What is a general solution to the differential equation #dy/dx=10-2y#?

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Answer 1 · 2016-07-19T19:48:11

#y = (e^(-2(x+C)) + 10)/(2)#

OR

#y = C e^(-2x) +5#

This differential equation is separable.

#dy/(dx) = 10-2y#

#dy = (10-2y) dx#

# 1/(10-2y) dy = dx#

#int dy/(10-2y) = int dx#

Let #u = 10-2y -> du = -2 dy -> -1/2 du = dy#

#-1/2 int (du)/(u) = int x dx#

#-1/2 ln(10-2y) = x + C#

#ln(10-2y) = -2(x+C)#

#10-2y = e^(-2(x+C))#

#-2y+10 = e^(-2(x+C))#

#-2y = e^(-2(x+C)) - 10#

#color(red)(-2y = e^(-2x)e^(-2C) - 10)#

#color(red)(-2y = C e^(-2x) - 10)#

#color(red)(y = C/(-2) e^(-2x) - 10/(-2))#

#color(red)(y = C e^(-2x) +5)#

#y = - (e^(-2(x+C)) -10)/(2)#

#y = (e^(-2(x+C)) + 10)/(2)#