MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 33

Lecture 6 Some Special Types of First-Order PDEs

We shall consider some special types of first-order partial differential equations whose

solutions may be obtained easily by Charpit’s method.

Type (a): (Equations involving only p and q)

If the equation is of the form

f(p, q) = 0 (1)

then Charpit’s equations take the form

dx

fp=dy

fq=

dz

pfp + qfq=dp

0=dq

0

An immediate solution is given by p = a, where a is an arbitrary constant. Substituting

p = a in (1), we obtain a relation

q = Q(a).

Then, integrating the expression

dz = adx+Q(a)dy

we obtain

z = ax+Q(a)y + b, (2)

where b is a constant. Thus, (2) is a complete integral of (1).

Note: Instead of taking dp = 0, we can take dq = 0 ⇒ q = a. In some problems, taking

dq = 0 the amount of computation involved may be reduced considerably.

EXAMPLE 1. Find a complete integral of the equation pq = 1.

Solution. If p = a then pq = 1 ⇒ q = 1a . In this case, Q(a) = 1/a. From (2), we

obtain a complete integral as

z = ax+y

a+ b

=⇒ a2x+ y − az = c,

where a and c are arbitrary constants.

Type (b) (Equations not involving the independent variables):

For the equation of the type

f(z, p, q) = 0, (3)

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 34

Charpit’s equation becomes

dx

fp=dy

fq=

dz

pfp + qfq=

dp

−pfz=

dq

−qfz.

From the last two relation, we have

dp

−pfz=

dq

−qfz=⇒ dp

p=dq

q

=⇒ p = aq, (4)

where a is an arbitrary constant. Solving (3) and (4) for p and q, we obtain

q = Q(a, z) =⇒ p = aQ(a, z).

Now

dz = pdx+ qdy

=⇒ dz = aQ(a, z)dx+Q(a, z)dy

=⇒ dz = Q(a, z) [adx+ dy] .

It gives complete integral as ∫dz

Q(a, z)= ax+ y + b, (5)

where b is an arbitrary constant.

EXAMPLE 2. Find a complete integral of the PDE p2z2 + q2 = 1.

Solution. Putting p = aq in the given PDE, we obtain

a2q2z2 + q2 = 1

=⇒ q2(1 + a2z2) = 1

=⇒ q = (1 + a2z2)−1/2.

Now,

p2 = (1− q2)/z2 =

(1− 1

(1 + a2z2)

)(1

z2

)=⇒ p2 =

a2

1 + a2z2

=⇒ p = a(1 + a2z2)−1/2.

Substituting p and q in dz = pdx+ qdy, we obtain

dz = a(1 + a2z2)−1/2dx+ (1 + a2z2)−1/2dy

=⇒ (1 + a2z2)1/2dz = adx+ dy

=⇒ 1

2a

{az(1 + a2z2)1/2 − log[az + (1 + a2z2)1/2]

}= ax+ y + b,

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 35

which is the complete integral of the given PDE.

Type (c): (Separable equations)

A first-order PDE is separable if it can be written in the form

f(x, p) = g(y, q). (6)

That is, a PDE in which z is absent and the terms containing x and p can be separated

from those containing y and q. For this type of equation, Charpit’s equations become

dx

fp=

dy

−gq=

dz

pfp − qgq=

dp

−fx=

dq

−gy.

From the last two relation, we obtain an ODE

dp

−fx=dx

fp=⇒ dp

dx+fxfp

= 0 (7)

which may be solved to yield p as a function of x and an arbitrary constant a. Writing

(7) in the form fpdp+ fzdx = 0, we see that its solution is f(x, p) = a. Similarly, we get

g(y, q) = a. Determine p and q from the equation

f(x, p) = a, g(y, q) = a

and then use the relation dz = pdx+ qdy to determine a complete integral.

EXAMPLE 3. Find a complete integral of p2y(1 + x2) = qx2.

Solution. First we write the given PDE in the form

p2(1 + x2)

x2=q

y(separable equation)

It follows that

p2(1 + x2)

x2= a2 =⇒ p =

ax√1 + x2

,

where a is an arbitrary constant. Similarly,

q

y= a2 =⇒ q = a2y.

Now, the relation dz = pdx+ qdy yields

dz =ax√1 + x2

dx+ a2ydy =⇒ z = a√

1 + x2 +a2y2

2+ b,

where a and b are arbitrary constant, a complete integral for the given PDE.

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 36

Type (d): (Clairaut’s equation)

A first-order PDE is said to be in Clairaut form if it can be written as

z = px+ qy + f(p, q). (8)

Charpit’s equations take the form

dx

x+ fp=

dy

y + fq=

dz

px+ qy + pfp + qfq=dp

0=dq

0.

Now, dp = 0 =⇒ p = a, where a is an arbitrary constant.

dq = 0 =⇒ q = b, where b is an arbitrary constant.

Substituting the values of p and q in (8), we obtain the required complete integral

z = ax+ by + f(a, b).

EXAMPLE 4. Find a complete integral of (p+ q)(z − xp− yq) = 1.

Solution. The given PDE can be put in the form

z = xp+ yq +1

p+ q, (9)

which is of Clairaut’s type. Putting p = a and q = b in (9), a complete integral is given

by

z = ax+ by +1

a+ b,

where a and b are arbitrary constants.

Practice Problems

Find complete integrals of the following PDEs.

1. p+ q = pq

2.√p+

√q = 1

3. z = p2 − q2

4. p(1 + q) = qz

5. p2 + q2 = x+ y

6. z = px+ qy +√

1 + p2 + p2

7. zpq = p2(xq + p2) + q2(yp+ q2)