题目
Problem
(a) Find the general solution of the differential equation
d 2 y d x 2 + d y d x + y = sin 2 x + cos 2 x . \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+
\frac{\mathrm{d}y}{\mathrm{d}x}
+
y
=
\sin 2x + \cos 2x. d x 2 d 2 y + d x d y + y = sin 2 x + cos 2 x . [7]
(b) For large positive values of x x x
y ≈ R sin ( 2 x − ϕ ) , y \approx R\sin(2x - \phi), y ≈ R sin ( 2 x − ϕ ) , where R R R ϕ \phi ϕ
[3]
题目中文翻译
(a) 求微分方程
d 2 y d x 2 + d y d x + y = sin 2 x + cos 2 x . \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+
\frac{\mathrm{d}y}{\mathrm{d}x}
+
y
=
\sin 2x + \cos 2x. d x 2 d 2 y + d x d y + y = sin 2 x + cos 2 x . 的通解。
(7)
(b) 对于较大的正 x x x
y ≈ R sin ( 2 x − ϕ ) , y \approx R\sin(2x - \phi), y ≈ R sin ( 2 x − ϕ ) , 其中 R R R ϕ \phi ϕ
(3)
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
展开
先解辅助方程得到互补函数。特解试作 p sin 2 x + q cos 2 x p\sin 2x+q\cos 2x p sin 2 x + q cos 2 x
答题过程
展开
The auxiliary equation m 2 + m + 1 = 0 m^2+m+1=0 m 2 + m + 1 = 0 − 1 2 ± 3 2 i -\frac12\pm\frac{\sqrt3}{2}\mathrm{i} − 2 1 ± 2 3 i
e − x / 2 ( A cos 3 x 2 + B sin 3 x 2 ) . e^{-x/2}\bigg(A\cos\frac{\sqrt3 x}{2}+B\sin\frac{\sqrt3 x}{2}\bigg). e − x /2 ( A cos 2 3 x + B sin 2 3 x ) . For a particular integral take y = p sin 2 x + q cos 2 x y=p\sin2x+q\cos2x y = p sin 2 x + q cos 2 x
y ′ = 2 p cos 2 x − 2 q sin 2 x , y ′ ′ = − 4 p sin 2 x − 4 q cos 2 x . \begin{align*}
y'=&\,2p\cos2x-2q\sin2x,\\
y''=&\,-4p\sin2x-4q\cos2x.
\end{align*} y ′ = y ′′ = 2 p cos 2 x − 2 q sin 2 x , − 4 p sin 2 x − 4 q cos 2 x . Substitution gives
( − 3 p − 2 q ) sin 2 x + ( 2 p − 3 q ) cos 2 x = sin 2 x + cos 2 x . (-3p-2q)\sin 2x+(2p-3q)\cos 2x=\sin 2x+\cos 2x. ( − 3 p − 2 q ) sin 2 x + ( 2 p − 3 q ) cos 2 x = sin 2 x + cos 2 x . Solving − 3 p − 2 q = 1 -3p-2q=1 − 3 p − 2 q = 1 2 p − 3 q = 1 2p-3q=1 2 p − 3 q = 1 p = − 1 13 p=-\frac1{13} p = − 13 1 q = − 5 13 q=-\frac5{13} q = − 13 5
y = e − x / 2 ( A cos 3 x 2 + B sin 3 x 2 ) − 1 13 sin 2 x − 5 13 cos 2 x . \begin{align*}
y=&\,e^{-x/2}\bigg(A\cos\frac{\sqrt3 x}{2}+B\sin\frac{\sqrt3 x}{2}\bigg)\\
&\,\hspace{2pt}-\frac1{13}\sin2x-\frac5{13}\cos2x.
\end{align*} y = e − x /2 ( A cos 2 3 x + B sin 2 3 x ) − 13 1 sin 2 x − 13 5 cos 2 x .
(b)
解法一
思路
展开
对任意固定的初始条件，含 e − x / 2 e^{-x/2} e − x /2 cos ϕ < 0 \cos\phi<0 cos ϕ < 0 sin ϕ > 0 \sin\phi>0 sin ϕ > 0 ϕ \phi ϕ
答题过程
展开
The complementary function tends to zero as x → + ∞ x\to+\infty x → + ∞ A , B A,B A , B R sin ( 2 x − ϕ ) R\sin(2x-\phi) R sin ( 2 x − ϕ )
R cos ϕ = − 1 13 , R sin ϕ = 5 13 . R\cos\phi=-\frac1{13},\qquad R\sin\phi=\frac5{13}. R cos ϕ = − 13 1 , R sin ϕ = 13 5 . The amplitude satisfies R 2 = ( 1 + 25 ) / 169 = 26 / 169 R^2=(1+25)/169=26/169 R 2 = ( 1 + 25 ) /169 = 26/169 R cos ϕ < 0 R\cos\phi<0 R cos ϕ < 0 R sin ϕ > 0 R\sin\phi>0 R sin ϕ > 0 ϕ \phi ϕ
R = 26 13 , ϕ = π − arctan 5 ≈ 1.77. R=\frac{\sqrt{26}}{13}, \qquad \phi=\pi-\arctan5\approx1.77. R = 13 26 , ϕ = π − arctan 5 ≈ 1.77. Hence, for large positive x x x
y ≈ 26 13 sin ( 2 x − ( π − arctan 5 ) ) . y\approx\frac{\sqrt{26}}{13}\sin\big(2x-(\pi-\arctan5)\big). y ≈ 13 26 sin ( 2 x − ( π − arctan 5 ) ) . 