JEE MAIN - Mathematics Hindi (2020 - 9th January Evening Slot - No. 7)
यदि $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; तब x का एक मान
जो y(x) = e को संतुष्ट करता है :
$$\sqrt 2 e$$
$${1 \over 2}\sqrt 3 e$$
$${e \over {\sqrt 2 }}$$
$$\sqrt 3 e$$
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