What is the solution for y^{''}+2y^'+3y=0,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+2y^'+3y=0,y(0)=1,y^'(0)=3 is y=e^{-t}(cos(sqrt(2)t)+2sqrt(2)sin(sqrt(2)t))

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y^{''}+2y^'+3y=0,y(0)=1,y^'(0)=3

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y = e^{- t} (cos (2 t) + 22 sin (2 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = e^{- t} (cos (2 t) + 22 sin (2 t))

y = e^{- t} (cos (2 t) + 22 sin (2 t))

What is the solution for y^{''}+2y^'+3y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+2y^'+3y=0,y(0)=1,y^'(0)=3 is y=e^{-t}(cos(sqrt(2)t)+2sqrt(2)sin(sqrt(2)t))