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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y + 5 = 0

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

y^{^{''}} + y - 5 = e^{2 x} - 3 cos (3 x)

y^{^{'''}} - 18 y^{^{''}} + 864 y = e^{12 x}

(D^{3} - D) y = x

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