What is the solution for y^'=-6.2-1.8y ?

The solution for y^'=-6.2-1.8y is y=(e^{-1.79999…t-c_{1}})/(1.8)-(6.2)/(1.8)

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y^'=-6.2-1.8y

y^{^{'}} = - 6.2 - 1.8 y

y = \frac{e ^{- 1.79999 \dots t - c_{1}}}{1.8} - \frac{6.2}{1.8}

Solution steps

y^{'} (t) = - 6.2 - 1.8 y

Solve separable ODE:

y = \frac{e ^{- 1.79999 \dots t - c_{1}}}{1.8} - \frac{6.2}{1.8}

y = \frac{e ^{- 1.79999 \dots t - c_{1}}}{1.8} - \frac{6.2}{1.8}

y = \frac{18}{x ^{4}}

\frac{d}{d x} (cos (e^{x^{4} c o s (x)}))

x \to 0 lim (\frac{2 ∣ x ∣}{3 x})

y = e^{9 - x}

\frac{\partial}{\partial x} (10 y + x)

What is the solution for y^'=-6.2-1.8y ?
The solution for y^'=-6.2-1.8y is y=(e^{-1.79999…t-c_{1}})/(1.8)-(6.2)/(1.8)