What is the solution for 20y^{''}+200y^'+520y=0 ?

The solution for 20y^{''}+200y^'+520y=0 is y=e^{-5t}(c_{1}cos(t)+c_{2}sin(t))

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20y^{''}+200y^'+520y=0

20 y^{^{''}} + 200 y^{^{'}} + 520 y = 0

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

20 y^{''} (t) + 200 y^{'} (t) + 520 y = 0

Solve linear ODE:

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

What is the solution for 20y^{''}+200y^'+520y=0 ?
The solution for 20y^{''}+200y^'+520y=0 is y=e^{-5t}(c_{1}cos(t)+c_{2}sin(t))