What is the solution for y^'+8y=t+e^{-3t} ?

The solution for y^'+8y=t+e^{-3t} is y=(8t-1)/(64)+1/(5e^{3t)}+(c_{1})/(e^{8t)}

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y^'+8y=t+e^{-3t}

y^{^{'}} + 8 y = t + e^{- 3 t}

y = \frac{8 t - 1}{64} + \frac{1}{5 e ^{3 t}} + \frac{c _{1}}{e ^{8 t}}

Solution steps

y^{'} (t) + 8 y = t + e^{- 3 t}

Solve linear ODE:

y = \frac{8 t - 1}{64} + \frac{1}{5 e ^{3 t}} + \frac{c _{1}}{e ^{8 t}}

y = \frac{8 t - 1}{64} + \frac{1}{5 e ^{3 t}} + \frac{c _{1}}{e ^{8 t}}

\int (8 x^{7} + x^{5}) 6 x^{8} + x^{6} d x

\int \frac{sinh ( x )}{1 - 4 cosh ( x )} d x

x \to 0 lim ((x - 1)^{\frac{1}{2}})

(2 x y^{2} + 3 x^{2} y) d x + (2 x^{2} y + x^{3}) d y = 0

(- 5, 3), (5, - 3)

What is the solution for y^'+8y=t+e^{-3t} ?
The solution for y^'+8y=t+e^{-3t} is y=(8t-1)/(64)+1/(5e^{3t)}+(c_{1})/(e^{8t)}