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

The solution for y^{''}+y^'-2y=1+t is y=c_{1}e^t+c_{2}e^{-2t}-t/2-3/4

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

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

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t}{2} - \frac{3}{4}

y^{^{'''}} + 25 y^{^{'}} = tan (5 x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (x + 1)^{2} e^{2 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8

y^{^{''}} + y^{^{'}} = tan^{2} (x)

\frac{d ^{2} x}{d t ^{2}} + 81 x = 2 sin (9 t)

What is the solution for y^{''}+y^'-2y=1+t ?
The solution for y^{''}+y^'-2y=1+t is y=c_{1}e^t+c_{2}e^{-2t}-t/2-3/4