y^{''}+12y^'+4y=0,y(0)=-3,y^'(0)=0

Lösung

y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 0

y = (- 3 - \frac{3 ( 3 - 2 2 )}{4 2}) e^{2 (2^{1 + \frac{1}{2}} - 3) t} + \frac{3 ( 3 - 2 2 ) e ^{- 2 (3 + 22) t}}{4 2}

Schritte zur Lösung

y^{''} (t) + 12 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (- 3 - \frac{3 ( 3 - 2 2 )}{4 2}) e^{2 (2^{1 + \frac{1}{2}} - 3) t} + \frac{3 ( 3 - 2 2 ) e ^{- 2 (3 + 22) t}}{4 2}

y = (- 3 - \frac{3 ( 3 - 2 2 )}{4 2}) e^{2 (2^{1 + \frac{1}{2}} - 3) t} + \frac{3 ( 3 - 2 2 ) e ^{- 2 (3 + 22) t}}{4 2}

y^{^{''}} + 16.2 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 4 y = 0

y^{^{''}} + 10 y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{'''}} + 9 y^{^{''}} + 8 y^{^{'}} = 0

y^{^{''}} - y^{^{'}} + cy = 0