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Mathematics · 2020

JEE Main · 9 January 2020, Shift 2 · Q64

If (d y)/(d x)=(x y)/(x^2+y^2); y(1)=1; then a value of x satisfying y(x)= e is:

If $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x y}{x^2+y^2} ; y(1)=1$; then a value of $\displaystyle x$ satisfying $\displaystyle y(x)=\mathrm{e}$ is :
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