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

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

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

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

y = e^{t - 2} (e - \frac{e ^{3}}{e ^{2} - 1}) + \frac{e ^{3}}{e ^{2} - 1} e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{t - 2} (e - \frac{e ^{3}}{e ^{2} - 1}) + \frac{e ^{3}}{e ^{2} - 1} e^{- t}

y = e^{t - 2} (e - \frac{e ^{3}}{e ^{2} - 1}) + \frac{e ^{3}}{e ^{2} - 1} e^{- t}

y^{^{''}} + 4 y^{^{'}} + y 5 = 0

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = - 3, y^{^{'}} (0) = - 4

y^{^{''}} + 3 y = 0, y^{0} = 1, y^{^{'}} 0 = 3

\frac{1}{2} y^{^{''}} + 6 y^{^{'}} + 26 y = 0

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