JEE Advance - Mathematics (2021 - Paper 2 Online - No. 4)
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?
$$f(x) = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}$$
$$f(x) = - {{{x^2}} \over 2}{e^{ - x}} + \left( {e + {1 \over 2}} \right){e^{ - x}}$$
$$f(x) = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}$$
$$f(x) = {{{e^x}} \over 2}\left( {{1 \over 2} - x} \right) + \left( {e + {{{e^2}} \over 4}} \right){e^{ - x}}$$
Explanation
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.
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.
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