Introduction to Linear Fractional Differential Equations (Pt.2)

Case II: bg    -  c b   =   0  

Since c and g cannot both be zero, let us assume c  ≠  0.  If we then let bx + cy = v then:

b dx + c dy = dv

Or:

dy/dx = - b/c +  1/ c (dv/dx)

By substituting this into our original DE template, we get:

 - b/c +  1/ c (dv/dx) =  (a  + v)/ (a +   bx +  g y)

And since: bg    -  c b  = 0   we have:

g / c =   b/b = k   or:   g = kc and    b = kb

So that:

 b  x   +   g y   = kbx +  kcy = k (bx + cy) = kv

And our starting DE becomes:

- b/c +  1/ c (dv/dx) =  a + v/ a  +  kv

We see the variables are now separable and the equation can be solved by simple integration.  If  c = 0 then  g  ≠  0 and if we then make the substitution:

b  x   +   g y   =  v 

The result will again be an equation in which variables are separable.

Sample Problem:

Solve:   (1 + x + y) dx + (3 + 2x + 2y) dy = 0

Solution:

First, note that:  bg    -  c b = 2 - 2 = 0

Now, let x + y = v, then:

dy/dx = dv/dx - 1  =  1 + v/ 3 +  2v  

Simplify to get:  (3 + 2v/ 2 + v) dv = dx

Integrating:  2v  - ln (v + 2) = x + c

Or, in terms of x, y:

x + 2y - ln (x + y + 2) = c

Suggested Problem:

Find the general solution:

(x - y - 3) dx +  (3x - 3y + 1) dy = 0