Question

Let $y=y\left(x\right)$be the solution of the differential equation $\frac{dy}{dx}=(y+1)\left[(y+1)e^{\frac{x^2}{2}}-x\right],0<x<2.1,$ with $y\left(2\right)=0$. Then the value of $\frac{dy}{dx}$ at$x=1$ is equal to:

• A. $\frac{\left[e^{5/2}\right]}{{[1+e^2]}^2}$
• B. $\frac{\left[5e^{1/2}\right]}{{[1+e^2]}^2}$
• C. $\frac{-2e^2}{{[1+e^2]}^2}$
• D. $\frac{-e^{3/2}}{{[1+e^2]}^2}$

Answer 1

The correct option is D

$\frac{-e^{3/2}}{{[1+e^2]}^2}$

Explanation for the correct option:

Given: $\frac{dy}{dx}/dx=(y+1)\left[(y+1)e^{\frac{x^2}{2}}-x\right],0<x<2.1,$

Replace $y+1=Y$

$\frac{dy}{dx}=\frac{dY}{dx}$

$\Rightarrow$ $\frac{dY}{dx}=Y\left[Y{e}^{\frac{x^2}{2}}-x\right]$

$$\begin{gathered} \Rightarrow \frac{dY}{dx}=Y^2{e}^{\frac{x^2}{2}}-xY \\ \Rightarrow \frac{dY}{dx}+xY=Y^2{e}^{\frac{x^2}{2}} \\ \Rightarrow \frac{1}{Y^2}\frac{dY}{dx}+\frac{x}{Y}=e^{\frac{x^2}{2}} \end{gathered}$$

Let $k=-\frac{1}{y}$

then $\frac{dk}{dx}=\frac{1}{Y^2}\frac{dY}{dx}$

$\frac{dk}{dx}-kx=e^{x^2/2}$

Find the integration factor

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

$$\begin{gathered} \Rightarrow k{e}^{-x^2/2}=\int e^{-x^2/2}{e}^{x^2/2}dx \\ \Rightarrow k{e}^{-x^2/2}=\int 1dx \\ \Rightarrow k{e}^{-x^2/2}=x+c \\ \Rightarrow k=\left(x+c\right)e^{x^2/2} \end{gathered}$$

Now put $k=\frac{-1}{y+1}\because Y=y+1$

$\frac{-1}{y+1}=\left(x+c\right)e^{x^2/2}....\left(i\right)$

Since $y\left(2\right)=0$, $x=2,y=0$

$$\begin{gathered} \Rightarrow \frac{-1}{0+1}=\left(2+c\right)e^{2^2/2} \\ \Rightarrow -1=\left(2+c\right)e^2 \\ \Rightarrow -1=2e^2+c{e}^2 \\ \Rightarrow c{e}^2=-1-2e^2 \\ \Rightarrow c=\frac{-1}{e^2}-2\frac{e^2}{e^2} \\ \Rightarrow c=\frac{-1}{e^2}-2 \end{gathered}$$

Put $c=-2-\frac{1}{e^2}$ in equation $\left(i\right)$

$$\begin{gathered} \Rightarrow \frac{-1}{y+1}=\left[x+\left(-2-\frac{1}{e^2}\right)\right]e^{x^2/2} \\ \Rightarrow \frac{-1}{y+1}=\left[x-\frac{2e^2-1}{e^2}\right]e^{x^2/2} \\ \Rightarrow \frac{-1}{y+1}=\left[\frac{x{e}^2-2e^2-1}{e^2}\right]e^{x^2/2} \\ \Rightarrow y+1=\left(\frac{e^2}{x{e}^2-2e^2-1}\right)\frac{-1}{e^{x^2/2}} \\ \Rightarrow {\frac{dy}{dx}}_{x=1}=-\left(\frac{e^2}{e^2-2e^2-1}\right)\frac{1}{e^{1/2}} \\ =\frac{-e^{2-1/2}}{2e^2-1} \\ =\frac{-e^{3/2}}{{\left(1+e^2\right)}^2} \end{gathered}$$

Therefore, $\frac{dy}{dx}$ at$x=1$ is equal to $\frac{-e^{3/2}}{{\left(1+e^2\right)}^2}$

Hence, option (D) is the correct answer.