What is the solution for y^{''}+2y^'+6y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+2y^'+6y=0,y(0)=0,y^'(0)=1 is y= 1/(sqrt(5))e^{-t}sin(sqrt(5)t)

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y^{''}+2y^'+6y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{5} e^{- t} sin (5 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = \frac{1}{5} e^{- t} sin (5 t)

y = \frac{1}{5} e^{- t} sin (5 t)

What is the solution for y^{''}+2y^'+6y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+2y^'+6y=0,y(0)=0,y^'(0)=1 is y= 1/(sqrt(5))e^{-t}sin(sqrt(5)t)