微分算子法解三角函数幂函数型微分方程

写出特解

\begin{align} & y^* = \frac{1}{D + D^2} x \cos{x} & \end{align}

cos x 转化为指数函数

\begin{align} & e^{ix} = \cos{x} + i \sin{x} & \end{align}

演算

\begin{align} & y^* = \frac{1}{D + D^2} x \cos{x} &\\& \rightarrow \frac{1}{D+D^2} x e^{ix} &\\& \text{(此处把虚部添加到表达式了，最终结果要舍弃虚部)}&\\& = e^{ix} \frac{1}{(D+i)+(D+i)^2} x &\\& = e^{ix} \frac{1}{i-1 + (2i+1)D + D^2} x & \end{align}

\begin{align} & &\\& \text{分母可以提取 i-1 或者乘以 i-1 的倒数，乘法比除法简单，选择乘法} &\\& \frac{1}{i-1} = \frac{i+1}{(i-1)(i+1)} = -\frac{1}{2} (i+1) & \end{align}

\begin{align} & y^* \rightarrow &\\& -\frac{1}{2}(i+1)e^{ix} \frac{1}{1 - \frac{1}{2}(i+1)(2i+1)D -\frac{1}{2}(i+1)D^2} x &\\& = -\frac{1}{2}(i+1)e^{ix} \frac{1}{1-\frac{1}{2}(3i -1)D - \frac{1}{2}(i+1)D^2} x &\\& & \end{align}

\begin{align} & \text{泰勒展开, 再演算} &\\& y^* \rightarrow &\\& -\frac{1}{2}(i+1)e^{ix} [1 + \frac{1}{2}(3i -1)D + \frac{1}{2}(i+1)D^2] x &\\& = -\frac{1}{2}(i+1)e^{ix} [1 + \frac{1}{2}(3i -1)D] x &\\& = -\frac{1}{2}(i+1)e^{ix} [x + \frac{1}{2}(3i -1)] &\\& = -\frac{1}{2}(i+1)(\cos{x} + i\sin{x})[x-\frac{1}{2} + i \frac{3}{2} ] &\\& = -\frac{1}{4} (\cos{x} + i\sin{x}) (i+1) (2x -1 + i3) &\\& = -\frac{1}{4} (\cos{x} + i\sin{x}) (i 2x -i -3 +2x -1 + i 3) &\\& = -\frac{1}{4} (\cos{x} + i\sin{x}) [2x-4 + i (2x+2)] &\\& = -\frac{1}{2} (\cos{x} + i\sin{x}) [x-2 + i (x+1)] &\\& \rightarrow -\frac{1}{2}[\cos{x} (x-2) - \sin{x} (x+1)] &\\& \text{(舍弃虚部了)} & \end{align}

\begin{align} \text{最终得到：} &\\& & y^* = \frac{1}{2} [ (x+1) \sin{x} - (x-2) \cos{x}] & \end{align}

\begin{align} & \text{验证：} &\\& y' = \frac{1}{2}[\sin{x} + (x+1)\cos{x} - \cos{x} + (x-2)\sin{x}] &\\& = \frac{1}{2}[x\cos{x} + (x-1)\sin{x}] &\\& y'' = \frac{1}{2}[\cos{x} - x\sin{x} + \sin{x} + (x-1)\cos{x}] &\\& = \frac{1}{2}[x\cos{x} + (1-x)\sin{x}] &\\& &\\& \Longrightarrow y'' + y' = x\cos{x} & \end{align}