x^{''}+2.8x^'+4x=10,x(0)=2,x^'(0)=3

Lösung

x^{^{''}} + 2.8 x^{^{'}} + 4 x = 10, x (0) = 2, x^{^{'}} (0) = 3

x = e^{- \frac{7 t}{5}} (- \frac{1}{2} cos (\frac{51 t}{5}) + \frac{23}{2 51} sin (\frac{51 t}{5})) + \frac{5}{2}

Schritte zur Lösung

x^{''} (t) + 2.8 x^{'} (t) + 4 x = 10

Solve linear ODE:

x = e^{- \frac{7 t}{5}} (- \frac{1}{2} cos (\frac{51 t}{5}) + \frac{23}{2 51} sin (\frac{51 t}{5})) + \frac{5}{2}

x = e^{- \frac{7 t}{5}} (- \frac{1}{2} cos (\frac{51 t}{5}) + \frac{23}{2 51} sin (\frac{51 t}{5})) + \frac{5}{2}

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{- 6.5 e ^{- 2 t}}{t ^{2} + 1}

y^{^{''}} + 49 y = cos (7 t), y (0) = 8, y^{^{'}} (0) = 7

y^{^{''}} - 4 y^{^{'}} - 5 y = 7 e^{2} t

y^{^{''}} + 6 \cdot y^{^{'}} + 25 y = e^{- 3 t} cos (4 t)

y^{^{''}} + y = e^{- x} + 2 x^{2} + 3