What is the solution for y^{''}+16y^'=0,y(0)=2,y^'(0)=-2 ?

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

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

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

y = 2 - \frac{1}{8} + \frac{1}{8} e^{- 16 t}

Solution steps

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

Solve linear ODE:

y = 2 - \frac{1}{8} + \frac{1}{8} e^{- 16 t}

y = 2 - \frac{1}{8} + \frac{1}{8} e^{- 16 t}

y^{^{'''}} + 7 y^{^{''}} - 17 y^{^{'}} + 9 y = 0

y^{^{''}} + y = 0, y (0) = 2, y^{^{'}} (0) = 1

x^{^{''}} + 3 x^{^{'}} + 2 x = 0, x^{^{'}} (0) = 2

y^{^{''}} + 3 y + 2 y = 0

y^{^{''}} + 20 y^{^{'}} + 10 y = 0

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