What is the solution for y^{''}+8y^'=0,y(0)=3,y^'(0)=16 ?

The solution for y^{''}+8y^'=0,y(0)=3,y^'(0)=16 is y=-2e^{-8t}+5

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y^{''}+8y^'=0,y(0)=3,y^'(0)=16

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 16

y = - 2 e^{- 8 t} + 5

Solution steps

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

Solve linear ODE:

y = - 2 e^{- 8 t} + 5

y = - 2 e^{- 8 t} + 5

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{2}

\frac{1}{2} y^{^{''}} - \frac{1}{3} y^{^{'}} - 2 y = 0

y^{^{''}} + y^{^{'}} + 9 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

What is the solution for y^{''}+8y^'=0,y(0)=3,y^'(0)=16 ?
The solution for y^{''}+8y^'=0,y(0)=3,y^'(0)=16 is y=-2e^{-8t}+5