Solving Differential Equations Using Clairaut's Equation

 Consider the equation:

  p²   +2xp - 2y = 0

How would you solve it for y, i.e. obtain general solutions for y but with no p in the answer?  Note that it cannot be solved in direct algebraic fashion.  For one thing there are 3 variables (x, y, p) but only one equation.  The only algebraic step is the first in which we write:

y  =   ½ p²   + xp 

Then, we need to differentiate with respect to x:

dy/dx = p =   p  dp/dx  + x dp/dx +   p   

But note: 

p  dp/dx  + x dp/dx  = (p + x) dp/dx = 0 

 It follows from this that either:

dp/dx = 0   or  p + x = 0

Hence:   p = c    or p =  -x  

Eliminate p between the last equations and the soln. for y yields:

y =  ½ c²   + xc

y =   ½ x²    -    x²  =  -½ x²  

Both of which satisfy the original DE.  Note, however, that the latter solution contains no arbitrary constant and therefore is not the general solution.  It is, however, a singular solution.

Fortunately, a method exists to solve differential equations which do not contain x or y explicitly.  These are of the form:

y = px + f(p)

The technique was first developed by Alexis Clairaut (1713- 1765) and is called Clairaut's equation.  For the foregoing equation 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.

Now consider eqn. (ii). If we solve this for x we obtain:

 (iii) x = - f'(p)

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

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

The equations (iii) and (iv) 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 

 ^{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     =  - x² / 8

Suggested Problems for Math Mavens:

1) Solve:  p² x - y  = 0

2) Solve: p² + 2y  - 2x  = 0