JEE Advanced 2021 Paper 2 Q04 Mathematics Integration & Differential Equations Differential Equations Hard

JEE Advanced 2021 Paper 2 · Q04 · Differential Equations

For any real numbers $\alpha$ and $\beta$, let $y_{\alpha,\beta}(x)$, $x\in\mathbb{R}$, be the solution of the differential equation $$\dfrac{dy}{dx} + \alpha y = xe^{\beta x},\quad y(1)=1.$$ Let $S = \{y_{\alpha,\beta}(x) : \alpha,\beta\in\mathbb{R}\}$. Then which of the following functions belong(s) to the set $S$?

  1. A. $f(x) = \dfrac{x^2}{2}e^{-x} + \left(e - \dfrac{1}{2}\right)e^{-x}$
  2. B. $f(x) = -\dfrac{x^2}{2}e^{-x} + \left(e + \dfrac{1}{2}\right)e^{-x}$
  3. C. $f(x) = \dfrac{e^x}{2}\left(x - \dfrac{1}{2}\right) + \left(e - \dfrac{e^2}{4}\right)e^{-x}$
  4. D. $f(x) = \dfrac{e^x}{2}\left(\dfrac{1}{2} - x\right) + \left(e + \dfrac{e^2}{4}\right)e^{-x}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

Linear ODE: integrating factor $e^{\alpha x}$ gives $\dfrac{d}{dx}(ye^{\alpha x}) = xe^{(\alpha+\beta)x}$. Case $\alpha+\beta\ne 0$: integrate to get $ye^{\alpha x} = \dfrac{xe^{(\alpha+\beta)x}}{\alpha+\beta} - \dfrac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$. Choosing $\alpha=1,\beta=1$ and applying $y(1)=1$ gives option (C): $y = \dfrac{e^x}{2}(x-\tfrac{1}{2}) + (e - \tfrac{e^2}{4})e^{-x}$. Case $\alpha+\beta=0$: $\dfrac{d}{dx}(ye^{\alpha x}) = x$, so $ye^{\alpha x} = \dfrac{x^2}{2} + C$, i.e. $y = \dfrac{x^2}{2}e^{-\alpha x} + Ce^{-\alpha x}$. With $\alpha=1$ (so $\beta=-1$) and $y(1)=1$: $C = e - 1/2$, giving option (A): $y = \dfrac{x^2}{2}e^{-x} + (e-\tfrac{1}{2})e^{-x}$. Hence A and C.

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