A: $\left(1+e^{-\frac{x}{y}}\right)y\text{d}x+(y-x)\text{d}y=0$
B: $x\left(\ln
x-\ln y\right) \text{d}x-y\text{d}y=0$
C: $x
\dfrac{\text{d}y}{\text{d}x}-y+\sqrt{x^2-y^2}=0$
D: $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1+y^2}{xy+x^3y}$
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: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$
- 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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
- 下列函数是多元初等函数的是( ) A: $f(x,y)=\left|x+y\right|$; B: $f(x,y)=\text{sgn}(x+y)$; C: $f(x,y)=\dfrac{\arcsin<br/>x-e^{y}}{~\ln(x^2+y^2)~}$; D: $f(x,y)=\left\{\begin{array}{cc}\dfrac{xy}{~x^2+y^2~},<br/>&x^2+y^2\neq 0; \\0, &x^2+y^2= 0. \end{array}\right.$
- 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: $\text{d}y=\frac{{{\text{e}}^{x+y}}+x}{y+{{\text{e}}^{x+y}}}\text{d}x$
- 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$y''=p'\dfrac{\text{d}p}{\text{d}y}$
内容
- 0
函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( ) A: $\left\{(x,y)\left|~x+y\geq<br/>0\right.\right\}$; B: $\left\{(x,y)\left|~x+y\geq<br/>1~\text{或}~x+y\leq<br/>-1 \right.\right\}$; C: $\left\{(x,y)\left|~x+y\geq<br/>1\right.\right\}$; D: $\left\{(x,y)\left|~x+y\geq<br/>\dfrac{4}{~\pi^2~}\right.\right\}$．
- 1
下列函数中( )不是方程\( y' + xy = 0 \)的解。 A: \( y = {e^{ - { { {x^2}} \over 2}}} \) B: \( \ln \left| y \right| = - { { {x^2}} \over 2} \) C: \( y = {e^{ - { { {x^2}} \over 2}}} + 2 \) D: \( \ln \left| y \right| = - { { {x^2}} \over 2} +2\)
- 2
下列微分方程中，( )是齐次方程。 A: \( xy' = y(\ln y - \ln x) \) B: \( xy' + {y \over x} - x = 0 \) C: \( y' + {y \over x} = {1 \over { { x^2}}} \) D: \( y - y' = 1 + xy' \)
- 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
函数$f(x,y)={{\text{e}}^{-x}}\ln (1+y)$在点$(0,0)$处2次Taylor多项式为 A: $y+\frac{1}{2}(-2xy-{{y}^{2}})$ B: $y+\frac{1}{2}(-2xy+{{y}^{2}})$ C: $y+\frac{1}{2}(2xy-{{y}^{2}})$ D: $y+\frac{1}{2}(-xy-{{y}^{2}})$