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Beliebt Rechnen >

y^{''}+40y^'+490y=0

Lösung

y^{^{''}} + 40 y^{^{'}} + 490 y = 0

Lösung

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

Schritte zur Lösung

y^{''} (t) + 40 y^{'} (t) + 490 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

y = e^{- 20 t} (c_{1} cos (310 t) + c_{2} sin (310 t))

Graph

Beliebte Beispiele

y^{''}-9y=0,y(0)=3,y^'(0)=-9

y^{^{''}} - 9 y = 0, y (0) = 3, y^{^{'}} (0) = - 9

y^{''}-25y^'+144y=0,y^'(0)=5

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y^{^{'}} (0) = 5

y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.455

3y^{''}+4y^'+5y=0

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

y^{''}-5y^{''}+24y^'-20y=0

y^{^{''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0