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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

49 y^{^{''}} - 49 y = 0

y^{^{''}} + 16 y = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 4 (\frac{d y}{d x}) - 32 y = 0

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1

x^{^{''}} + 0.01 x^{^{'}} + x = 0

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