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

Given differential equation,

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

Put

$$u=y+1$$

Then,

$$\frac{du}{dx}=u(ue^{x^2/2}-x)$$

$$\frac{du}{dx}+xu=u^2e^{x^2/2}$$

Now put

$$v=\frac1u$$

Then,

$$\frac{dv}{dx}=-\frac1{u^2}\frac{du}{dx}$$

Substituting,

$$-\frac{dv}{dx}+xv=e^{x^2/2}$$

$$\frac{dv}{dx}-xv=-e^{x^2/2}$$

This is a linear differential equation.

Integrating factor:

$$IF=e^{\int -x\,dx}$$

$$=e^{-x^2/2}$$

Multiplying throughout,

$$e^{-x^2/2}\frac{dv}{dx}-xe^{-x^2/2}v=-1$$

Hence,

$$\frac{d}{dx}\left(ve^{-x^2/2}\right)=-1$$

Integrating,

$$ve^{-x^2/2}=-x+C$$

$$v=e^{x^2/2}(C-x)$$

Since

$$v=\frac1{y+1},$$

$$\frac1{y+1}=e^{x^2/2}(C-x)$$

Using condition

$$y(2)=0,$$

$$1=e^2(C-2)$$

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

Therefore,

$$\frac1{y+1}=e^{x^2/2}(2+e^{-2}-x)$$

At

$$x=1,$$

$$y+1=\frac1{e^{1/2}(1+e^{-2})}$$

Now,

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

At

$$x=1,$$

$$(y+1)e^{1/2}=\frac1{1+e^{-2}}$$

Hence,

$$\left.\frac{dy}{dx}\right|_{x=1} =\frac1{e^{1/2}(1+e^{-2})} \left(\frac1{1+e^{-2}}-1\right)$$

$$=\frac1{e^{1/2}(1+e^{-2})} \left(\frac{1-(1+e^{-2})}{1+e^{-2}}\right)$$

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

Multiply numerator and denominator by $$e^4$$:

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

Therefore,

$$\boxed{-\frac{e^{3/2}}{(e^2+1)^2}}$$