Complex Variables and Partial Differential Equations (3130005)

Question-4b

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

Q4) (b)

Solve : $x^{2} p + y^{2} q = (x + y) z$

Auxiliary equation:

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

Group - I $\frac{dx}{x^{2}} = \frac{dy}{y^{2}}$

\int \frac{1}{x^{2}} dx = \int \frac{1}{y^{2}} dy \Rightarrow - \frac{1}{x} = - \frac{1}{y} + c_{1} \Rightarrow \frac{1}{y} - \frac{1}{x} = c_{1}

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

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

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

\int \frac{1}{(x - y)} d (x - y) = \int 1 dz \Rightarrow \log (x - y) = z + c_{2} \Rightarrow \log (x - y) - z = c_{2}

Solution:

f (\frac{1}{y} - \frac{1}{x}, \log (x - y) - z) = 0

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