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

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

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

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

y = c_{2} e^{t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = c_{2} e^{t} + c_{2} e^{- t}

y = c_{2} e^{t} + c_{2} e^{- t}

y = \frac{d ^{2} y}{d x ^{2}}

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 8 y = 0

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

y^{^{''}} - 16 y^{^{'}} + 113 y = 0

y^{^{'''}} - 9 y^{^{''}} + 8 y^{^{'}} = 0

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