What is the solution for 10y^{''}+40y^'+240y=0 ?

The solution for 10y^{''}+40y^'+240y=0 is y=e^{-2t}(c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t))

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

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

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

Solution steps

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))

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

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

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

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

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

What is the solution for 10y^{''}+40y^'+240y=0 ?
The solution for 10y^{''}+40y^'+240y=0 is y=e^{-2t}(c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t))