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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 26 x^{^{'}} + 25 x = 0, x (0) = - 1, x^{^{'}} (0) = 2

D^{2} (D + 4)^{3} y = 0

y^{^{'''}} - 8 y^{^{''}} + 15 y^{^{'}} = 0

y^{^{'''}} + 15 y^{^{''}} + 58 y^{^{'}} + 44 y = 0

y^{^{''}} - 6 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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