Consider the equation:

^{ }p^{2 }+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 p^{2 }+ 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 c^{2 }+ xc

y = ½ x^{2 }-^{ } x^{2 }= -½ x^{2}

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 + 2p^{2 }

^{ 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} = - x^{2 }/ 8

__Suggested Problems for Math Mavens:__

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

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

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