Let y = y(x) be the solution of the differential equation [dy dx = (y + 1)( (y + 1)e^x^2/2…

Question

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

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

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

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

Solutions · 1