3x^{''}-7x^'+4x=0

Lösung

3 x^{^{''}} - 7 x^{^{'}} + 4 x = 0

x = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

x = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{t}

x = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{t}

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (ln (2)) = 3, y (0) = 33

y^{^{'''}} - 5 y^{^{''}} + 5 y^{^{'}} + 11 y = 0

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0, y^{^{'}} (1) = 3, y (1) = 4

x^{^{''}} - x^{^{'}} - 6 x = 0, x (0) = 2, x^{^{'}} (0) = - 1

y^{^{'''}} + 3 y^{^{''}} - 6 y^{^{'}} - 8 y = 0