Complex Variables and Partial Differential Equations (3130005)

Question-4b-OR

BE | Semester-3 Winter-2019 | 26-11-2019

Q4) (b)

Solve $px + qy = pq$ using Charpit's method.

∎ px + qy - pq = 0 - - - - - (1)

\Rightarrow f = px + qy - pq

Auxiliary equation:

\frac{dp}{f_{x} + p f_{z}} = \frac{dq}{f_{y} + q f_{z}} = \frac{dz}{- p f_{p} - q f_{q}} = \frac{dx}{- f_{p}} = \frac{dy}{- f_{q}}

\Rightarrow \frac{dp}{p} = \frac{dq}{q} = \frac{dz}{- p (x - q) - q (y - p)} = \frac{dx}{- (x - q)} = \frac{dy}{- (y - p)}

\Rightarrow \frac{dp}{p} = \frac{dq}{q} & dz = p dx + q dy - - - - - [A]

∎ \frac{dp}{p} = \frac{dq}{q} \Rightarrow \int \frac{1}{p} dp = \int \frac{1}{q} dq \Rightarrow \ln p = \ln q + \ln a

\Rightarrow p = q a - - - - - [2]

→ By [1] and [2], We have

px + qy - pq = 0 \Rightarrow qax + qy - qaq = 0 \Rightarrow ax + y - aq = 0

\Rightarrow q = \frac{ax + y}{a}

→ By [2], We have

\Rightarrow p = ax + y

→ By 2^nd equation of [A], We have

\Rightarrow dz = (ax + y) dx + \frac{ax + y}{a} dy

\Rightarrow \frac{1}{ax + y} dz = dx + \frac{1}{a} dy

\Rightarrow \int \frac{1}{ax + y} dz = \int dx + \int \frac{1}{a} dy

\Rightarrow \ln (ax + y) = x + \frac{y}{a} + b

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