y^{''}-4y^'+4y=(e^{2x})/(1+x)

解答

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{1 + x}

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

求解步骤

y^{''} (x) - 4 y^{'} (x) + 4 y = \frac{e ^{2 x}}{1 + x}

求解线性方程:

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

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

q^{^{''}} + 3 q^{^{'}} + 2 q = e^{- t} + e^{- 2 t}

y^{^{''}} - 14 y^{^{'}} + 40 y = 26 + 40 x

\frac{d ^{2} y}{d x ^{2}} - 4 y = 8 x^{2} - x + 12

y^{^{'''}} + 5 y^{^{''}} - 14 = 0

y^{^{''}} + y^{^{'}} - 2 y = 6 x - 3