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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} = 0, y (0) = 2 - 2 e^{7}, y (1) = 0

\frac{d ^{2} y}{d x ^{2}} + 15 \frac{d y}{d x} + 56 y = 0

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

y^{^{''}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

y^{^{''}} + 7 y^{^{'}} + 20 y = 0

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