(Partial) Solution to Linear Fractional Differential Equations (1)

Find the general solution of:

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

 

Recall:   dy/dx  =  M(x,y)/ N(x,y)

Or:  M(x,y) dx = N(x,y) dy

M(x,y) =  2 - x - y =  a + bx + cy  

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

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

a = 2, b = -1, c = -1

a =  3,  b=  1,  g = 1

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

The two straight lines defined by setting the two equations (for M, N) equal to zero and intersecting at a point (h,k). We translate the origin to this point using:

x = X + h   and: y = Y + k

Then we obtain:  dy/dx = dY/dX

Find the point of intersection (h,k) by solving the simultaneous pair:

2 -   x -    y  = 0

3  + x  + y = 0

 ->

6 – 3x – 3y = 0

6 + 2x + 2y = 0

-x – y = 0   or -x = y  so x = -1

Then: x = h = -1  and: y = k = 1

Then the transformation:

x = X + h =  X - 1 and y = Y + k = Y + 1, yields:

(2 – X – 1 – Y + 1)dX  + (3 + X – 1 + Y + 1)dY = 0

dY/dX = (3 + X + Y)/ (2 – X – Y) 

Or:

(2 – X – Y) dX + (3 + X + Y)dY = 0

The resulting DE is homogeneous, so we let:

Y = vX  and dY = v dX + X dv, to obtain:

(2 – X – vX) dX + (3 + X + vX)(v dX + X dv)

After working through the algebra and then using the reverse transformations:  

X  = x + 1  and Y = y - 1, 

And integrating,  we obtain:

Ans. 2y  -  2x + 5 ln (2x + 2y + 1) = c

(Detailed working left to industrious readers.)