ISI2019-MMA-18

Question

For the differential equation  $$\frac{dy}{dx} + xe^{-y}+2x=0$$

It is given that $y=0$ when $x=0$. When $x=1$, $\:y$ is given by

• A. $\text{ln} \bigg(\frac{3}{2e} – \frac{1}{2} \bigg)$
• B. $\text{ln} \bigg(\frac{3e}{2} – \frac{1}{4} \bigg)$
• C. $\text{ln} \bigg(\frac{3}{e} – \frac{1}{2} \bigg)$
• D. $\text{ln} \bigg(\frac{3}{2e} – \frac{1}{4} \bigg)$

Answer 1

Given Differential Equation is :-

$\frac{dy}{dx} + xe^{-y} + 2x = 0$

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

$\Rightarrow \frac{-dy}{\left ( \frac{1}{e^{y}}+2 \right )}= xdx$

Now, separating terms in $y$ and $dy$ one side and terms in $x$ and $dx$ another side, 

$ -(\frac{e^{y}}{1 + 2e^{y}})dy = xdx$

On Integrating both sides,

$\int (\frac{-e^{y}}{1+2e^{y}})dy = \int xdx$

$ \frac{-log|(1 + 2e^{y})|}{2} = \frac{x^{2}}{2} + c$

When $x=0$ then $y=0$

So, $c= -\frac{log 3}{2}$ ----------- Equation $(1)$

After putting value of '$c$' in equation $(1)$, we get

$ \frac{-log|(1 + 2e^{y})|}{2} = \frac{x^{2}}{2} – \frac{log 3}{2}$

$\Rightarrow log|1+2e^{y}|=-x^{2}+log3$

Now, When $x=1,$

$\Rightarrow log|1+2e^{y}|=-1+log3$

$\Rightarrow log|1+2e^{y}|-log3=-1$

$\Rightarrow log\frac{|1+2e^{y}|}{3}= -1$

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

$\Rightarrow e^{y} = \frac{3e^{-1}-1}{2}$

$y = ln(\frac{3e^{-1}-1}{2})$

$y = ln(\frac{3}{2e}-\frac{1}{2})$

option A will be correct