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: $y=x+\sin \frac{1}{x}$
内容
- 0
求方程$y\frac{{{d}^{2}}y}{d{{x}^{2}}}-(\frac{dy}{dx})^{2}=0$的通解： A: $y={{C}_{1}}{{e}^{-{{C}_{2}}x}}$ B: $y={{C}_{1}}{{e}^{-{{C}_{2}}{{x}^{2}}}}$ C: $y={{C}_{1}}x{{e}^{-{{C}_{2}}{{x}^{2}}}}$ D: $y={{C}_{1}}{{e}^{{{C}_{2}}x}}$
- 1
方程$(x^2+1)(y^2-1) + xy y' = 0$的通解为 A: $y^2 = C \frac{e^{-x^2}}{x^2}$ B: $y = C \frac{e^{-x^2}}{x^2}$ C: $y^2 = C \frac{e^{-x^2}}{x^2}+1$ D: $y=C \frac{e^{-x^2}}{x^2}+1$
- 2
以下方程不属于齐次方程类型的是( ) A: $\left(1+e^{-\frac{x}{y}}\right)y\text{d}x+(y-x)\text{d}y=0$ B: $x\left(\ln<br/>x-\ln y\right) \text{d}x-y\text{d}y=0$ C: $x<br/>\dfrac{\text{d}y}{\text{d}x}-y+\sqrt{x^2-y^2}=0$ D: $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1+y^2}{xy+x^3y}$
- 3
方程\(\left( {1 - {x^2}} \right)y - xy' = 0\)的通解是（ ）。 A: \(y = C\sqrt {1 - {x^2}} \) B: \(y = - {1 \over 2}{x^3} + Cx\) C: \(y = {C \over {\sqrt {1 - {x^2}} }}\) D: \(y = Cx{e^{ - {1 \over 2}{x^2}}}\)
- 4
【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )