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

(A) $f(\mathrm{x})=\frac{\mathrm{x}^{2}}{2} \mathrm{e}^{-\mathrm{x}}+\left(\mathrm{e}-\frac{1}{2}\right) \mathrm{e}^{-\mathrm{x}}$; (C) $f(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}}{2}\left(\mathrm{x}-\frac{1}{2}\right)+\left(\mathrm{e}-\frac{\mathrm{e}^{2}}{4}\right) \mathrm{e}^{-\mathrm{x}}$

(B) $f(\mathrm{x})=-\frac{\mathrm{x}^{2}}{2} \mathrm{e}^{-\mathrm{x}}+\left(\mathrm{e}+\frac{1}{2}\right) \mathrm{e}^{-\mathrm{x}}$

(D) $f(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}}{2}\left(\frac{1}{2}-\mathrm{x}\right)+\left(\mathrm{e}+\frac{\mathrm{e}^{2}}{4}\right) \mathrm{e}^{-\mathrm{x}}$

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Differential Equations

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For any real numbers

α

and

β

, let

y_{α, β} (x), x \in R

, be the solution of the differential equation

\frac{d y}{d x} + α y = x e^{β x}, y (1) = 1

Let

S = {y_{α, β} (x) : α, β \in R}

. Then which of the following functions belong(s) to the set S?

(A) $f (x) = \frac{x ^{2}}{2} e^{- x} + (e - \frac{1}{2}) e^{- x}$ ✓ Correct
(B) $f (x) = - \frac{x ^{2}}{2} e^{- x} + (e + \frac{1}{2}) e^{- x}$
(C) $f (x) = \frac{e ^{x}}{2} (x - \frac{1}{2}) + (e - \frac{e ^{2}}{4}) e^{- x}$ ✓ Correct
(D) $f (x) = \frac{e ^{x}}{2} (\frac{1}{2} - x) + (e + \frac{e ^{2}}{4}) e^{- x}$

Answer: (A), (C)

Solution

The integrating factor is

e^{α x}

. Multiplying:

ye^{α x} = \int xe^{(α + β) x} dx

.

Case I: If

α + β = 0

, then

ye^{α x} = \frac{x ^{2}}{2} + c

. Using

y (1) = 1

e^{α} = \frac{1}{2} + c

, so

c = e^{α} - \frac{1}{2}

. Thus

y = \frac{x ^{2}}{2} e^{- α x} + (e^{α} - \frac{1}{2}) e^{- α x}

. For

α = 1

y = \frac{x ^{2}}{2} e^{- x} + (e - \frac{1}{2}) e^{- x}

, which is option (A).

Case II: If

α + β \neq = 0

, integrating by parts gives the general solution. For

α = β = 1

: after applying the initial condition and simplifying, we get

y = \frac{e ^{x}}{2} (x - \frac{1}{2}) + (e - \frac{e ^{2}}{4}) e^{- x}

, which is option (C).

Options (B) and (D) do not satisfy the differential equation with the given initial condition for any real

α, β

.

Correct answers: (A), (C).

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Differential Equations

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