What is the solution for y^'+5y=t+e^{-4t} ?

The solution for y^'+5y=t+e^{-4t} is y=(5t-1)/(25)+1/(e^{4t)}+(c_{1})/(e^{5t)}

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y^'+5y=t+e^{-4t}

y^{^{'}} + 5 y = t + e^{- 4 t}

y = \frac{5 t - 1}{25} + \frac{1}{e ^{4 t}} + \frac{c _{1}}{e ^{5 t}}

Solution steps

y^{'} (t) + 5 y = t + e^{- 4 t}

Solve linear ODE:

y = \frac{5 t - 1}{25} + \frac{1}{e ^{4 t}} + \frac{c _{1}}{e ^{5 t}}

y = \frac{5 t - 1}{25} + \frac{1}{e ^{4 t}} + \frac{c _{1}}{e ^{5 t}}

arctan (5 x)

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What is the solution for y^'+5y=t+e^{-4t} ?
The solution for y^'+5y=t+e^{-4t} is y=(5t-1)/(25)+1/(e^{4t)}+(c_{1})/(e^{5t)}