QJEEM21Mae52b: JEE Main 2021 Mathematics - Class 12 JEEM21Mae52b Solution: Let y = y(x) be the solution of the differential equation , 0 < …

Question

Verified Solution • 2025–26 Session moderate

Let y = y(x) be the solution of the differential equation

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

, 0 < x < 2.1, with y(2) = 0. Then the value of

{{dy} \over {dx}}

at x = 1 is equal to :

Accepted Answer

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

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

Put,

{1 \over {y + 1}} = z

- {1 \over {{{(y + 1)}^2}}}.{{dy} \over {dx}} = {{dz} \over {dx}}

\therefore

{{dz} \over {dx}} + z( - x) = - {e^{{{{x^2}} \over 2}}}

I.F = {e^{\int { - xdx} }} = {e^{{{ - {x^2}} \over 2}}}

z.\left( {{e^{ - {{{x^2}} \over 2}}}} \right) = - \int {{e^{ - {{{x^2}} \over 2}}}.{e^{{{{x^2}} \over 2}}}dx} = - \int {1.dx = - x + C}

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

.... (1)

Given y = 0 at x = 2 Put in (1)

{{{e^{ - 2}}} \over {0 + 1}} = - 2 + C

C = {e^{ - 2}} + 2

.... (2)

From (1) and (2)

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

Again, at x = 1

\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}

\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}

\therefore

{\left. {{{dy} \over {dx}}} \right|_{x = 1}} = {{{e^{3/2}}} \over {{e^2} + 1}}\left( {{{{e^{3/2}}} \over {{e^2} + 1}} \times {e^{1/2}} - 1} \right)

= - {{{e^{3/2}}} \over {{{({e^2} + 1)}^2}}}

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