二阶非齐次线性方程：两题

齐次解同 $y''+y'+y=0$：

$$y_g=e^{-x/2}\left[c_1\cos\left(\tfrac{\sqrt3}{2}x\right)+c_2\sin\left(\tfrac{\sqrt3}{2}x\right)\right].$$

1. 常数特解取 $y_p=1$，故 $y=y_g+1$。

2. 取线性特解 $y_p=Ax+B$，代入得 $A=1,B=-1$，故 $y=y_g+x-1$。

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Original Explanation

First, the homogeneous solution is $$y_g(x) = e^{-\tfrac12x} \Bigl[ c_1\cos\!\bigl(\tfrac{\sqrt{3}}{2}x\bigr) + c_2\sin\!\bigl(\tfrac{\sqrt{3}}{2}x\bigr) \Bigr].$$

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

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