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二阶非齐次线性方程：两题

Second-Order Nonhomogeneous Linear ODE

专题
General / 综合
难度
L4

题目详情

解：

  1. y′′+y′+y=1y''+y'+y=1；

  2. y′′+y′+y=xy''+y'+y=x。

英文原题

For
a y′′+b y′+c y=d(x),a\,y'' + b\,y' + c\,y = d(x),
find a particular solution yp.y_p. The general solution is
y=yp+yg,y = y_p + y_g,
where ygy_g solves the homogeneous equation a y′′+b y′+c y=0.a\,y'' + b\,y' + c\,y=0.

Particular solutions are found by an “educated guess” method (polynomial, exponential, or trigonometric forms, possibly multiplied by xx or x2x^2 if certain roots appear in the homogeneous solution).

Question: Solve

  1. y′′+y′+y=1,y'' + y' + y = 1,
  2. y′′+y′+y=x.y'' + y' + y = x.
解析

齐次解同 y′′+y′+y=0y''+y'+y=0：

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

  1. 常数特解取 yp=1y_p=1，故 y=yg+1y=y_g+1。

  2. 取线性特解 yp=Ax+By_p=Ax+B，代入得 A=1,B=−1A=1,B=-1，故 y=yg+x−1y=y_g+x-1。

英文解析

First, the homogeneous solution is
yg(x)=e−12x[c1cos⁡ ⁣(32x)+c2sin⁡ ⁣(32x)].y_g(x) = e^{-\tfrac12x} \Bigl[ c_1\cos\!\bigl(\tfrac{\sqrt{3}}{2}x\bigr) + c_2\sin\!\bigl(\tfrac{\sqrt{3}}{2}x\bigr) \Bigr].

  1. For d(x)=1,d(x)=1, a constant guess: yp=1.y_p=1.
    Hence
    y=yg+1.y = y_g + 1.

  2. For d(x)=x,d(x)=x, try a linear polynomial yp=Ax+B.y_p=Ax+B.
    Then yp′=A,y_p'=A, yp′′=0.y_p''=0.
    Plugging into y′′+y′+y=x,y''+y'+y=x, we get
    0+A+(Ax+B)=x.0 + A + (Ax+B) = x.
    Comparing coefficients gives A=1,  A+B=0⇒B=−1.A=1,\; A+B=0 \Rightarrow B=-1.
    Thus yp=x−1.y_p=x-1. Finally,
    y=yg+x−1.y = y_g + x - 1.