What is the solution for y^'=b-b*y-a*y ?

The solution for y^'=b-b*y-a*y is y=-(e^{-bt-at-c_{1}b-c_{1}a})/(b+a)+b/(b+a)

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y^'=b-by-ay

y^{^{'}} = b - b \cdot y - a \cdot y

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

Solution steps

y^{'} (t) = b - b \cdot y - a \cdot y

Solve separable ODE:

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

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

\frac{d x}{d y} = \frac{e ^{- 2 y}}{x ^{2} + 4}

\frac{d y}{d x} = \frac{x y}{ln ( y )}, y > 1

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

y^{^{'}} = \frac{cos ( x )}{2 y}, y (0) = - 1

(x^{3} + 5 x^{3} y^{4}) d x + e^{x^{4}} y^{3} d y = 0

What is the solution for y^'=b-b*y-a*y ?
The solution for y^'=b-b*y-a*y is y=-(e^{-bt-at-c_{1}b-c_{1}a})/(b+a)+b/(b+a)