Page 289 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Substituting in (6) we have
2 2
2
p z + a , q 2 = 1
1
p 2 =
z 2 a 2
1
1/2
or p = z 2 a 2 ... (8)
1
Substituting in
dz = p dx + q dy

dx a dy
1
dz = 1/2 1/2
z 2 a 2 1 z 2 a 2 1
we have

z 2 a 1 2 1/2 dz = dx + a dy
1

so z 2 a 1 2 1/2 dz = x + a y + a 2 ... (9)
1
It can be shown that

z 2 a 2 1 1/2 dz z z 2 a 2 1 1/2 a 2 1 log z z 1 2 a 2 1 ... (10)
2 2 a 1

So the solution is (9) with integral (10).
(c) Separable equation
Let the equation be of the form

f (x, p) = g (y, q) ... (11)
instead of
F (x, y, z, p, q) = 0 ... (12)
Then from the subsidiary equations, we have

dp dq dx dy dz
f x g y f p g q (pf p qg q )

dp f x
So dx f = 0
p
or fp dp fx dx = 0 ... (13)

which can be solved for p. Similarly we can solve for q and the complete integral is obtained.

Example 4: Solve
2
2
p y (1 + x ) = q x 2 ... (14)

282 LOVELY PROFESSIONAL UNIVERSITY