Verify that $a x^{2} + b y^{2} = 1$ is a solution of the differential equation $x (y y_{2} + y_{1}^{2}) = y y_{1}$ .

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Solution

Here $a x^{2} + b y^{2} = 1$ $\Rightarrow 2 a x + 2 b y y_{1} = 0$ $\Rightarrow \frac{y y_{1}}{x} = - \frac{a}{b}$
$\Rightarrow \frac{x (y y_{2} + y_{1} y_{1}) - y y_{1} \times 1}{x^{2}} = 0$ $\Rightarrow x (y y_{2} + y_{1}^{2}) = y y_{1},$ hence verified.

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