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Beliebt Rechnen >

2y^{''}+4y^'+20y=0

Lösung

2 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

Lösung

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Schritte zur Lösung

2 y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Graph

Beliebte Beispiele

y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y^{''}+25y^'+1/(0.046)y=0

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

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

(d^2y)/(dx^2)+2(dy)/(dx)=0

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

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

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