What is the solution for y^'=4-0.75y ?

The solution for y^'=4-0.75y is y=-(e^{-0.75t-c_{1}})/(0.75)+4/(0.75)

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y^'=4-0.75y

y^{^{'}} = 4 - 0.75 y

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} + \frac{4}{0.75}

Solution steps

y^{'} (t) = 4 - 0.75 y

Solve separable ODE:

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} + \frac{4}{0.75}

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} + \frac{4}{0.75}

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What is the solution for y^'=4-0.75y ?
The solution for y^'=4-0.75y is y=-(e^{-0.75t-c_{1}})/(0.75)+4/(0.75)