What is the solution for y^'=ay+be^{-b} ?

The solution for y^'=ay+be^{-b} is y=(e^{ta+c_{1}a})/a-(be^{-b})/a

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y^'=ay+be^{-b}

y^{^{'}} = a y + b e^{- b}

y = \frac{e ^{t a + c_{1} a}}{a} - \frac{b e ^{- b}}{a}

Solution steps

y^{'} (t) = a y (t) + b e^{- b}

Solve separable ODE:

y = \frac{e ^{t a + c_{1} a}}{a} - \frac{b e ^{- b}}{a}

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What is the solution for y^'=ay+be^{-b} ?
The solution for y^'=ay+be^{-b} is y=(e^{ta+c_{1}a})/a-(be^{-b})/a