JEE Main 2021 (Online) 18th March Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)\left((y+1)e^{x^{2}/2}-x\right)$, $0<x<2.1$, with $y(2)=0$. Then the value of $\frac{dy}{dx}$ at $x=1$ is equal to :
Differential Equation
Solution
Concept. With $u=y+1$ the equation is a Bernoulli equation; $w=\frac1u$ makes it linear. Formulas used. (1) $w=\frac1u\Rightarrow w'=-\dfrac{u'}{u^2}$(2) $w'+Pw=Q$: $w\,e^{\int P\,dx}=\int Q\,e^{\int P\,dx}\,dx+C$Step 1 - Linear form. $u'=u^2e^{x^2/2}-xu$. By (1), $$w'=-e^{x^2/2}+xw\ \Longrightarrow\ w'-xw=-e^{x^2/2},\qquad\text{IF}=e^{-x^2/2}.$$ Step 2 - Integrate. By (2), $\big(w\,e^{-x^2/2}\big)'=-1$, so $w\,e^{-x^2/2}=-x+C$. Step 3 - Use $y(2)=0$. $u=w=1$: $e^{-2}=-2+C$, so $C=2+e^{-2}$. Step 4 - At $x=1$. $w\,e^{-1/2}=1+e^{-2}$, so $$w=\frac{e^{1/2}(e^2+1)}{e^2},\qquad u=\frac{e^{3/2}}{e^2+1},\qquad u\,e^{1/2}=\frac{e^2}{e^2+1}.$$ Step 5 - Slope. $$\frac{dy}{dx}=u\left(u\,e^{1/2}-1\right)=u\cdot\frac{-1}{e^2+1}=-\frac{e^{3/2}}{(e^2+1)^2}.$$ Answer: (D)