y^{''}-3y^'=0,y(0)=2-2e^3,y(1)=0

解答

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = 2 - 2 e^{3}, y (1) = 0

y = (2 - 2 e^{3} - \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}) e^{3 t} + \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}

求解步骤

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

求解线性方程:

y = (2 - 2 e^{3} - \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}) e^{3 t} + \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}

y = (2 - 2 e^{3} - \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}) e^{3 t} + \frac{- 2 e ^{3} + 2 e ^{6}}{- e ^{3} + 1}

y^{^{''}} - 19 y^{^{'}} + 90 y = 0

y^{^{''}} + 5 y^{^{'}} + \frac{29}{4} y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{7}{2}

y^{^{''''}} - 2 y^{^{''}} = 0

- 10 y^{^{''}} + y = 0

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 2