 4 x + 1 4 x 2 + 16 \frac{1}{4}x+\frac{1}{4}\sqrt{x^2+16} 4 1 x + 4 1 x 2 + 16 (4)
(b) 据此求微分方程
x 2 + 16 d y d x + y = x x 2 + 16 \sqrt{x^2+16}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x\sqrt{x^2+16} x 2 + 16 d x d y + y = x x 2 + 16 的解，使得当 x = 3 x=3 x = 3 y = 6 y=6 y = 6
(6)
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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先把 y ′ y' y ′ 1 1 1 y y y arsinh ( x / 4 ) \operatorname{arsinh}(x/4) arsinh ( x /4 )
答题过程
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Dividing through by x 2 + 16 \sqrt{x^2+16} x 2 + 16 y ′ + y / x 2 + 16 = x y'+y/\sqrt{x^2+16}=x y ′ + y / x 2 + 16 = x arsinh ( x / 4 ) = ln ( ( x + x 2 + 16 ) / 4 ) \operatorname{arsinh}(x/4)=\ln\big((x+\sqrt{x^2+16})/4\big) arsinh ( x /4 ) = ln ( ( x + x 2 + 16 ) /4 )
exp ( ∫ 1 x 2 + 16 d x ) = e arsinh ( x / 4 ) = x + x 2 + 16 4 . \exp\bigg(\int\frac1{\sqrt{x^2+16}}\,\mathrm{d}x\bigg)
=e^{\operatorname{arsinh}(x/4)}
=\frac{x+\sqrt{x^2+16}}4. exp ( ∫ x 2 + 16 1 d x ) = e arsinh ( x /4 ) = 4 x + x 2 + 16 .
解法二
思路
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也可直接验证题给函数。求它的导数后除以函数本身，检查是否等于标准形式中 y y y
答题过程
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Let μ = ( x + x 2 + 16 ) / 4 > 0 \mu=(x+\sqrt{x^2+16})/4>0 μ = ( x + x 2 + 16 ) /4 > 0
μ ′ = 1 4 ( 1 + x x 2 + 16 ) = μ x 2 + 16 . \mu'=\frac14\bigg(1+\frac{x}{\sqrt{x^2+16}}\bigg)
=\frac{\mu}{\sqrt{x^2+16}}. μ ′ = 4 1 ( 1 + x 2 + 16 x ) = x 2 + 16 μ . Thus μ ′ / μ \mu'/\mu μ ′ / μ y y y y ′ + y / x 2 + 16 = x y'+y/\sqrt{x^2+16}=x y ′ + y / x 2 + 16 = x μ \mu μ
(b)
解法一
思路
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乘上积分因子后，左边变成乘积的导数。右边分别积分 x 2 x^2 x 2 x x 2 + 16 x\sqrt{x^2+16} x x 2 + 16
答题过程
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Multiplying the standard form by x + x 2 + 16 x+\sqrt{x^2+16} x + x 2 + 16
d d x [ y ( x + x 2 + 16 ) ] = x 2 + x x 2 + 16 . \frac{\mathrm{d}}{\mathrm{d}x}\big[y(x+\sqrt{x^2+16})\big]
=x^2+x\sqrt{x^2+16}. d x d [ y ( x + x 2 + 16 ) ] = x 2 + x x 2 + 16 . For the second integral, set v = x 2 + 16 v=x^2+16 v = x 2 + 16 d v = 2 x d x \mathrm{d}v=2x\,\mathrm{d}x d v = 2 x d x
∫ x x 2 + 16 d x = 1 2 ∫ v 1 / 2 d v = 1 3 ( x 2 + 16 ) 3 / 2 . \int x\sqrt{x^2+16}\,\mathrm{d}x=\frac12\int v^{1/2}\,\mathrm{d}v=\frac13(x^2+16)^{3/2}. ∫ x x 2 + 16 d x = 2 1 ∫ v 1/2 d v = 3 1 ( x 2 + 16 ) 3/2 . Integrating,
y ( x + x 2 + 16 ) = x 3 3 + ( x 2 + 16 ) 3 / 2 3 + C . y(x+\sqrt{x^2+16})=\frac{x^3}{3}+\frac{(x^2+16)^{3/2}}{3}+C. y ( x + x 2 + 16 ) = 3 x 3 + 3 ( x 2 + 16 ) 3/2 + C . At x = 3 x=3 x = 3 y = 6 y=6 y = 6 48 = 27 + 125 3 + C 48=\frac{27+125}{3}+C 48 = 3 27 + 125 + C C = − 8 3 C=-\frac83 C = − 3 8
y = x 3 + ( x 2 + 16 ) 3 / 2 − 8 3 ( x + x 2 + 16 ) . y=\frac{x^3+(x^2+16)^{3/2}-8}{3(x+\sqrt{x^2+16})}. y = 3 ( x + x 2 + 16 ) x 3 + ( x 2 + 16 ) 3/2 − 8 . 