Mathematics · 2018
JEE Main · 15 April 2018, Shift 2 · Q77
The curve satisfying the differential equation, (x^2-y^2) d x+2 x y d y=0 and passing through the point (1,1) is:
The curve satisfying the differential equation, $\displaystyle \left(x^2-y^2\right) \mathrm{d} x+2 x y \mathrm{~d} y=0$ and passing through the point $\displaystyle (1,1)$ is :
Official answer
(1)
a circle of radius one.
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.