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Beliebt Rechnen >

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

Lösung

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

Lösung

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

Graph

Beliebte Beispiele

(d^{4} + 4 d^{2}) y = 0

y^{''}-y^'-11y=0

y^{^{''}} - y^{^{'}} - 11 y = 0

y^{''}+2y^'+5y=0,y(0)=2,y(0)=4

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y (0) = 4

2y^{''}-7y^'-4y=0

2 y^{^{''}} - 7 y^{^{'}} - 4 y = 0

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

y^{^{''}} + y = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3