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Beliebt Rechnen >

10y^{''}+40y^'+240y=0

Lösung

10 y^{^{''}} + 40 y^{^{'}} + 240 y = 0

Lösung

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

Schritte zur Lösung

10 y^{''} (t) + 40 y^{'} (t) + 240 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

y = e^{- 2 t} (c_{1} cos (25 t) + c_{2} sin (25 t))

Graph

Beliebte Beispiele

x^{''}+2x^'+3x=0,x(0)=1,x^'(0)=0

x^{^{''}} + 2 x^{^{'}} + 3 x = 0, x (0) = 1, x^{^{'}} (0) = 0

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

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

y^{''}+9y=0,y(0)=2,y^'(0)=4

y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

(d^2y)/(dx^2)-(dy)/(dx)=0

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

u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48

u^{^{''}} + 144 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 48