# Exercise 8.10 – Differential Equations

#### Charpit's method

We will now consider a general method of solving a non-linear partial differential equation of the first order due to Charpit. This method is to be applied when the given equation cannot be reduced to any of the standard forms discussed earlier.

Let the equation to be solved be denoted by

Since z is a function of x and y we have

Now if we can find another relation

satisfied by p and q we could solve Eqs. (8.565) and (8.567) for p and q and substitute in (8.566). This will give the solution provided (8.566) is integrable.

In order to obtain φ, differentiate (8.565) and (8.567) w.r.t. x and y. We get

Eliminating between Eqs. (8.568) and (8.569) we obtain

Similarly, eliminating between the Eqs. (8.570) and (8.571) we obtain

Since the last term in (8.572) is same as that in (8.573) except for a minus sign. Hence, if we add (8.572) and (8.573) they cancel out. The other terms, on rearrangement, give

(i.e.)

This is Lagrange's linear partial differential equation of the first order with x, y, z, p, q as independent variables and φ as the dependent variable. Its solution will depend on the solution of the subsidiary equations

An integral of these equations which involves p or q or both will serve along with the main equation for finding p and q. We should take the simplest of the integrals so that it may be easier to solve for p and q.

Example 8.13.23 Solve px + qy = pq.

Solution Let

Then

Charpit's auxiliary equations are

on integrating the last two terms we get

Solving Eqs. (8.576) and (8.577) for p and q we get

putting these values in

dz = pdx + qdy

we get

or adz = (y + ax) d(y + ax)

Integrating we get the complete solution

Example 8.13.24 Solve (p2 + q2)y = qz.

Solution Charpit's auxiliary equations give

Integrating we get p2 + q2 = a2.

With the help of the given equation we get

Now

Integrating we get

∴ The complete integral is

z2 = a2 y2 + (ax + b)2.

Example 8.13.25 Find the complete integral of

Solution Charpit's auxiliary equations give

p = aq

Substituting in the given equation we get

or

Hence we get

dz = pdx + qdy = (adx + dy)q

or

Putting u = vz we have

Hence the complete integral is

where .

Example 8.13.26 Find the complete integral of p = (qy + z)2.

Solution Charpit's auxiliary equations are

Putting this value in the given equation

Substituting in dz = pdx + qdy we get

Integrating we get the complete integral

Example 8.13.27 Solve z2 = pqxy.

Solution Charpit's auxiliary equations are

These give

Integrating

log px = log qy + log b2 or px = b2 qy

Solving this and the given equation for p and q we get

Substituting in dz = pdx + qdy we get

Integrating we get which is the required complete integral.

Example 8.13.28 Solve z = p2x + q2y.

Solution Charpit's auxiliary equations are

These give

Integrating we get

log p2 x = log q2 y + log ap2 x = aq2 y

Solving this with the given equation for p and q we get

Putting these values in dz = pdx + qdy we get

Integrating we get

which is the required complete integral.

##### EXERCISE 8.10

Solve the following equations:

1. q = 3 p2.

Ans. z = ax + 3a2 y + b

2. zpg = p + q.

Ans. z2 = 2(l + a)(x + y/a) + b

3. 2(z + xp + yq) = p2 y.

Ans.

4. (p2 + q2) y = qz.

Ans. z2 = (a + cx)2 + c2 y2

5. 2zxpx2 − 2qxy + pq = 0.

Ans. z = ay + b (x2a)

6. p = (ay + z)2.

Ans.

7. q = 2y p2.

Ans. z = ax + a2 y2 + b

8. z = pq.

Ans.