u^{''}+3u^'+5\H(t)=0

Lösung

u^{''} + 3 u^{'} + 5 H (t (x)) = 0

0 + 3 \cdot 0 + 5 H (t (x)) = 0

Schritte zur Lösung

u^{''} + 3 u^{'} + 5 H (t (x)) = 0

u^{''} = 0

0 + 3 u^{^{'}} + 5 H (t (x)) = 0

u^{'} = 0

0 + 3 \cdot 0 + 5 H (t (x)) = 0

y^{^{''}} - 6 y = 0, y (0) = - 23, y^{^{'}} (0) = 6

(D^{3} - D^{2}) y = 0

4 y^{^{'''}} + 5 y^{^{''}} - 23 y^{^{'}} - 6 y = 0

y^{^{''}} - 8 y^{^{'}} - 7 y = 0

x^{^{''}} + 6 x^{^{'}} + 9 x = 0, x (0) = a, x^{^{'}} (0) = b