What is the solution for 0.2y^{''}+14y^'+1/(0.004)y=0,y(0)=0 ?

The solution for 0.2y^{''}+14y^'+1/(0.004)y=0,y(0)=0 is y=c_{2}e^{-35t}sin(5t)

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0.2y^{''}+14y^'+1/(0.004)y=0,y(0)=0

0.2 y^{^{''}} + 14 y^{^{'}} + \frac{1}{0.004} y = 0, y (0) = 0

y = c_{2} e^{- 35 t} sin (5 t)

Solution steps

0.2 y^{''} (t) + 14 y^{'} (t) + \frac{1}{0.004} y = 0

Solve linear ODE:

y = c_{2} e^{- 35 t} sin (5 t)

y = c_{2} e^{- 35 t} sin (5 t)

2 \frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 3 y = 0

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 1, y^{^{'}} (0) = 25

2500 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

y^{^{''}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 12

y^{^{''}} + 3 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for 0.2y^{''}+14y^'+1/(0.004)y=0,y(0)=0 ?
The solution for 0.2y^{''}+14y^'+1/(0.004)y=0,y(0)=0 is y=c_{2}e^{-35t}sin(5t)