Page 248 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 16: Classifications of Integrals of the First Order Partial Differential Equations

To find differentiate (8) with respect to , i.e. Notes
2
2
0 = (x + y ) + 2
(x 2 y 2 )
or = ...(9)
2
Eliminating from (8) we have
2 2
2
4Z = (x + y ) ...(10)
Self Assessment

5. Eliminate the arbitrary function from the equation

y 2 2 2
,(x y z )/z 0
2

16.4 Singular Integrals

The complete integral of a partial differential equation represents a family of surfaces. If these
surfaces have an envelope, its equation is called a singular integral. To see that this is really an
integral we have merely to notice that at any point of the envelope there is a surface of the
family touching it. Therefore the normals to the envelope and this surface coincide, so the values
of p and q at any point of the envelope are the same as that of some surface of the family and
therefore it satisfies the same equation.
The working rule for finding out the singular integral is to start with the complete integral of
the form

f(x, y, z, p, q, a, b) = 0 ...(1)
Differentiate (1) with respect to a and b i.e.
f
= 0 ...(2)
a
f
= 0 ...(3)
b
and eliminate a, b, from (1), (2) and (3) to get the envelope.
or by eliminating p and q from the differential equation.
F(x, y, z, p, q) = 0 ...(4)
And two derived equations
F
= 0 ...(5)
p
F
= 0 ...(6)
q

One should test whether the singular integral obtained really satisfies the differential equation.

Example: Verify that

Z = ax + by + a b ab ...(7)

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