What is the solution for y^'+7y=t+e^{-6t} ?

The solution for y^'+7y=t+e^{-6t} is y=(7t-1)/(49)+1/(e^{6t)}+(c_{1})/(e^{7t)}

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y^'+7y=t+e^{-6t}

y^{^{'}} + 7 y = t + e^{- 6 t}

y = \frac{7 t - 1}{49} + \frac{1}{e ^{6 t}} + \frac{c _{1}}{e ^{7 t}}

Solution steps

y^{'} (t) + 7 y = t + e^{- 6 t}

Solve linear ODE:

y = \frac{7 t - 1}{49} + \frac{1}{e ^{6 t}} + \frac{c _{1}}{e ^{7 t}}

y = \frac{7 t - 1}{49} + \frac{1}{e ^{6 t}} + \frac{c _{1}}{e ^{7 t}}

2 x d x + d y = 0

n = 0 \sum \infty 5 (0.7)^{n - 1}

\int e^{4 x} d x

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

\frac{d x}{d t} = x + t

What is the solution for y^'+7y=t+e^{-6t} ?
The solution for y^'+7y=t+e^{-6t} is y=(7t-1)/(49)+1/(e^{6t)}+(c_{1})/(e^{7t)}