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

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

y^{^{''}} + 3 y^{^{'}} + 6 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 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})

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1

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

y^{^{''}} + 0.5 y^{^{'}} + y = 0

y^{^{''}} + 64 π^{2} y = 0

y^{^{''}} + 16 y^{^{'}} + 73 y = 0, y^{^{'}} (0) = 8

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