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举一反三
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- 已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$，则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
- Skim the text and answer the following questions. 1) What type of writing is the text?
- Part 1 ： The text begins with an _________.
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