What is the solution for 8y^{''}+4y^'+y=0,y(0)=0,y^'(0)=1 ?

The solution for 8y^{''}+4y^'+y=0,y(0)=0,y^'(0)=1 is y=4e^{-t/4}sin(t/4)

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8y^{''}+4y^'+y=0,y(0)=0,y^'(0)=1

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

y = 4 e^{- \frac{t}{4}} sin (\frac{t}{4})

Solution steps

8 y^{''} (t) + 4 y^{'} (t) + y = 0

Solve linear ODE:

y = 4 e^{- \frac{t}{4}} sin (\frac{t}{4})

y = 4 e^{- \frac{t}{4}} sin (\frac{t}{4})

What is the solution for 8y^{''}+4y^'+y=0,y(0)=0,y^'(0)=1 ?
The solution for 8y^{''}+4y^'+y=0,y(0)=0,y^'(0)=1 is y=4e^{-t/4}sin(t/4)