u^{''}+2u^'+2\H(t)=0

Lösung

u^{''} + 2 u^{'} + 2 H (t (x)) = 0

0 + 2 \cdot 0 + 2 H (t (x)) = 0

Schritte zur Lösung

u^{''} + 2 u^{'} + 2 H (t (x)) = 0

u^{''} = 0

0 + 2 u^{^{'}} + 2 H (t (x)) = 0

u^{'} = 0

0 + 2 \cdot 0 + 2 H (t (x)) = 0

55 x^{^{''}} + 220 x = 0, x (0) = 0, x^{^{'}} (0) = - 6

y^{^{'''}} + 3.1 y^{^{''}} + 3.1 y^{^{'}} + y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = \frac{2}{5}, y^{^{'}} (0) = \frac{2}{5}

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y^{^{'}} (0) = 0, y (0) = 5

- 2 y^{^{''}} + 8 y^{^{'}} = 0, y (- 1) = 2, y^{^{'}} (2) = 2