What is the solution for 0.1y^{''}+4y^'+250y=0 ?

The solution for 0.1y^{''}+4y^'+250y=0 is y=e^{-20t}(c_{1}cos(10sqrt(21)t)+c_{2}sin(10sqrt(21)t))

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0.1y^{''}+4y^'+250y=0

0.1 y^{^{''}} + 4 y^{^{'}} + 250 y = 0

y = e^{- 20 t} (c_{1} cos (1021 t) + c_{2} sin (1021 t))

Solution steps

0.1 y^{''} (t) + 4 y^{'} (t) + 250 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (1021 t) + c_{2} sin (1021 t))

y = e^{- 20 t} (c_{1} cos (1021 t) + c_{2} sin (1021 t))

125 y^{^{''}} + 30 y^{^{'}} + 500 y = 0

x^{^{''}} + 1 0^{6} x^{^{'}} - 1.6 \cdot 1 0^{11} x = 0

t^{''} - a t (x) = 0

y^{^{''}} + 4 y^{^{'}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 1

2 y^{^{''}} + 16 y = 0, y (0) = 0.125, y^{^{'}} (0) = 0

What is the solution for 0.1y^{''}+4y^'+250y=0 ?
The solution for 0.1y^{''}+4y^'+250y=0 is y=e^{-20t}(c_{1}cos(10sqrt(21)t)+c_{2}sin(10sqrt(21)t))