QJEEAdv21Maa1be: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Maa1be Solution: For any real numbers and , let , x R, be the solution of the…

Question

Verified Solution • 2025–26 Session moderate

For any real numbers

\alpha

and

\beta

, let

{y_{\alpha ,\beta }}(x)

, x

\in

R, be the solution of the differential equation

{{dy} \over {dx}} + \alpha y = x{e^{\beta x}},y(1) = 1

. Let

S = \{ {y_{\alpha ,\beta }}(x):\alpha ,\beta \in R\}

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

A $f(x) = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}$
B $f(x) = - {{{x^2}} \over 2}{e^{ - x}} + \left( {e + {1 \over 2}} \right){e^{ - x}}$
C $f(x) = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}$
D $f(x) = {{{e^x}} \over 2}\left( {{1 \over 2} - x} \right) + \left( {e + {{{e^2}} \over 4}} \right){e^{ - x}}$

Accepted Answer

Given,

{{dy} \over {dx}} + \alpha y = x\,.\,{e^{\beta x}}

which is a linear differential equation.

Integrating factor

(IF) = {e^{\int {\alpha dx} }} = {e^{\alpha x}}

So, the solution is

y \times {e^{\alpha x}} = \int {x{e^{\beta x}}\,.\,{e^{\alpha x}}dx}

\Rightarrow y \times {e^{\alpha x}} = \int {x{e^{(\alpha + \beta )x}}dx}

.... (i)

Case (I) If

\alpha

+

\beta

= 0

From Eq. (i), we get

\Rightarrow y{e^{\alpha x}} = \int {x{e^{0.x}}dx = \int {xdx = {{{x^2}} \over 2} + C} }

.... (ii)

Given, y(1) = 1 i.e. when x = 1, then y = 1

From Eq. (ii), we get

1.{e^\alpha } = {1 \over 2} + C \Rightarrow C = {e^\alpha } - {1 \over 2}

From Eq. (ii), we get

y{e^{\alpha x}} = {{{x^2} - 1} \over 2} + {e^\alpha }

For

\alpha

= 1

y{e^x} = {{{x^2} - 1} \over 2} + e \Rightarrow y = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}

Option (a) is correct.

Case (II) If

\alpha

+

\beta

\ne

0

\Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - \int {1 \times {{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}}dx}

\Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - {{{e^{(\alpha + \beta )x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}

\Rightarrow y = {{x\,.\,{e^{\beta x}}} \over {(\alpha + \beta )}} - {{{e^{\beta x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}{e^{ - \alpha x}}

(Cancelling e^{$\alpha$ x} from both sides)

\Rightarrow y = {{{e^{\beta x}}} \over {\alpha + \beta }}\left( {x - {1 \over {\alpha + \beta }}} \right) + {c_1}{e^{ - \alpha x}}

.... (iii)

Putting

\alpha

=

\beta

= 1 in Eq. (iii), we get

y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + {c_1}{e^{ - x}}

Given, y(1) = 1

\therefore

1 = {e \over 2} \times {1 \over 2} + {{{c_1}} \over e} \Rightarrow {c_1} = e - {{{e^2}} \over 4}

So,

y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}

\to

option (c) is correct.

Step-by-Step Verified Solution