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.