JEE MAIN - Mathematics Hindi (2022 - 24th June Morning Shift - No. 11)
$$y \frac{\mathrm{d} x}{\mathrm{~d} y}=2 x+y^{3}(y+1) \mathrm{e}^{y}, x(1)=0$$ का हल $$x=x(y)$$ है, तो $$x(\mathrm{e})$$ बराबर है :
$$e^{3}\left(e^{e}-1\right)$$
$$e^{e}\left(e^{3}-1\right)$$
$$e^{2}\left(e^{e}+1\right)$$
$$e^{e}\left(e^{2}-1\right)$$
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