What is a solution to the differential equation #dy/dx = 1 - 0.2y#?

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Answer 1 · 2016-07-02T22:12:31

#y = alpha e^ (- 1/5 x) + 5#

#dy/dx = 1 - 0.2y#

# = 1/5 (5 - y)#

#int 1/(5 - y) dy = 1/5 int dx #

#int 1/(y-5) dy = -1/5 int dx #

#ln(y-5) = - 1/5 x + C#

#y-5 = exp (- 1/5 x + C)#

#= alpha e^ (- 1/5 x)# [where #alpha = e^C#]

#y = alpha e^ (- 1/5 x) + 5#