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

The solution for y^{''}+y^'-2=e^{-t} is y=c_{1}+c_{2}e^{-t}-e^{-t}t+2t

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

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

y = c_{1} + c_{2} e^{- t} - e^{- t} t + 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - e^{- t} t + 2 t

y = c_{1} + c_{2} e^{- t} - e^{- t} t + 2 t

\int \frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} d y

y^{^{'}} + (1 + 3 x^{2}) y = 3 + 9 x^{2}

\int \frac{1 + 2 t ^{3}}{t} d t

\frac{\partial}{\partial x} (2 e^{2 x})

f (x) = x (x^{3} + 2 x + 3)

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