y^{''}-4y^'+4y=(e^{2x})/(x^2)

解答

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{x ^{2}}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

求解步骤

y^{''} (x) - 4 y^{'} (x) + 4 y = \frac{e ^{2 x}}{x ^{2}}

求解线性方程:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

x \to a lim (sin (x))

\int_{6}^{8} \frac{1}{x ^{2} - 1} d x

x \to 0 lim (x sec (x) csc (x))

t an g e n t y = \frac{7}{x}, (4, \frac{7}{2})

\frac{d}{d x} (50 - x^{2} - x)