What is the solution for y^'+0.15y=1 ?

The solution for y^'+0.15y=1 is y=-(e^{-0.15t-c_{1}})/(0.15)+1/(0.15)

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y^'+0.15y=1

y^{^{'}} + 0.15 y = 1

y = - \frac{e ^{- 0.15 t - c_{1}}}{0.15} + \frac{1}{0.15}

Solution steps

y^{'} (t) + 0.15 y = 1

Solve separable ODE:

y = - \frac{e ^{- 0.15 t - c_{1}}}{0.15} + \frac{1}{0.15}

y = - \frac{e ^{- 0.15 t - c_{1}}}{0.15} + \frac{1}{0.15}

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

\frac{d y}{d x} = \frac{9 x}{5 y}

\int \frac{1}{2 x ^{2} + 8} d x

x \to 0 - lim (\frac{1}{x} + \frac{1}{∣ x ∣})

\frac{d y}{d x} = \frac{x}{6 y}

What is the solution for y^'+0.15y=1 ?
The solution for y^'+0.15y=1 is y=-(e^{-0.15t-c_{1}})/(0.15)+1/(0.15)