What is the solution for y^'+10y=t+e^{-5t} ?

The solution for y^'+10y=t+e^{-5t} is y=(10t-1)/(100)+1/(5e^{5t)}+(c_{1})/(e^{10t)}

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y^'+10y=t+e^{-5t}

y^{^{'}} + 10 y = t + e^{- 5 t}

y = \frac{10 t - 1}{100} + \frac{1}{5 e ^{5 t}} + \frac{c _{1}}{e ^{10 t}}

Solution steps

y^{'} (t) + 10 y = t + e^{- 5 t}

Solve linear ODE:

y = \frac{10 t - 1}{100} + \frac{1}{5 e ^{5 t}} + \frac{c _{1}}{e ^{10 t}}

y = \frac{10 t - 1}{100} + \frac{1}{5 e ^{5 t}} + \frac{c _{1}}{e ^{10 t}}

\frac{d}{d x} (x^{2} \cdot cos (x^{2}))

\int (e^{- x} + 4^{x}) d x

G (r) = r + 4 r

\int \frac{1}{x ^{2} - 8 x} d x

\frac{\partial}{\partial x} (2 y cos (x y))

What is the solution for y^'+10y=t+e^{-5t} ?
The solution for y^'+10y=t+e^{-5t} is y=(10t-1)/(100)+1/(5e^{5t)}+(c_{1})/(e^{10t)}