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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 x^{^{''}} + 15 x^{^{'}} + 14 x = 0

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} - 2 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

3 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = - 1

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

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