What is the solution for 10y^'=10*9.81-5y ?

The solution for 10y^'=10*9.81-5y is y=-0.2e^{(-t-c_{1})/2}+19.62

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10y^'=10*9.81-5y

10 y^{^{'}} = 10 \cdot 9.81 - 5 y

y = - 0.2 e^{\frac{- t - c _{1}}{2}} + 19.62

Solution steps

10 y^{'} (t) = 10 \cdot 9.81 - 5 y

Solve separable ODE:

y = - 0.2 e^{\frac{- t - c _{1}}{2}} + 19.62

y = - 0.2 e^{\frac{- t - c _{1}}{2}} + 19.62

y = \frac{1}{x - 1}, at x = 6

y^{^{''}} - 4 y^{^{'}} + 3 y = 0

\frac{d}{d x} (- 10 sin (2 x + 10))

y^{^{''}} + 25 y = - 20 sin (5 t)

\frac{d}{d x} (3 cos (3 x + 1))

What is the solution for 10y^'=10*9.81-5y ?
The solution for 10y^'=10*9.81-5y is y=-0.2e^{(-t-c_{1})/2}+19.62