What is the solution for y^{''}+40y^'+6250y=0 ?

The solution for y^{''}+40y^'+6250y=0 is y=e^{-20t}(c_{1}cos(15sqrt(26)t)+c_{2}sin(15sqrt(26)t))

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y^{''}+40y^'+6250y=0

y^{^{''}} + 40 y^{^{'}} + 6250 y = 0

y = e^{- 20 t} (c_{1} cos (1526 t) + c_{2} sin (1526 t))

Solution steps

y^{''} (t) + 40 y^{'} (t) + 6250 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (1526 t) + c_{2} sin (1526 t))

y = e^{- 20 t} (c_{1} cos (1526 t) + c_{2} sin (1526 t))

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

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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

y^{^{''}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 14 y^{^{'}} + 53 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

What is the solution for y^{''}+40y^'+6250y=0 ?
The solution for y^{''}+40y^'+6250y=0 is y=e^{-20t}(c_{1}cos(15sqrt(26)t)+c_{2}sin(15sqrt(26)t))