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}