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二阶齐次方程

Second-Order Homogeneous Linear ODE

专题
General / 综合
难度
L4

题目详情

解微分方程：y′′+y′+y=0y''+y'+y=0。

英文原题

For a 2nd-order homogeneous linear ODE
>a d2ydx2>  +  >b dydx>  +  >c y>=>0,>> a\,\frac{d^2y}{dx^2} > \;+\; > b\,\frac{dy}{dx} > \;+\; > c\,y > = > 0, >
the characteristic equation is
>a r2+b r+c=0.>> a\,r^2 + b\,r + c = 0. >

  1. Two real distinct roots r1, r2.r_1,\,r_2. The general solution is
    y=c1 e r1x+c2 e r2x.y = c_1\,e^{\,r_1 x} + c_2\,e^{\,r_2 x}.
  2. A repeated real root r.r. Then
    y=c1 e rx+c2 x e rx.y = c_1\,e^{\,r x} + c_2\,x\,e^{\,r x}.
  3. Complex roots α±iβ.\alpha\pm i\beta. Then
    y=e αx [c1 cos⁡(βx)+c2 sin⁡(βx)].y = e^{\,\alpha x}\, \bigl[ c_1\,\cos(\beta x) + c_2\,\sin(\beta x) \bigr].

Question: Solve
y′′+y′+y=0.y'' + y' + y = 0.

解析

特征方程 r2+r+1=0r^2+r+1=0，根为 r=−12±i32r=-\frac12\pm i\frac{\sqrt3}{2}。

因此通解为

y=e−x/2[c1cos⁡(32x)+c2sin⁡(32x)].y=e^{-x/2}\left[c_1\cos\left(\tfrac{\sqrt3}{2}x\right)+c_2\sin\left(\tfrac{\sqrt3}{2}x\right)\right].

英文解析

Characteristic equation:
r2+r+1=0    ⟹    r=−12  ±  32 i.r^2 + r + 1 = 0 \;\;\Longrightarrow\;\; r = -\frac{1}{2} \;\pm\; \frac{\sqrt{3}}{2}\,i.
Hence
y=e−x/2[c1cos⁡ ⁣(32 x)+c2sin⁡ ⁣(32 x)].y = e^{-x/2} \Bigl[ c_1 \cos\!\bigl(\tfrac{\sqrt{3}}{2}\,x\bigr) + c_2 \sin\!\bigl(\tfrac{\sqrt{3}}{2}\,x\bigr) \Bigr].