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Let \(y=y(x)\) be the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{y}+1\right)\…

Q1

Let \(y=y(x)\) be the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{y}+1\right)\left(\left(\mathrm{y}+1\right){\mathrm{e}}^{{\mathrm{x}}^{2}/2}-\mathrm{x}\right),0<\mathrm{x}<2.1,\mathrm{with}\mathrm{y}\left(2\right)=0,\) Then the value of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\mathrm{x}=1\) is equal to:

[JEE Main 2021, 18 Mar (Shift 2)]

a

\(\frac{-e^{3 / 2}}{\left(e^2+1\right)^2}\)

b

\(-\frac{2 e^2}{\left(1+e^2\right)^2}\)

c

\(\frac{e^{5 / 2}}{\left(1+e^2\right)^2}\)

d

\(\frac{5 e^{1 / 2}}{\left(e^2+1\right)^2}\)

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