What is the solution for y^{''}+8y^'+16=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+8y^'+16=0,y(0)=1,y^'(0)=0 is y=1+1/4-1/4 e^{-8t}-2t

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y^{''}+8y^'+16=0,y(0)=1,y^'(0)=0

y^{^{''}} + 8 y^{^{'}} + 16 = 0, y (0) = 1, y^{^{'}} (0) = 0

y = 1 + \frac{1}{4} - \frac{1}{4} e^{- 8 t} - 2 t

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 = 0

Solve linear ODE:

y = 1 + \frac{1}{4} - \frac{1}{4} e^{- 8 t} - 2 t

y = 1 + \frac{1}{4} - \frac{1}{4} e^{- 8 t} - 2 t

y^{^{''}} - 2 y^{^{'}} = 6 x + 6 - 2 e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 15 sin (x) - 35 cos (x)

y^{^{''}} + 9 y = cos (2 t), y (\frac{π}{2}) = - 1, y (0) = 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 x + x^{2}

y^{^{''}} - a^{2} \cdot y + b = 0

What is the solution for y^{''}+8y^'+16=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+8y^'+16=0,y(0)=1,y^'(0)=0 is y=1+1/4-1/4 e^{-8t}-2t