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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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