What is the solution for y^'+6y=t+e^{-2t} ?

The solution for y^'+6y=t+e^{-2t} is y=(6t-1)/(36)+1/(4e^{2t)}+(c_{1})/(e^{6t)}

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y^'+6y=t+e^{-2t}

y^{^{'}} + 6 y = t + e^{- 2 t}

y = \frac{6 t - 1}{36} + \frac{1}{4 e ^{2 t}} + \frac{c _{1}}{e ^{6 t}}

Solution steps

y^{'} (t) + 6 y = t + e^{- 2 t}

Solve linear ODE:

y = \frac{6 t - 1}{36} + \frac{1}{4 e ^{2 t}} + \frac{c _{1}}{e ^{6 t}}

y = \frac{6 t - 1}{36} + \frac{1}{4 e ^{2 t}} + \frac{c _{1}}{e ^{6 t}}

\frac{d}{d x} (\frac{5 x}{1 + 2 x ^{2}})

x \to 0 + lim (\frac{sin ^{2} ( x )}{x})

x^{2} + x y - y^{2} = - 1, (1, 2)

(x - 7) (2 x + 4)

x \to 1 lim (\frac{x}{x ^{2} + 4})

What is the solution for y^'+6y=t+e^{-2t} ?
The solution for y^'+6y=t+e^{-2t} is y=(6t-1)/(36)+1/(4e^{2t)}+(c_{1})/(e^{6t)}