Introduction to Linear Fractional Differential Equations (Pt.1)

 Back to differential equations!

The linear fractional version of differential equations is usually encountered later in the typical DE course. Starting with the basics, a first order, first degree differential equation may be written - as we've seen in earlier posts - as:

dy/dx  =  M(x,y)/ N(x,y)

And if M(x,y) and N(x,y) are linear functions of x and y the equation is called a linear fractional equation. The general form of M and N are then given by:

M(x,y) = a + bx + cy   and: N(x,y) =   a + bx + gy

For which it is assumed that: b  and b  (c  and g )  are not both zero at the same time. If they were the equation would be one in which the variables are immediately separable.  Two cases can be considered:

Case I: bg    -  c b  ≠   0

In this case the two straight lines defined by setting the two equations (for M, N) equal to zero are not parallel and will intersect at a point (h,k). If we translate the origin to this point using:

x = X + h   and: y = Y + k

Then we obtain:  dy/dx = dY/dX

And the initial differential equation reduces to: 

dY/dX = [a + b(X + h) + c(Y + k)]/ [a + b(X  + h)  + g (Y + k)]

= (a + bh + ck + bX + cY)/ (a + bh  +  gk +  b X  +  g Y) = 

bX  + cY/ bX  + g Y

By virtue of:

a + bh + ck  = 0   and: a + bh  +  gk  = 0

Since (h,k) is the point of intersection of the 2 lines M(x,y) = 0 and N(x,y) = 0, and must therefore satisfy the above pair of equations.

Example Problem:  Find the general solution of:

(x - 2y + 1)dx +  (2x - y - 1) dy = 0

Solution:

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

x - 2y + 1 = 0

2x -y - 1  = 0

---------------

3x - 3y = 0

3x = 3y,   so x = 1, y = 1

From which: x = h = 1  and y = k = 1

Then the transformation:  x = X + 1 and y = Y + 1, yields:

(X - 2Y) dX  +  (2X - Y) dY = 0

This equation is obtained from the given differential equation by simply replacing a (= 1)  and  a (= -1) by zero and changing the unknowns to capital letters.  The resulting DE is homogeneous, so let:

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

dX/X +  (2 - v) dv/ 1 - v²  =  0

Integration yields:

ln X + ln (1 + v)/ (1 - v) +  ½  ln (1 - v² ) = ln c   

Using:  X  = x - 1  and Y = y - 1, we obtain:

(x + y - 2)³  = c (x - y)

Suggested Problem:

Find the general solution of:

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