Q21Mathematics II

Question

Q.4 Apply Charpit's method to solve -

p x + q y = pq

Answer 1

Given the non-linear partial differential equation: $p x + q y = pq$ . We systematically write it strictly as $f (x, y, z, p, q) = p x + q y - pq = 0$ .

Step 1: Partial Derivatives

We rigorously calculate the fundamental partial derivatives of $f$ :

$f_{x} = p$
$f_{y} = q$
$f_{z} = 0$
$f_{p} = x - q$
$f_{q} = y - p$

Step 2: Charpit's Auxiliary Equations

The standard Charpit's auxiliary differential equations are rigidly formulated as:

\frac{d p}{f _{x} + p f _{z}} = \frac{d q}{f _{y} + q f _{z}} = \frac{d z}{- p f _{p} - q f _{q}} = \frac{d x}{- f _{p}} = \frac{d y}{- f _{q}}

Substituting the exact calculated derivatives entirely into the first two fractions:

\frac{d p}{p + 0} = \frac{d q}{q + 0} ⟹ \frac{d p}{p} = \frac{d q}{q}

Step 3: Finding $p$ and $q$

Integrating this extremely simple relation mathematically yields:

ln p = ln q + ln a ⟹ p = a q

We systematically substitute $p = a q$ directly back into the original PDE ( $p x + q y = pq$ ) to solve entirely for $q$ :

(a q) x + q y = (a q) q ⟹ q (a x + y) = a q^{2}

Assuming strictly $q \neq = 0$ , we mathematically divide by $q$ :

a x + y = a q ⟹ q = \frac{a x + y}{a} = x + \frac{y}{a}

Correspondingly, we find $p$ :

p = a (x + \frac{y}{a}) = a x + y

Step 4: Final Integration

We rigidly substitute these precise expressions directly into the total differential $d z = p d x + q d y$ :

d z = (a x + y) d x + (x + \frac{y}{a}) d y

We carefully expand and strictly rearrange the exact terms to form perfect differentials:

d z = a x d x + (y d x + x d y) + \frac{y}{a} d y

d z = a x d x + d (x y) + \frac{y}{a} d y

Integrating exactly the entire equation term by term mathematically yields the absolute complete integral:

z = \frac{a x ^{2}}{2} + x y + \frac{y ^{2}}{2 a} + c

Q21Mathematics II

Question

Answer

Step 1: Partial Derivatives

Step 2: Charpit's Auxiliary Equations

Step 3: Finding p and q

Step 4: Final Integration

Step 3: Finding $p$ and $q$