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Complete Solution of Charpit’s method

Solution: Consider the non-linear partial differential equation f(x, y, z, p, q)= 0 —————————-(1)
Since z is a function of x and y, it follows that
dz = pdx + qdy —————————————–(2)
Let us assume p=u (x, y, z, a), where a is arbitrary constant, substitute in (1) solve to obtain q= v(x, y, z, a). For these value (2) becomes
dz = udx + vdy —————————————-(3)
Now if (3) can be integrated, yielding
g(x, y, z, a, b)=0———(4)
This is a complete solution of (1).
Since the success of the above procedure depends upon making a fortunate choice for p, it cannot be suggested as a standard procedure. We turn now to a general method for solving (1). This consists in finding an equation
F(x, y, z, p, q) = 0 ————–(5)
Such that (1) and (10) may be solved for p = P(x, y, z) and q= Q(x, y, z), that is, such that

and such that for these values of p and q the total differential equation.

dz = pdx + qdy = P(x, y, z)+Q(x, y, z) ——————————(7)

is integrable, that is

Differentiating (1) and (5) partially with respect to x and y, we find

Multiplying (8)by ∂F/∂p, (9) by ∂F/∂q, (10) by -∂f/∂p and (11) by -∂f/∂q and adding we obtain

This is a linear partial differential equation in F, considered as a function of the independent variables x, y, z, p, q. The auxiliary system is

Thus, we may take for (5) any solution of this system which involves p or, q, or both, which contains an arbitrary constant, and for which (6) holds.

Solve the following equation using charpits method

Solution: Given that, Let, f(x, y, z, p, q) = p – 3x^2 – q^2 ...