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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{3} x^{2} d x

\int \frac{2 x ^{3} + x ^{2} - x}{x ^{2}} d x

\int (e^{x} + x^{e}) d x

f (x) = - arctan (x)

x 5^{2 x}

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