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Mathematics · 2023

JEE Main · 12 April 2023, Shift 1 · Q11

Let y=y(x), y>0, be a solution curve of the differential equation (1+x^2) d y=y(x-y) d x. If y(0)=1 and y(2 √ 2 )=β, then

Let $\displaystyle y=y(x), y>0$, be a solution curve of the differential equation $\displaystyle \left(1+x^2\right) \mathrm{d} y=y(x-y) \mathrm{d} x$. If $\displaystyle y(0)=1$ and $\displaystyle y(2 \sqrt{2})=\beta$, then
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.