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}}

x \to - 3 lim (\frac{1}{( x + 3 ) ^{2}})

y^{^{'}} - e^{x} e^{y} = - 1

\frac{\partial}{\partial x} (e^{1 + x^{2} - y^{2}})

\frac{d}{d x} (6 cos (2 x) - e^{- x})

f (x) = 2 x^{3} + 7 x^{2} - 8 x - 3, at x = - 1

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)}