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Beliebt Rechnen >

2y^{''}+2y^'+7y=0

Lösung

2 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

Lösung

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

Schritte zur Lösung

2 y^{''} (t) + 2 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

Graph

Beliebte Beispiele

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

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

(d^2y)/(dx^2)+4(dy)/(dx)+3y=0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

2y^{''''}+11y^{'''}+18y^{''}+4y^'-8y=0

2 y^{^{''''}} + 11 y^{^{'''}} + 18 y^{^{''}} + 4 y^{^{'}} - 8 y = 0

3y^{''}-5y^'+2y=0

3 y^{^{''}} - 5 y^{^{'}} + 2 y = 0

u^{'''}+u^{''}-2\H(t)=0

u^{^{'''}} + u^{^{''}} - 2 H (t) = 0