y^{'''}-y^{''}+9y^'-145y=0,y(0)=e^{5x}

解答

y^{^{'''}} - y^{^{''}} + 9 y^{^{'}} - 145 y = 0, y (0) = e^{5 x}

y = c_{1} e^{5 x} + e^{- 2 x} (c_{2} cos (5 x) + c_{3} sin (5 x))

求解步骤

y^{'''} (x) - y^{''} (x) + 9 y^{'} (x) - 145 y = 0

求解线性方程:

y = c_{1} e^{5 x} + e^{- 2 x} (c_{2} cos (5 x) + c_{3} sin (5 x))

y = c_{1} e^{5 x} + e^{- 2 x} (c_{2} cos (5 x) + c_{3} sin (5 x))

f^{^{''}} (x) - f^{^{'}} (x) - 2 f (x) = 0, f (0) = 2

100 (1 0^{- 6}) y^{^{''}} + \frac{1}{500} y^{^{'}} + y = 0

y^{^{''}} + 3 y = 0, y^{^{'}} (1) = 1

16 y^{^{''}} + 40 y^{^{'}} + 50 y = 0

y^{^{''}} - 16 y = 0, y (0) = e^{4 x}, y (0) = 9, y^{^{'}} (0) = 3