Question #d88e8

1 Answer
Jan 19, 2017

#y(x)= a-c/abs(x)#

Explanation:

In the equation:

#x(dy)/(dx) + y = a#

we can separate the variables in this way:

#x(dy)/(dx) = a-y#

#(dy)/(a-y) = (dx)/x#

then integrate both sides:

#int (dy)/(a-y) = int (dx)/x#

#-int (d(a-y))/(a-y) = int (dx)/x#

#-lnabs(a-y) = lnabs(x) + C#

#lnabs(a-y) = -lnabs(x) + C#

We can now take the exponential of both sides posing #c=e^C > 0#:

#abs(a-y) = e^(-lnabs(x) + C) = e^C/ e^(lnabs(x)) = c/abs(x)#

If we make the hypothesis #y<a# we have:

#a-y = c/abs(x)#

#y= a-c/abs(x)#

which is in fact always less than #a#