What is the solution for y^{''}-16y=0,y(0)=1,y^{''}(0)=2 ?

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

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y^{''}-16y=0,y(0)=1,y^{''}(0)=2

y^{^{''}} - 16 y = 0, y (0) = 1, y^{^{''}} (0) = 2

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 4 t}

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 4 t}

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

u^{^{''}} + 29 H (t) = 0

y^{^{''''}} - 7 y^{^{'''}} + 18 y^{^{''}} - 20 y^{^{'}} + 8 y = 0

- y^{^{''}} - 2 \cdot π^{2} y = 0

y^{^{''}} + 64 y^{^{'}} + 16 y = 0

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