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
No comments:
Post a Comment