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

The solution for y^{''}-y=0,y(0)=2,y^'(0)=-1 is y=(2-3/2)e^t+3/2 e^{-t}

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

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

y = (2 - \frac{3}{2}) e^{t} + \frac{3}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{3}{2}) e^{t} + \frac{3}{2} e^{- t}

y = (2 - \frac{3}{2}) e^{t} + \frac{3}{2} e^{- t}

y^{^{''}} + y = 0, y (\frac{π}{2}) = 0, y (0) = 1

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} = 0

64 y^{^{''}} + 144 y^{^{'}} + 81 y = 0

y^{^{'''}} + 3 y^{^{''}} - 22 y^{^{'}} - 24 y = 0

2 y^{^{''''}} + 3 y^{^{'''}} + 2 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

What is the solution for y^{''}-y=0,y(0)=2,y^'(0)=-1 ?
The solution for y^{''}-y=0,y(0)=2,y^'(0)=-1 is y=(2-3/2)e^t+3/2 e^{-t}