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Beliebt Rechnen >

4y^{''}+20y^'+29y=0,y(0)=8

Lösung

4 y^{^{''}} + 20 y^{^{'}} + 29 y = 0, y (0) = 8

Lösung

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

Schritte zur Lösung

4 y^{''} (t) + 20 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

Graph

Beliebte Beispiele

9y^{''}+24y^'-33y=0

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y^{'''}+3y^{''}+4y^'+12y=0

y^{^{'''}} + 3 y^{^{''}} + 4 y^{^{'}} + 12 y = 0

y^{''}+4y=0,y(pi/4)=-2

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

(d^2x)/(dt^2)=-k(dx)/(dt)

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}