What is the solution for y^'+7y=t+e^{-5t} ?

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

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y^'+7y=t+e^{-5t}

y^{^{'}} + 7 y = t + e^{- 5 t}

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

Solution steps

y^{'} (t) + 7 y = t + e^{- 5 t}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} ((n - 1) (\frac{1}{x} - \frac{1}{y}))

\int \frac{1}{( x + 4 ) ^{2} ( x - 1 )} d x

x \to 0 lim (\frac{e ^{x}}{x ^{3}})

\int (3 x - 1)^{5} d x

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

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