题目
Problem
Find the solution of the differential equation
d y d x − x + 5 x 2 + 10 x + 61 y = 1 , \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
- \frac{x + 5}{x^2 + 10x + 61}y
=&\, 1,
\end{align*} d x d y − x 2 + 10 x + 61 x + 5 y = 1 , given that y = 0 y = 0 y = 0 x = 3 x = 3 x = 3
[10]
题目中文翻译
求微分方程
d y d x − x + 5 x 2 + 10 x + 61 y = 1 , \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
- \frac{x + 5}{x^2 + 10x + 61}y
=&\, 1,
\end{align*} d x d y − x 2 + 10 x + 61 x + 5 y = 1 , 的解，已知当 x = 3 x = 3 x = 3 y = 0 y = 0 y = 0
(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
解法一
思路
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分子是分母导数的一半，可直接求积分因子。将分母配方后，右边积分化为反双曲正弦，再用 x = 3 x=3 x = 3
答题过程
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Write Q = x 2 + 10 x + 61 = ( x + 5 ) 2 + 36 Q=x^2+10x+61=(x+5)^2+36 Q = x 2 + 10 x + 61 = ( x + 5 ) 2 + 36 y y y − Q ′ / ( 2 Q ) -Q'/(2Q) − Q ′ / ( 2 Q ) Q − 1 / 2 Q^{-1/2} Q − 1/2
∫ − x + 5 Q d x = − 1 2 ln Q , e ∫ − ( x + 5 ) / Q d x = Q − 1 / 2 . \begin{align*}
\int-\frac{x+5}{Q}\,\mathrm{d}x=&\,-\frac12\ln Q,\\
e^{\int-(x+5)/Q\,\mathrm{d}x}=&\,Q^{-1/2}.
\end{align*} ∫ − Q x + 5 d x = e ∫ − ( x + 5 ) / Q d x = − 2 1 ln Q , Q − 1/2 . For the integral, putting u = ( x + 5 ) / 6 u=(x+5)/6 u = ( x + 5 ) /6 d x = 6 d u \mathrm{d}x=6\,\mathrm{d}u d x = 6 d u
∫ 1 ( x + 5 ) 2 + 36 d x = ∫ 1 1 + u 2 d u = arsinh u + C . \int\frac1{\sqrt{(x+5)^2+36}}\,\mathrm{d}x
=\int\frac1{\sqrt{1+u^2}}\,\mathrm{d}u
=\operatorname{arsinh}u+C. ∫ ( x + 5 ) 2 + 36 1 d x = ∫ 1 + u 2 1 d u = arsinh u + C . Therefore
d d x ( y Q ) = 1 ( x + 5 ) 2 + 36 . \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{y}{\sqrt Q}\bigg) =\frac1{\sqrt{(x+5)^2+36}}. d x d ( Q y ) = ( x + 5 ) 2 + 36 1 . Integrating,
y Q = arsinh ( x + 5 6 ) + C . \frac{y}{\sqrt Q}=\operatorname{arsinh}\bigg(\frac{x+5}{6}\bigg)+C. Q y = arsinh ( 6 x + 5 ) + C . At x = 3 x=3 x = 3 arsinh ( 4 3 ) = ln ( 4 3 + 5 3 ) = ln 3 \operatorname{arsinh}(\frac43)=\ln(\frac43+\frac53)=\ln3 arsinh ( 3 4 ) = ln ( 3 4 + 3 5 ) = ln 3 y = 0 y=0 y = 0 C = − ln 3 C=-\ln3 C = − ln 3
y = x 2 + 10 x + 61 [ arsinh ( x + 5 6 ) − ln 3 ] . y=\sqrt{x^2+10x+61}\bigg[\operatorname{arsinh}\bigg(\frac{x+5}{6}\bigg)-\ln3\bigg]. y = x 2 + 10 x + 61 [ arsinh ( 6 x + 5 ) − ln 3 ] . 