What is the solution for 2= 1/10 y+(100*10^{-3})y^' ?

The solution for 2= 1/10 y+(100*10^{-3})y^' is y=e^{-t-c_{1}}+20

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2= 1/10 y+(100*10^{-3})y^'

2 = \frac{1}{10} y + (100 \cdot 1 0^{- 3}) y^{^{'}}

y = e^{- t - c_{1}} + 20

Solution steps

2 = \frac{1}{10} y + (100 \cdot 1 0^{- 3}) y^{'} (t)

Solve separable ODE:

y = e^{- t - c_{1}} + 20

y = e^{- t - c_{1}} + 20

\frac{d y}{d x} = x^{3} y^{2} - x y^{2}

\frac{d y}{d t} = 2 t y^{2} + 3 t^{2} y^{2}

\frac{d}{d t} (v (t) e^{k t}) = g e^{k t}

d x = x^{7} y d y

\frac{d y}{d x} = (\frac{( ln ( x ^{6} ) )}{x y ^{2}}), y (1) = 5

What is the solution for 2= 1/10 y+(100*10^{-3})y^' ?
The solution for 2= 1/10 y+(100*10^{-3})y^' is y=e^{-t-c_{1}}+20