What is the solution for y^'+y/a = b/a ,y(0)=1 ?

The solution for y^'+y/a = b/a ,y(0)=1 is y=-be^{-t/a}+e^{-t/a}+b

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y^'+y/a = b/a ,y(0)=1

y^{^{'}} + \frac{y}{a} = \frac{b}{a}, y (0) = 1

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

Solution steps

y^{'} (t) + \frac{y}{a} = \frac{b}{a}

Solve separable ODE:

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

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

\frac{d y}{d x} = - \frac{x}{y}, y (4) = - 3

(1 + x) y^{^{'}} = 4 y

x d y + 1 + y^{2} d x = 0

\frac{d y}{d x} + x y = \frac{x}{y ^{2}}, y (1) = 0.5

\frac{1}{x} \frac{d y}{d x} + 4 y = 2

What is the solution for y^'+y/a = b/a ,y(0)=1 ?
The solution for y^'+y/a = b/a ,y(0)=1 is y=-be^{-t/a}+e^{-t/a}+b