(Partial) Solution to Linear Fractional Differential Equations (2)

 Find the general solution:

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

Recall:   dy/dx  =  M(x,y)/ N(x,y)

Or:  M(x,y) dx = N(x,y) dy

M(x,y) =  x - y - 3 =  a + bx + cy  

N(x,y) = (3x - 3y + 1) = =   a + bx + gy 

Then: ( a + bx + cy) dx =  (a + bx + gy)   dy

a = -3, b = 1, c = -1

a =  1,  b=  3,  g = -3

We demand (for Case I): bg    -  c b =   0

Check: (1)(-3) - (-1)(3) = -3 - (-3) = -3 + 3 = 0

Let bx + cy = v then:

b dx + c dy = dv

Or:

dy/dx = - b/c +  1/ c (dv/dx)

Substituting values:

dy/dx = - (1)/(-1) + (-1) (dv/dx) = 1 - dv/dx

1 - 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

Or:   1  - dv/dx =  -3 + v/ 1 + (b/b )v

Or:

dv/dx -  1   =  (3 + v)/ 1 + 3v

Simplifying:

dx = [(2 + 3v)/ (3 + v)] dv

Replace  (v = x + y) and integrate;

Solution shown from Mathcad 14:

And including integration const.

3x + 3y - 7 ln(x + y + 3) = c