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Beliebt Rechnen >

2y^{''}+4y^'+5y=0

Lösung

2 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

Lösung

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

Schritte zur Lösung

2 y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

Graph

Beliebte Beispiele

y^{''}-10y^'-24y=0

y^{^{''}} - 10 y^{^{'}} - 24 y = 0

(d^2y)/(dx^2)+100y=0

\frac{d ^{2} y}{d x ^{2}} + 100 y = 0

(d^2y)/(dx^2)-(dy)/(dx)-12y=0

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 12 y = 0

4y^{''}+17y=0,y(0)=3,y^'(0)=5

4 y^{^{''}} + 17 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y^{''}+2y^'+4y=0,y(0)=1

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1