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

The solution for y^{''}-y^'-2y=((e^t+e^{-t}))/2 is y=c_{1}e^{2t}+c_{2}e^{-t}-1/4 e^t-(e^{-t}t)/6

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} e^{t} - \frac{e ^{- t} t}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} e^{t} - \frac{e ^{- t} t}{6}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} e^{t} - \frac{e ^{- t} t}{6}

What is the solution for y^{''}-y^'-2y=((e^t+e^{-t}))/2 ?
The solution for y^{''}-y^'-2y=((e^t+e^{-t}))/2 is y=c_{1}e^{2t}+c_{2}e^{-t}-1/4 e^t-(e^{-t}t)/6