What is the solution for linear y^{''}-y=t,y(0)=0,y^'(0)=1 ?

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

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linear y^{''}-y=t,y(0)=0,y^'(0)=1

linear

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 4 sin (x))

y^{^{''}} - y^{^{'}} = 5 e^{2 x}

y^{^{''}} - y - 2 y = 1 + t

y^{^{''}} + 5 y = g (t), y (0) = - 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 e^{2 x}

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