Mathematical Physics Vol 1

Chapter 7. Partial differential equations

386

Problem 251 Solve the equation

∂ z ∂ x

∂ z ∂ y

1 + √ z

− x − y

= 2 .

+

Solution The corresponding system of ordinary equations in this case is d x ( 1 + √ z − x − y ) = d y 1 = d z 2 . From the properties of proportionality, it follows that − d x − ( 1 + √ z − x − y ) = − d y − 1 = d z 2 = d z − d x − d y 2 − 1 − 1 − √ z − x − y from where we obtain − d x − ( 1 + √ z − x − y ) = − d y − 1 = d z 2 = d ( z − x − y ) − √ z − x − y . The first integrals are obtained from d y 1 = d z 2 ⇒ 2 y = z + c o i d y 1 = d ( z − x − y ) − √ z − x − y ⇒ y = − 2 √ z − x − y + c 1 , that is y + 2 √ z − x − y = c 1 , and thus the general solution is f 2 y − z , y + 2 √ z − x − y = 0 . f is an arbitrary differentiable function of corresponding arguments.

∂ z ∂ x

∂ z ∂ y

Note 15 that a solution is also the function z = x + y . Namely, given that = 1, it is obvious that this function z also satisfies the initial partial equation. This solution is the so-called singular solution . = 1and

R

Problem 252 Determine the integration factor v = v ( x , y ) so that the expression 2 x 3 y − y 2 d x − 2 x 4 + xy d y = 0 15 Prof. Arpad Takacˇi pointed out to this solution.