What is the solution for y^'+5y=t+e^{-2t} ?

The solution for y^'+5y=t+e^{-2t} is y=(5t-1)/(25)+1/(3e^{2t)}+(c_{1})/(e^{5t)}

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y^'+5y=t+e^{-2t}

y^{^{'}} + 5 y = t + e^{- 2 t}

y = \frac{5 t - 1}{25} + \frac{1}{3 e ^{2 t}} + \frac{c _{1}}{e ^{5 t}}

Solution steps

y^{'} (t) + 5 y = t + e^{- 2 t}

Solve linear ODE:

y = \frac{5 t - 1}{25} + \frac{1}{3 e ^{2 t}} + \frac{c _{1}}{e ^{5 t}}

y = \frac{5 t - 1}{25} + \frac{1}{3 e ^{2 t}} + \frac{c _{1}}{e ^{5 t}}

y = 7 x - x^{2}, y = 2 x

\int_{1}^{4} (3 x + 5) d x

\int (3 sin (x))^{2} d x

\int \frac{1}{3 x - 2} d x

\frac{d}{d x} (12 cos (3 x))

What is the solution for y^'+5y=t+e^{-2t} ?
The solution for y^'+5y=t+e^{-2t} is y=(5t-1)/(25)+1/(3e^{2t)}+(c_{1})/(e^{5t)}