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

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

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y^{''}-y^'=2e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} - y^{^{'}} = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = e^{t} - 2 + e^{- t}

Solution steps

y^{''} (t) - y^{'} (t) = 2 e^{- t}

Solve linear ODE:

y = e^{t} - 2 + e^{- t}

y = e^{t} - 2 + e^{- t}

y^{^{''}} + 9 y = 12 cos (3 t)

y^{^{''}} + 3 y = e^{2 x}

y^{^{'''}} + 3 y^{^{'}} = sin (x) + 1

y^{^{''}} - 4 y = 12 t^{2} cos^{2} (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 2 x} + x

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