y^{''}-3y^'-10y=-4e^{4t},y(0)=0,y^'(0)=0

解答

y^{^{''}} - 3 y^{^{'}} - 10 y = - 4 e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{2}{21} - \frac{2}{3}) e^{5 t} - \frac{2}{21} e^{- 2 t} + \frac{2}{3} e^{4 t}

求解步骤

y^{''} (t) - 3 y^{'} (t) - 10 y = - 4 e^{4 t}

求解线性方程:

y = (\frac{2}{21} - \frac{2}{3}) e^{5 t} - \frac{2}{21} e^{- 2 t} + \frac{2}{3} e^{4 t}

y = (\frac{2}{21} - \frac{2}{3}) e^{5 t} - \frac{2}{21} e^{- 2 t} + \frac{2}{3} e^{4 t}

\int 12 e^{4 x} d x

l a pl a ce t r an s f or m 10 t^{3} e^{- 2 t}

y^{^{''}} - y^{^{'}} = 6 x e^{x}

x \to 1 lim (\frac{1 - x}{3 8 x ^{3} - 8})

\frac{d}{d x} (- 12 x^{2})