• 2022-06-19
    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: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
  • D

    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

      微分方程\(2y''+5y'=5x^2-2x-1\)的通解是( )。 A: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3-\frac{3}{5}x^2+\frac{7}{25}x\) B: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3-\frac{3}{5}x^2\) C: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3+\frac{7}{25}x\) D: \(y=C_1+C_2e^{-\frac{5}{2}x}-\frac{3}{5}x^2+\frac{7}{25}x\)

    • 2

      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: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$

    • 3

      下列函数在点$(0,0)$的重极限存在的是 A: $f(x,y)=\frac{y^2}{x^2+y^2}$ B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$ C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$

    • 4

      \(\int { { {\sec }^{3}}xdx}\)=( ) A: \(\frac{1}{2}\sec x\cot x-\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) B: \(\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) C: \(-\frac{1}{2}\csc x\tan x+\frac{1}{2}\ln \left| \sec x-\cot x \right|+C\) D: \(-\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln \left| \csc x+\tan x \right|+C\)