What is the solution for y^'+3y=t+e^{-2t} ?

The solution for y^'+3y=t+e^{-2t} is y=(3t-1)/9+1/(e^{2t)}+(c_{1})/(e^{3t)}

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y^'+3y=t+e^{-2t}

y^{^{'}} + 3 y = t + e^{- 2 t}

y = \frac{3 t - 1}{9} + \frac{1}{e ^{2 t}} + \frac{c _{1}}{e ^{3 t}}

Solution steps

y^{'} (t) + 3 y = t + e^{- 2 t}

Solve linear ODE:

y = \frac{3 t - 1}{9} + \frac{1}{e ^{2 t}} + \frac{c _{1}}{e ^{3 t}}

y = \frac{3 t - 1}{9} + \frac{1}{e ^{2 t}} + \frac{c _{1}}{e ^{3 t}}

\int \frac{x ^{2} arcsin ( x )}{1 - x ^{2}} d x

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

\int_{0}^{4} 16 - x^{2} d x

\frac{1}{x ^{19}}

\frac{t + 2 t ^{2}}{1 + t ^{2}}

What is the solution for y^'+3y=t+e^{-2t} ?
The solution for y^'+3y=t+e^{-2t} is y=(3t-1)/9+1/(e^{2t)}+(c_{1})/(e^{3t)}