Reduce the following differential equation to the variables separable form and hence solve.
$(x - y)^{2} \frac{d y}{d x} = a^{2}$ .

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Solution

${(x - y)}^{2} \frac{d y}{d x} = a^{2}$
Let $x - y = v \Rightarrow 1 - \frac{d y}{d x} = \frac{d v}{d x}$
$\frac{d y}{d x} = 1 - \frac{d v}{d x} {(x - y)}^{2} \frac{d y}{d x} = a^{2} \Rightarrow v^{2} (1 + \frac{d v}{d x}) = a^{2} \Rightarrow 1 + \frac{d v}{d x} = \frac{a^{2}}{v^{2}} \Rightarrow \frac{d v}{d x} = \frac{a^{2}}{v^{2}} - 1 = \frac{a^{2} - v^{2}}{v^{2}} \Rightarrow \int \frac{v^{2}}{a^{2} - v^{2}} d v = \int d x + C = - [\int \frac{{(a}^{2} - v^{2}) - a^{2}}{a^{2} - v^{2}} d v] = \int d x + C \Rightarrow [\int (1 - \frac{a^{2}}{a^{2} - v^{2}}) d v] = \int d x + C \Rightarrow - [v - a^{2} \times \frac{1}{2 a} ln ∣ ∣ ∣ \frac{a + v}{a - v} ∣ ∣ ∣] = x + C \Rightarrow - v + \frac{a}{2} ln ∣ ∣ ∣ \frac{a + v}{a - v} ∣ ∣ ∣ = x + C v = x - y \Rightarrow - (x - y) + \frac{a}{2} ln ∣ ∣ ∣ \frac{a + x - y}{a - x + y} ∣ ∣ ∣ = x + C \Rightarrow - y - x + \frac{a}{2} ln ∣ ∣ ∣ \frac{a + x - y}{a - x + y} ∣ ∣ ∣ = x + C \frac{a}{2} ln ∣ ∣ ∣ \frac{a + x - y}{a - x + y} ∣ ∣ ∣ + y = 2 x + C \Rightarrow a ln ∣ ∣ ∣ \frac{a + x - y}{a - x + y} ∣ ∣ ∣ + 2 y = 4 x + 2 C$

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