What is the solution for y^'+7y=t+e^{-4t} ?

The solution for y^'+7y=t+e^{-4t} is y=(7t-1)/(49)+1/(3e^{4t)}+(c_{1})/(e^{7t)}

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y^'+7y=t+e^{-4t}

y^{^{'}} + 7 y = t + e^{- 4 t}

y = \frac{7 t - 1}{49} + \frac{1}{3 e ^{4 t}} + \frac{c _{1}}{e ^{7 t}}

Solution steps

y^{'} (t) + 7 y = t + e^{- 4 t}

Solve linear ODE:

y = \frac{7 t - 1}{49} + \frac{1}{3 e ^{4 t}} + \frac{c _{1}}{e ^{7 t}}

y = \frac{7 t - 1}{49} + \frac{1}{3 e ^{4 t}} + \frac{c _{1}}{e ^{7 t}}

\frac{d}{d x} (4 (3 - 9 x)^{3})

\frac{2}{3 s ^{2} + 6 s + 2}

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

y^{^{'}} = - 5 y + 5 t^{2} + 2 t

\int_{0}^{1} (x - x^{3}) d x

What is the solution for y^'+7y=t+e^{-4t} ?
The solution for y^'+7y=t+e^{-4t} is y=(7t-1)/(49)+1/(3e^{4t)}+(c_{1})/(e^{7t)}