What is a solution to the differential equation #y'-y=5#?

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Answer 1 · 2016-08-07T08:58:53

#y= C e^x - 5#

separate it!

#y' - y = 5#

#y' = y + 5#

#1/(y+5) y' = 1#

#int 1/(y+5) y' \ dx=int \dx#

#int d/dx (int 1/(y+5) \dy) \ dx=int \dx#

#int 1/(y+5) \dy=int \dx#

#ln(y+5) =x + C#

#y+5 =e^(x + C) = C e^x#

#y= C e^x - 5#