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Beliebt Rechnen >

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

Lösung

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

Lösung

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

Graph

Beliebte Beispiele

16y^{''''}+48y^{''}+36y=0

16 y^{^{''''}} + 48 y^{^{''}} + 36 y = 0

y^{''}+4y^'+4y=0,y(0)=11

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 11

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

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

x^{^{''}} + 6 x = 0

solvefor u,(d^2u)/(dt^2)=au(t)

so l v e f or u, \frac{d ^{2} u}{d t ^{2}} = a u (t)