Mathematical Physics Vol 1

Chapter 7. Partial differential equations

388

Given that this is true for an arbitrary constant c ( c > 0), we shall assume that c = 1, i.e. v = x − 2 y − 3 . Thus, the expression x − 2 y − 3 2 x 3 y − y 2 d x − x − 2 y − 3 2 x 4 + xy d y = 0 represents the total differential of a function u , i.e. d u = x − 2 y − 3 2 x 3 y − y 2 d x − x − 2 y − 3 2 x 4 + xy d y = 0 . (7.301) From here we obtain

y Z y 0

x Z x 0

2 xy − 2 − x − 2 y − 1 d x −

2 x 2 y − 3 + x − 1 y − 2 d y =

u =

= x 2 y − 2 + y − 1 x − 1 x = 2 x 2 y − 2 + 2 x − y − 1 + c .

x 0 − −

x 2 y − 2 − y − 1 x − 1 y y 0

=

Check.

∂ u ∂ x

∂ u ∂ y d y = 4 xy − 2 − 2 x − 2 y − 1 d x + − 4 x 2 y − 3 − 2 x − 1 y − 2 d y ,

d u =

d x +

and, according to (7.301) this is equal to zero, i.e. d u = 0.

Problem 253 Determine the complete solution and singular solution (if it exists) for equation

z = xp + yq + pq .

Solution This is a non-linear partial differential equation. We shall solve it using the Lagrange– Charpit method. We shall us the conditions (7.80). In our case, given that f = xp + yq + pq − z = 0, it follows ∂ f ∂ p = x + q ∂ f ∂ q = y + p ∂ f ∂ x = p ∂ f ∂ y = q ∂ f ∂ z = − 1

and the equation (7.80) becomes d x x + q = d y y + p =

d z xp + yq + pq

d p − p + p

d q − q + q

=

=

that is

d x x + q

d y y + p

d z z

d p 0

d q 0

=

=

=

=