What is the solution for y^'+0.003y-0.105=0 ?

The solution for y^'+0.003y-0.105=0 is y=(e^{-0.003t-c_{1}})/(0.003)+(0.105)/(0.003)

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y^'+0.003y-0.105=0

y^{^{'}} + 0.003 y - 0.105 = 0

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

Solution steps

y^{'} (t) + 0.003 y - 0.105 = 0

Solve separable ODE:

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

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What is the solution for y^'+0.003y-0.105=0 ?
The solution for y^'+0.003y-0.105=0 is y=(e^{-0.003t-c_{1}})/(0.003)+(0.105)/(0.003)