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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{'''}} - 5 y^{^{''}} + 7 y^{^{'}} - 3 y = 0

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 3

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