y^{''}+16y=g(t),y(0)=5,y^'(0)=0

Lösung

y^{^{''}} + 16 y = g (t), y (0) = 5, y^{^{'}} (0) = 0

y = 5 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

Schritte zur Lösung

y^{''} (t) + 16 y = g (t)

Solve linear ODE:

y = 5 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

y = 5 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 t} ln (t)

y^{^{''}} - y^{^{'}} - 2 y = 1, y (0) = 1, y^{^{'}} (0) = 3

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 6 y = x e^{x}

(D^{2} - 4 D + 4) y = e^{- 4 x} + 5 cos (3 x)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 10 y = 50 x