23.7 The Case Where M,N Are Affine Linear
Something which often occurs is an equation of the form
It doesn’t quite fit anything in the earlier discussion. It won’t be exact, homogeneous, or separable or
linear. However, one can massage it to get something which is homogeneous. This is illustrated in some
Example 23.7.1 Find the integral curves for
Of course the problem is those constants 3,1 so it is reasonable to change variables. Let
u = x−a,v = y −b where we choose a,b in an auspicious manner to get the constants to disappear. First,
dx = du,dy = dv. Then in terms of the new variables,
and we want
Hence we should let a = −1 and b = −1. Then with this, the equations reduce to
This is now a homogeneous equation, or we could use the integrating factor described earlier, but, in this
case, it is also an exact equation. A scalar potential is
and so the integral curves for the original equation would be
The example illustrates what to do in general. You just change the variables to remove those
constant terms and then obtain a homogeneous equation which can be solved by a variety of
Example 23.7.2 Find the integral curves for
As before, let u = x − a,v = y − b and write in terms of these new variables. Thus
Then you need
Thus a = −2,b = 0. The new equation then is
This can be considered as a homogeneous equation.
Then let z =
and do the usual substitution. This yields
and so, separating the variables,
Then after much work one obtains integral curves of the form
Then you plug in what u,v are in terms of x,y.
Actually, it was real easy to do this. The computer algebra system did it for me. Here is one which is
not so ugly.
Example 23.7.3 Find the integral curve which contains the given ordered pair.
The equation is
Now let x = u + a,y = v + b. Then we choose a,b such that in terms of the new variables the equation
becomes homogeneous. Thus we need
Thus we let a = 1,b = 2. Then the equation is
This is a homogeneous equation. Let z =
Separating the variables,
This is easily solved,
The in terms of the original variables,
Then to contain the ordered pair, you need
Procedure 23.7.4 To solve affine linear equations of the form
do the following:
- Change the variables u = x − a, v = y − b, plug in and choose a,b to make the resulting
- Solve the resulting homogeneous equation. Then substitute back in x−a for u and y −b for v.
Pick the constant to satisfy initial conditions.