JEE MAIN - Mathematics Hindi (2023 - 12th April Morning Shift - No. 3)
अगर $$y=y(x), y > 0$$, एक विभेदी समीकरण का समाधान वक्र हो $$\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$$। अगर $$y(0)=1$$ और $$y(2 \sqrt{2})=\beta$$, तो
$${e^{{\beta ^{ - 1}}}} = {e^{ - 2}}\left( {3 + 2\sqrt 2 } \right)$$
$${e^{3{\beta ^{ - 1}}}} = e\left( {5 + \sqrt 2 } \right)$$
$${e^{3{\beta ^{ - 1}}}} = e\left( {3 + 2\sqrt 2 } \right)$$
$${e^{{\beta ^{ - 1}}}} = {e^{ - 2}}\left( {5 + \sqrt 2 } \right)$$
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