Solving Math Problems (Part 14)

In doing the problems in section 1.4, I had noticed that a few of the problems seemed to fit a larger classification.  See, for example, problems 1 and 2  In this post, I am going to solve a generalized equation of one sort:

xy' = ax + by

Because there is a point where we divide by b-1, we have to consider two situations.  The simpler is where b = 1.  In this case, we divide everything by x and then use a change of variable where u = y/x and y' = u + u'x.  From this we get:

u + u'x = a + u

u' = a/x

Integrating we get:

u = a ln |x| + C

And converting back to y from u:

y = x (a ln |x| + C )

In the case where b ≠ 1, and using the same change of variable as above, we get:

u + u'x = a + bu

u'x = a + (b-1)u

Integrating both sides, and to be explicit, we will change the variable again by z = a + (b-1)u and dz = (b-1)du to get:

Taking e to the power of both sides, we get:

Taking both sides to the power of (b-1) so as to get z to the first power and then changing back to u, we get:

Rearranged to isolate u, we get:

And converting back to y, we get:

Where c* is defined as:

Summary

Where the differential equation takes the form of

xy' = ax + by

The solution is