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Beliebt Rechnen >

y^{''}+(4)y^'+25y=0

Lösung

y^{^{''}} + (4) y^{^{'}} + 25 y = 0

Lösung

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

Schritte zur Lösung

y^{''} (t) + (4) y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

Graph

Beliebte Beispiele

(d^2y)/(dt^2)+ky=0

\frac{d ^{2} y}{d t ^{2}} + k y = 0

(d^2y)/(dx^2)-9y=0,y(0)=1,y(1)=-1

\frac{d ^{2} y}{d x ^{2}} - 9 y = 0, y (0) = 1, y (1) = - 1

y^{''}+6y^'+73y=0,y(0)=-5,y^'(0)=-25

y^{^{''}} + 6 y^{^{'}} + 73 y = 0, y (0) = - 5, y^{^{'}} (0) = - 25

y^{'''}=-3y^{''}+y^'

y^{^{'''}} = - 3 y^{^{''}} + y^{^{'}}

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

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