JEE Advanced 2021 · Paper 2 · Mathematics Q4

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MathematicsMultiple correct+4 / -2Question 42 of 57

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

f(x) = \dfrac{x^2}2e^{-x} + \left(e - \dfrac12\right)e^{-x}

f(x) = -\dfrac{x^2}2e^{-x} + \left(e + \dfrac12\right)e^{-x}

f(x) = \dfrac{e^x}2\left(x - \dfrac12\right) + \left(e - \dfrac{e^2}4\right)e^{-x}

f(x) = \dfrac{e^x}2\left(\dfrac12 - x\right) + \left(e + \dfrac{e^2}4\right)e^{-x}

Answer and worked solution

Official answer: A, C

General solution. With integrating factor $e^{\alpha x}$ :

If $\alpha+\beta\ne0$ : $y=e^{\beta x}\left(\dfrac{x}{\alpha+\beta}-\dfrac1{(\alpha+\beta)^2}\right)+Ce^{-\alpha x}$ .
If $\alpha=-\beta$ : $y=\left(\dfrac{x^2}2+C\right)e^{\beta x}$ .

Check each option.

(A) Take $\beta=-1$ , $\alpha=1$ : $y=e^{-x}\left(\frac{x^2}2+C\right)$ , and $y(1)=1$ gives $C=e-\frac12$ . This matches. In $S$ .
(B) The $x^2e^{-x}$ coefficient must be $+\frac12$ , not $-\frac12$ . Not in $S$ .
(C) Take $\beta=1$ , $\alpha=1$ : the particular part is $e^x\left(\frac x2-\frac14\right)=\frac{e^x}2\left(x-\frac12\right)$ . Then $y(1)=1$ gives $C=e-\frac{e^2}4$ , matching. In $S$ .
(D) The $-\frac12xe^x$ term needs $\alpha+\beta=-2$ with $\beta=1$ , so $\alpha=-3$ . The other term would then be $e^{3x}$ , not $e^{-x}$ . Not in $S$ .

(A), (C)

Topics:Linear first-order equations

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