2x^{''}+5x^'+4x=10r(t)

Lösung

2 x^{^{''}} + 5 x^{^{'}} + 4 x = 10 r (t)

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})) + \frac{5 r t}{2} - \frac{25 r}{8}

Schritte zur Lösung

2 x^{''} (t) + 5 x^{'} (t) + 4 x = 10 r (t)

Solve linear ODE:

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})) + \frac{5 r t}{2} - \frac{25 r}{8}

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})) + \frac{5 r t}{2} - \frac{25 r}{8}

x^{^{''}} + 6 x^{^{'}} - 4 x = 8 e^{t}

y^{^{''}} + 9 y = 2 cos (3 x) - 3 sin (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 = 0

y^{^{''}} + 3 y = 6 x

y^{^{''}} - 3 y^{^{'}} + 2 = 0