4y^{''}-7y^'+4y=0

解答

4 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

求解步骤

4 y^{''} (t) - 7 y^{'} (t) + 4 y = 0

求解线性方程:

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

y^{^{''}} + y^{^{'}} + 7 y = 0

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

(D^{3} + D^{2} + 4 D + 4) y = 0

\frac{d ^{4} y}{d x ^{4}} + y = 0

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