Mathematics · 2020
JEE Main · 3 September 2020, Shift 1 · Q62
The solution curve of the differential equation, (1+ e^-x)(1+y^2) (d y)/(d x)=y^2, which passes through the point (0,1), is:
The solution curve of the differential equation, $\displaystyle \left(1+\mathrm{e}^{-x}\right)\left(1+y^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=y^2$, which passes through the point $\displaystyle (0,1)$, is :
Official answer
(1)
$\displaystyle y^2=1+y \log _{\mathrm{e}}\left(\frac{1+\mathrm{e}^x}{2}\right)$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.