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 + 2p^{2 }

^{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 + 2p^{2 }= -4p^{2 }+ 2p^{2 }= - 2p^{2}

We can then eliminate the parameter p whereby:

y= -2p^{2 }= -2 (-x/4)^{2}

So: y = - x^{2 }/ 8

* Suggested Problems*:

1)Solve: p^{2} x - y = 0

2) Solve: p^{2} + 2y - 2x = 0

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