What is the solution for ((dQ)/(dt)+2/300 Q)=0.8 ?

The solution for ((dQ)/(dt)+2/300 Q)=0.8 is Q=-e^{(-t-c_{1})/(150)}+120

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((dQ)/(dt)+2/300 Q)=0.8

(\frac{d Q}{d t} + \frac{2}{300} Q) = 0.8

Q = - e^{\frac{- t - c _{1}}{150}} + 120

Solution steps

(\frac{d Q}{d t} + \frac{2}{300} Q) = 0.8

Solve separable ODE:

Q = - e^{\frac{- t - c _{1}}{150}} + 120

Q = - e^{\frac{- t - c _{1}}{150}} + 120

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(x^{2} + 1) \frac{d y}{d x} + x y = x

cos (3 y) d y + sin (2 x) d x = 0

\frac{d x}{d t} + cos (t) x = cos (t), x (0) = 2

\frac{d r}{d t} = - 4 r t, r = r_{0} (t), t = 0

What is the solution for ((dQ)/(dt)+2/300 Q)=0.8 ?
The solution for ((dQ)/(dt)+2/300 Q)=0.8 is Q=-e^{(-t-c_{1})/(150)}+120