题目
Problem
It is given that v = x y 3 v=xy^3 v = x y 3
x y 2 d 2 y d x 2 + 2 x y ( d y d x ) 2 + ( 3 x + 2 ) y 2 d y d x + ( x + 1 ) y 3 = x + 5. \begin{align*}
xy^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+2xy\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2
+&\,(3x+2)y^2\frac{\mathrm{d}y}{\mathrm{d}x}\\
&\,\hspace{2pt}+(x+1)y^3=x+5.
\end{align*} x y 2 d x 2 d 2 y + 2 x y ( d x d y ) 2 + ( 3 x + 2 ) y 2 d x d y + ( x + 1 ) y 3 = x + 5. (a) Show that
1 3 d 2 v d x 2 + d v d x + v = x + 5. \frac{1}{3}\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+\frac{\mathrm{d}v}{\mathrm{d}x}+v=x+5. 3 1 d x 2 d 2 v + d x d v + v = x + 5. (4)
(b) Find the general solution for y y y x x x
(7)
(c) Deduce that, for large positive values of x x x
y ≈ a + b x 3 , y\approx\sqrt[3]{a+\frac{b}{x}}, y ≈ 3 a + x b , where the constants a a a b b b
(2)
题目中文翻译
已知 v = x y 3 v=xy^3 v = x y 3
x y 2 d 2 y d x 2 + 2 x y ( d y d x ) 2 + ( 3 x + 2 ) y 2 d y d x + ( x + 1 ) y 3 = x + 5. \begin{align*}
xy^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+2xy\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2
+&\,(3x+2)y^2\frac{\mathrm{d}y}{\mathrm{d}x}\\
&\,\hspace{2pt}+(x+1)y^3=x+5.
\end{align*} x y 2 d x 2 d 2 y + 2 x y ( d x d y ) 2 + ( 3 x + 2 ) y 2 d x d y + ( x + 1 ) y 3 = x + 5. (a) 证明
1 3 d 2 v d x 2 + d v d x + v = x + 5. \frac{1}{3}\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+\frac{\mathrm{d}v}{\mathrm{d}x}+v=x+5. 3 1 d x 2 d 2 v + d x d v + v = x + 5. (4)
(b) 求 y y y x x x
(7)
(c) 推导出：当 x x x
y ≈ a + b x 3 , y\approx\sqrt[3]{a+\frac{b}{x}}, y ≈ 3 a + x b , 其中常数 a a a b b b
(2)
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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对 v = x y 3 v=xy^3 v = x y 3 v ′ ′ / 3 , v ′ , v v''/3,v',v v ′′ /3 , v ′ , v
答题过程
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Differentiating gives
v ′ = y 3 + 3 x y 2 y ′ , v ′ ′ = 3 y 2 y ′ + 3 y 2 y ′ + 3 x ( 2 y ( y ′ ) 2 + y 2 y ′ ′ ) = 6 y 2 y ′ + 6 x y ( y ′ ) 2 + 3 x y 2 y ′ ′ . \begin{align*}
v'=&\,y^3+3xy^2y',\\
v''=&\,3y^2y'+3y^2y'+3x\big(2y(y')^2+y^2y''\big)\\
=&\,6y^2y'+6xy(y')^2+3xy^2y''.
\end{align*} v ′ = v ′′ = = y 3 + 3 x y 2 y ′ , 3 y 2 y ′ + 3 y 2 y ′ + 3 x ( 2 y ( y ′ ) 2 + y 2 y ′′ ) 6 y 2 y ′ + 6 x y ( y ′ ) 2 + 3 x y 2 y ′′ . Consequently
1 3 v ′ ′ + v ′ + v = x y 2 y ′ ′ + 2 x y ( y ′ ) 2 + ( 3 x + 2 ) y 2 y ′ + ( x + 1 ) y 3 = x + 5. \begin{align*}
\frac13v''+v'+v=&\,xy^2y''+2xy(y')^2+(3x+2)y^2y'\\
&\,\hspace{2pt}+(x+1)y^3\\
=&\,x+5.
\end{align*} 3 1 v ′′ + v ′ + v = = x y 2 y ′′ + 2 x y ( y ′ ) 2 + ( 3 x + 2 ) y 2 y ′ + ( x + 1 ) y 3 x + 5.
(b)
解法一
思路
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先解关于 v v v v = x y 3 v=xy^3 v = x y 3 y y y
答题过程
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The auxiliary equation 1 3 m 2 + m + 1 = 0 \frac13m^2+m+1=0 3 1 m 2 + m + 1 = 0 − 3 2 ± 3 2 i -\frac32\pm\frac{\sqrt3}{2}\mathrm{i} − 2 3 ± 2 3 i v p = p x + q v_p=px+q v p = p x + q v p ′ = p v_p'=p v p ′ = p v p ′ ′ = 0 v_p''=0 v p ′′ = 0 p x + p + q = x + 5 px+p+q=x+5 p x + p + q = x + 5 p = 1 p=1 p = 1 q = 4 q=4 q = 4 v v v
v = x + 4 + T ( x ) , T ( x ) = e − 3 x / 2 ( A cos 3 x 2 + B sin 3 x 2 ) . \begin{align*}
v=&\,x+4+T(x),\\
T(x)=&\,e^{-3x/2}\bigg(A\cos\frac{\sqrt3\,x}{2}+B\sin\frac{\sqrt3\,x}{2}\bigg).
\end{align*} v = T ( x ) = x + 4 + T ( x ) , e − 3 x /2 ( A cos 2 3 x + B sin 2 3 x ) . Thus for x ≠ 0 x\ne0 x = 0
y = 1 + 4 x + T ( x ) x 3 . y=\sqrt[3]{1+\frac4x+\frac{T(x)}x}. y = 3 1 + x 4 + x T ( x ) .
(c)
解法一
思路
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初始条件只改变指数项中的常数。随着 x x x e − 3 x / 2 e^{-3x/2} e − 3 x /2 1 / x 1/x 1/ x
答题过程
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For any fixed A , B A,B A , B e − 3 x / 2 / x e^{-3x/2}/x e − 3 x /2 / x 1 / x 1/x 1/ x
y ≈ 1 + 4 x 3 , a = 1 , b = 4 . y\approx\sqrt[3]{1+\frac4x},
\qquad
\boxed{a=1,\quad b=4}. y ≈ 3 1 + x 4 , a = 1 , b = 4 . 