Revisiting Clairault's Equation

 The technique for solving differential equations which do not contain x or y explicitly was first developed by Alexis Clairaut (1713- 1765) and is called Clairaut's equation.  Such equations are of the form:

y = px + f(p)

If we differentiate with respect to x we get:

p  =  p + [x +  df/dp] p'

Or:   [x +  df/dp] p'  =  0

Then either:

i) p' = dp/dx = 0

or  (letting df/dp = f'(p):

ii) x + f'(p) = 0  

The solution of eqn. (i) is:  p = c (a constant)

After substituting this into the original equation:

y = cx +  f(c) 

 which is the general solution to Clairaut's equation.  In effect to obtain the general solution to Clairaut's equation it is only necessary to replace p by the arbitrary constant c.

Consider eqn. (ii) above. If we solve this for x we obtain:

 (iii) x = - f'(p)

Using this in tandem with Clairaut's equation we get:

(iv) y = f(p) - p f'(p)

The equations (iii) and (iv) then yield parametric equations for x and y in terms of the parameter p since f(p) and f'(p) are known.  (Note: If f(p) is not a linear function of p and not a constant then equations (iii) and (iv) form a solution of Clairaut's equation which can't be obtained from y = cx +  f(c)  and is therefore a singular solution.)

Example Problem:

Solve:   y = px + 2p² 

^{Solution:  We recognize the form of a Clairaut equation so we can write a general solution directly:}

^{y = cx + 2c2}

^{So the singular solution is found in parametric form:}

^{f '(p)= df/ dp = 4p} 

^{And by eqn. (iii):}

^{x = - 4p}  

^Hence:  

y = px + 2p²   = -4p²  +  2p²  =  -  2p²

We can then eliminate the parameter p whereby:

y= -2p²    =  -2 (-x/4)²     

So:   y =  - x² / 8

Suggested Problems:

1)Solve:  p² x - y  = 0

2) Solve: p² + 2y  - 2x  = 0