若函数$y=y(x)$由方程${{\text{e}}^{x+y}}=xy+1$确定，则 ( )。
A: $\text{d}x=\frac{{{\text{e}}^{x+y}}-x}{y-{{\text{e}}^{x+y}}}\text{d}y$
B: $\text{d}y=\frac{{{\text{e}}^{x+y}}-x}{y-{{\text{e}}^{x+y}}}\text{d}x$
C: $\text{d}x=\frac{{{\text{e}}^{x+y}}+x}{y+{{\text{e}}^{x+y}}}\text{d}y$
D: $\text{d}y=\frac{{{\text{e}}^{x+y}}+x}{y+{{\text{e}}^{x+y}}}\text{d}x$
A: $\text{d}x=\frac{{{\text{e}}^{x+y}}-x}{y-{{\text{e}}^{x+y}}}\text{d}y$
B: $\text{d}y=\frac{{{\text{e}}^{x+y}}-x}{y-{{\text{e}}^{x+y}}}\text{d}x$
C: $\text{d}x=\frac{{{\text{e}}^{x+y}}+x}{y+{{\text{e}}^{x+y}}}\text{d}y$
D: $\text{d}y=\frac{{{\text{e}}^{x+y}}+x}{y+{{\text{e}}^{x+y}}}\text{d}x$
举一反三
- 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: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$
- 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: $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1+y^2}{xy+x^3y}$
- 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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
- 从原点向曲线$$y=1-\ln x$$作切线，则由切线、曲线和$$x$$轴围成图形的面积为()． A: $$\frac{1}{2}{{\text{e}}^{2}}+\text{e}$$ B: $$\frac{1}{2}{{\text{e}}^{2}}-\text{e}$$ C: $${{\text{e}}^{2}}+\text{e}$$ D: $${{\text{e}}^{2}}-\text{e}$$
- 求方程$2yy''=(y')^2+y^2$的通解时,可令( ) A: $y'=p$,则$y''=p'$ B: $y'=p$,则$y''=p\dfrac{\text{d}p}{\text{d}y}$ C: $y'=p$,则$y''=p\dfrac{\text{d}p}{\text{d}x}$ D: $y'=p$,则$y''=p'\dfrac{\text{d}p}{\text{d}y}$