What is the solution for y^{''}+3y^'+6*y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+3y^'+6*y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(15))e^{-(3t)/2}sin((sqrt(15)t)/2)

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

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

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

Solution steps

y^{''} (t) + 3 y^{'} (t) + 6 \cdot y = 0

Solve linear ODE:

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 8.1, y^{^{'}} (0) = 3.9

y^{^{'''}} + y^{^{''}} - 24 y^{^{'}} + 36 y = 0

36 y^{^{''}} - 132 y^{^{'}} + 121 y = 0, y (0) = 5, y^{^{'}} (0) = 5

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} + 3 y = 0

625 y^{^{''}} - 9 y = 0, y (0) = 9, y^{^{'}} (0) = 8

What is the solution for y^{''}+3y^'+6*y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'+6*y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(15))e^{-(3t)/2}sin((sqrt(15)t)/2)