How do you find all solutions of the differential equation #dy/dx=B+ky#?

We have to first separate the variables, treating #dy/dx# as division and getting all #y# terms on one side and all #x# terms on the other.

First, we see that there are no #x# terms other than #dx#, so this will need to be on a side unto itself.

Further, we can't subtract #ky# to get #dy# and #ky# on the same side: it doesn't deal with the issues of multiplying later by #dx# and #dy# must be attached to all other #y# terms by multiplication.

So, what we can do is divide both sides by #B+ky# entirely. If we also multiply by #dx#, this should give us:

#dy/dx=B+ky#

#dy/(B+ky)=dx#

We then integrate:

#intdy/(B+ky)=intdx#

The left-hand integral can either be solved using the substitution #u=B+ky# or just by inspection. Remember #B# and #k# are constants (we'll also add a constant of integration).

#1/kint((k)dy)/(B+ky)=x+C#

#1/klnabs(B+ky)=x+C#

Multiply both sides by #k#. Even though #C# changes, we'll leave it as #C# since it's a catch-all for any constant.

#lnabs(B+ky)=kx+C#

#abs(B+ky)=e^(kx+C)#

Note that #e^(kx+C)=e^(kx)e^C=Ce^(kx)# where #C# again represents #e^C#, some other constant.

#abs(B+ky)=Ce^(kx)#

The absolute value bars are just a reminder that it's possible that #B+ky<0#.

#ky=Ce^(kx)-B#

#y=(Ce^(kx)-B)/k#