What is the solution for y^'+7y=t+e^{-2t} ?

The solution for y^'+7y=t+e^{-2t} is y=(7t-1)/(49)+1/(5e^{2t)}+(c_{1})/(e^{7t)}

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y^'+7y=t+e^{-2t}

y^{^{'}} + 7 y = t + e^{- 2 t}

y = \frac{7 t - 1}{49} + \frac{1}{5 e ^{2 t}} + \frac{c _{1}}{e ^{7 t}}

Solution steps

y^{'} (t) + 7 y = t + e^{- 2 t}

Solve linear ODE:

y = \frac{7 t - 1}{49} + \frac{1}{5 e ^{2 t}} + \frac{c _{1}}{e ^{7 t}}

y = \frac{7 t - 1}{49} + \frac{1}{5 e ^{2 t}} + \frac{c _{1}}{e ^{7 t}}

x \to 0 + lim (x^{3} + 3 x + 2)

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

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

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

\frac{d}{d x} (x^{3} (ln (x))^{c o s (x)})

What is the solution for y^'+7y=t+e^{-2t} ?
The solution for y^'+7y=t+e^{-2t} is y=(7t-1)/(49)+1/(5e^{2t)}+(c_{1})/(e^{7t)}