What is the solution for y^{''}+3y^'+8y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+3y^'+8y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(23))e^{-(3t)/2}sin((sqrt(23)t)/2)

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y^{''}+3y^'+8y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 3 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{2}{23} e^{- \frac{3 t}{2}} sin (\frac{23 t}{2})

Solution steps

y^{''} (t) + 3 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = \frac{2}{23} e^{- \frac{3 t}{2}} sin (\frac{23 t}{2})

y = \frac{2}{23} e^{- \frac{3 t}{2}} sin (\frac{23 t}{2})

y^{^{''}} - 14 y^{^{'}} + 49 y = 0

y^{^{'''}} - y^{^{''}} - 26 y^{^{'}} - 24 y = 0

y^{^{''}} - 2 y^{^{'}} + 20 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y^{^{''}} - 10 y^{^{'}} - 25 y = 0

What is the solution for y^{''}+3y^'+8y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'+8y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(23))e^{-(3t)/2}sin((sqrt(23)t)/2)