y^{''}-2y^'-3y=3te^{2t},y(0)=3,y^'(0)=0

解答

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

y = (- \frac{13}{6} + \frac{11}{3}) e^{3 t} + \frac{13}{6} e^{- t} - e^{2 t} t - \frac{2}{3} e^{2 t}

求解步骤

y^{''} (t) - 2 y^{'} (t) - 3 y = 3 t e^{2 t}

求解线性方程:

y = (- \frac{13}{6} + \frac{11}{3}) e^{3 t} + \frac{13}{6} e^{- t} - e^{2 t} t - \frac{2}{3} e^{2 t}

y = (- \frac{13}{6} + \frac{11}{3}) e^{3 t} + \frac{13}{6} e^{- t} - e^{2 t} t - \frac{2}{3} e^{2 t}

y^{^{''}} + 36 y = 10 cos (6 x) + 15 sin (6 x)

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 4 x^{2} - 3 x + 2

y^{^{''''}} - 2 y^{^{'''}} + 2 y^{^{''}} - 2 y^{^{'}} + y = e^{x}

y^{^{''}} + 6 y^{^{'}} + 5 y = 3 e^{- x}

y^{^{''}} + 4 y = tan (2 x) + 2