Complex Variables and Partial Differential Equations (3130005)

Question-5b

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

Q5) (b)

Solve $z (xp - yq) = y^{2} - x^{2}$ .

Auxiliary equation:

\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R} \Rightarrow \frac{dx}{zx} = \frac{dy}{- zy} = \frac{dz}{y^{2} - x^{2}}

Group - I $\frac{dx}{zx} = \frac{dy}{- zy}$

\int \frac{1}{x} dx = - \int \frac{1}{y} dy \Rightarrow \log x = - \log y + \log c_{1} \Rightarrow \log x + \log y = \log c_{1} \Rightarrow xy = c_{1}

Group - II $\frac{dx + dy}{zx - zy} = \frac{dz}{y^{2} - x^{2}}$

\Rightarrow \frac{dx + dy}{z (x - y)} = \frac{dz}{(x - y) (x + y)}

\Rightarrow \frac{dx + dy}{z} = \frac{dz}{(x + y)}

\int (x + y) d (x + y) = \int z dz \Rightarrow \frac{{(x + y)}^{2}}{2} = \frac{z^{2}}{2} + c \Rightarrow {(x + y)}^{2} - z^{2} = c_{2}

Solution:

f (xy, {(x + y)}^{2} - z^{2}) = 0

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