A: y=Ce-x
B: y=e-x+C
C: y=C1e-x+C2
D: y=e-x
举一反三
- 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: $y=C \frac{e^{-x^2}}{x^2}+1$
- 微分方程y'+y=0的通解为y=() A: e-x+C B: -e-x+C C: Ce-x D: Cex
- 3. 方程$x y' + xy = y $的通解为 A: \[y=\mathit{c}\,{{e}^{-x}}\] B: \[y=\mathit{c}x\,{{e}^{-x}}\] C: \[y=\mathit{c}x\,{{e}^{-x^2}}\] D: \[y=\mathit{c}x^2\,{{e}^{-x}}\]
- 微分方程y'=y的通解是( ). A: y=x B: y= Cx C: y=eˣ D: y= Ceˣ
- 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: $y={{C}_{1}}{{e}^{{{C}_{2}}x}}$
内容
- 0
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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
- 1
方程xy'-ylny=0的通解为( )。 A: y=ecx B: y=x C: y=e-x D: y=ex
- 2
设\(z = {e^ { { y \over x}}} + {x^y} + {y^x}\),则\({z_x} = \) A: \({1 \over x}{e^ { { y \over x}}} + {x^y}\ln x + x{y^{x - 1}}\) B: \(- {y \over { { x^2}}}{e^ { { y \over x}}} + {x^y}\ln x + x{y^{x - 1}}\) C: \({e^ { { y \over x}}} + y{x^{y - 1}} + {y^x}\ln y\) D: \( - {y \over { { x^2}}}{e^ { { y \over x}}} + y{x^{y - 1}} + {y^x}\ln y\)
- 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
下列不等式正确的是( ) A: \( { { {e^x} + {e^y}} \over 2} < {e^ { { {x + y} \over 2}}}\quad (x \ne y)\) B: \((x + y){e^{x + y}} < x{e^{2x}} + y{e^{2y}}\quad (x > 0,y > 0,x \ne y)\) C: \( { { {x^n} + {y^n}} \over 2} < {( { { x + y} \over 2})^n}\quad (x > 0,y > 0,x \ne y,n > 1)\) D: \(x\ln x + y\ln y < (x + y)ln { { x + y} \over 2}\quad (x > 0,y > 0,x \ne y)\)