The Solution of the differential equation ( x 2 + y 2 ) d x - 5 x y d y = 0 , y(1) = 0, is: [2024]

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Solution

Q.
The Solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0$ , y(1) = 0, is: [2024]

1 $| x^{2} - 2 y^{2} |^{5} = x^{2}$

2 $| x^{2} - 2 y^{2} |^{6} = x$

3 $| x^{2} - 4 y^{2} |^{5} = x^{2}$

4 $| \begin{matrix} x^{2} - 4 y^{2} \end{matrix} |^{6} = x$

Ans.
(3)

We have, $(x^{2} + y^{2}) d x - 5 x y d y = 0$

$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{5 x y}$

Let $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$∴ v + x \frac{d y}{d x} = \frac{1 + v^{2}}{5 v} \Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{5 v} - v$

$\Rightarrow \frac{5 v d v}{1 - 4 v^{2}} = \frac{d x}{x} \Rightarrow \frac{1}{8} \int \frac{8 \times 5 v d v}{1 - 4 v^{2}} = \int \frac{d x}{x}$

$\Rightarrow \frac{- 5}{8} \ln | 1 - 4 v^{2} | = \ln | x | + \ln c$

$\Rightarrow \frac{5}{8} \ln \frac{| x^{2} - 4 y^{2} |}{x^{2}} + \ln | x | + \ln c = 0$

$\Rightarrow | \frac{x^{2} - 4 y^{2}}{x^{2}} |^{5 / 8} | x | = k$

$\Rightarrow \frac{| x^{2} - 4 y^{2} |^{5 / 8}}{| x |^{\frac{1}{4}}} = k ∵ y (1) = 0 \Rightarrow k = 1$

$\Rightarrow | \begin{matrix} x^{2} - 4 y^{2} \end{matrix} |^{5 / 8} = | x |^{\frac{1}{4}} \Rightarrow | x^{2} - 4 y^{2} |^{5} = x^{2}$

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