What is the solution for y^'+8y=t+e^{-7t} ?

The solution for y^'+8y=t+e^{-7t} is y=(8t-1)/(64)+1/(e^{7t)}+(c_{1})/(e^{8t)}

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y^'+8y=t+e^{-7t}

y^{^{'}} + 8 y = t + e^{- 7 t}

y = \frac{8 t - 1}{64} + \frac{1}{e ^{7 t}} + \frac{c _{1}}{e ^{8 t}}

Solution steps

y^{'} (t) + 8 y = t + e^{- 7 t}

Solve linear ODE:

y = \frac{8 t - 1}{64} + \frac{1}{e ^{7 t}} + \frac{c _{1}}{e ^{8 t}}

y = \frac{8 t - 1}{64} + \frac{1}{e ^{7 t}} + \frac{c _{1}}{e ^{8 t}}

5 x^{3} - 6 cos (x)

f (x) = \frac{2}{x - 8}, (10, 1)

\int csc^{2} (2 x) d x

\int \frac{8}{x ^{2}} d x

e^{2 x - 3}, 1

What is the solution for y^'+8y=t+e^{-7t} ?
The solution for y^'+8y=t+e^{-7t} is y=(8t-1)/(64)+1/(e^{7t)}+(c_{1})/(e^{8t)}