What is the solution for y^'=9.8-(3y)/(3.5) ?

The solution for y^'=9.8-(3y)/(3.5) is y=-(e^{-0.85714…t-c_{1}})/3+(34.3)/3

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y^'=9.8-(3y)/(3.5)

y^{^{'}} = 9.8 - \frac{3 y}{3.5}

y = - \frac{e ^{- 0.85714 \dots t - c_{1}}}{3} + \frac{34.3}{3}

Solution steps

y^{'} (t) = 9.8 - \frac{3 y}{3.5}

Solve separable ODE:

y = - \frac{e ^{- 0.85714 \dots t - c_{1}}}{3} + \frac{34.3}{3}

y = - \frac{e ^{- 0.85714 \dots t - c_{1}}}{3} + \frac{34.3}{3}

x \to \infty lim (\frac{n ^{1.5}}{ln ( n )})

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(6 + x) (\frac{d y}{d x}) = 9 y

\frac{d}{d t} (e^{t} sin (t))

What is the solution for y^'=9.8-(3y)/(3.5) ?
The solution for y^'=9.8-(3y)/(3.5) is y=-(e^{-0.85714…t-c_{1}})/3+(34.3)/3