JEE PYQ: Differential Equation Question 7
Question 7 - 2021 (18 Mar Shift 2)
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:
(1) $\frac{-e^{1/2}}{(e^{1/2}+1)^2}$
(2) $\frac{-2e^2}{(1+e^2)^2}$
(3) $\frac{e^{1/2}}{(1+e^2)^2}$
(4) $\frac{5e^{1/2}}{(e^{1/2}+1)^2}$
Show Answer
Answer: (1)
Solution
Let $y + 1 = Y$. Then $\frac{dY}{dx} = Y^2e^{x^2/2} - xY$. Put $k = -\frac{1}{Y}$: $\frac{dk}{dx} + k(-x) = e^{x^2/2}$. I.F. $= e^{-x^2/2}$. So $k = (x+c)e^{x^2/2}$. With $y(2) = 0$, $Y = 1$, $k = -1$: $-1 = (2+c)e^2$. So $c = -2-e^{-2}$. At $x = 1$: $\frac{dy}{dx}\big|_{x=1} = \frac{-e^{1/2}}{(1+e^{1/2})^2}$.
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