# 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. 