y^{''}+9y=g(t),y(0)=2,y^'(0)=0

Lösung

y^{^{''}} + 9 y = g (t), y (0) = 2, y^{^{'}} (0) = 0

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

Schritte zur Lösung

y^{''} (t) + 9 y = g (t)

Solve linear ODE:

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

4 y^{^{''}} - 6 y^{^{'}} + 2 y = 24 e^{- x} - 10 sin (x)

10 y^{^{''}} + 60 y^{^{'}} + 50 y = 5 e^{- x}

y^{^{''}} - 16 y^{^{'}} + 64 y = \frac{2 e ^{8 x}}{3 x}

y^{^{''}} + 2 y^{^{'}} + 3 y = cos (2 t)

y^{^{''}} + y = cos (x), y (0) = 1, y^{^{'}} (0) = - 1