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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
举一反三
- 方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 以下关系式中，正确的是( )。 A: $2\arctan x+\arcsin \frac{2x}{1+{{x}^{2}}}=\text{ }\!\!\pi\!\!\text{ }$,$|x|\ge 1$ B: $\arctan x=\arcsin \frac{x}{\sqrt{1+{{x}^{2}}}}+\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$,$-\infty \lt x \lt \infty $ C: $\arcsin x+\arccos x=\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$，$|x|\le 1$ D: $\arcsin x=\arctan \frac{x}{\sqrt{1-{{x}^{2}}}}-\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$，$|x| \lt 1$
- 函数$y={{\ln }^{3}}{{x}^{2}}$的微分为（ ）。 A: $\text{d}y=6x{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ B: $\text{d}y=\frac{6}{x}{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ C: $\text{d}y=3{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ D: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$
- 函数$f(x,y)={{\text{e}}^{-x}}\cos y$在点$(0,0)$处2次Taylor多项式为 A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$ D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
- 4．下列曲线中有渐近线的是 A: $y={{x}^{2}}+\sin x$ B: $y=x+\sin x$ C: $y={{x}^{2}}+\sin \frac{1}{x}$ D: $y=x+\sin \frac{1}{x}$