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
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