Complex Variables and Partial Differential Equations (3130005)

Question-4c ( ii )-OR

BE | Semester-3 Summer-2020 | 27-10-2020

Q4) (c ( ii ))

Solve by Charpit’s method ${yzp}^{2} - q = 0$ .

{yzp}^{2} - q = 0 - - - - - [1]

\Rightarrow f = {yzp}^{2} - q

Auxiliary equation:

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

\Rightarrow \frac{dp}{{yp}^{3}} = \frac{dq}{{zp}^{2} + {yp}^{2} q} = \frac{dz}{- 2 {yzp}^{2} + q} = \frac{dx}{- 2 yzp} = \frac{dy}{1}

\Rightarrow \frac{dp}{{yp}^{3}} = \frac{dy}{1}

\Rightarrow \int \frac{1}{p^{3}} dp = \int y dy

\Rightarrow \frac{p^{- 2}}{- 2} = \frac{y^{2}}{2} - \frac{a}{2} \Rightarrow \frac{1}{p^{2}} = a - y^{2} \Rightarrow p^{2} = \frac{1}{a - y^{2}}

\Rightarrow p = \frac{1}{\sqrt{a - y^{2}}} - - - - - [2]

→ By

[1]

and

[2]

, We have

{yzp}^{2} - q = 0 \Rightarrow q = {yzp}^{2} \Rightarrow q = \frac{yz}{a - y^{2}}

Solution:

dz = \frac{1}{\sqrt{a - y^{2}}} dx + \frac{yz}{a - y^{2}} dy

\Rightarrow dz - \frac{yz}{a - y^{2}} dy = \frac{1}{\sqrt{a - y^{2}}} dx

\Rightarrow \sqrt{a - y^{2}} dz - \frac{yz}{\sqrt{a - y^{2}}} dy = dx

\Rightarrow d [z \sqrt{a - y^{2}}] = dx

\Rightarrow \int d [z \sqrt{a - y^{2}}] = \int dx

\Rightarrow z \sqrt{a - y^{2}} = x + b

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