Let y = y x be the solution of the differential equation d y d x = 2 x x + y 3 − x x + y − 1 , y 0 = 1 .…

Given: $\frac{dy}{dx}=2x{\left(x+y\right)}^3-x\left(x+y\right)-1$

Let, $x+y=t$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1$

$\Rightarrow \frac{dt}{dx}-1=2x{t}^3-xt-1$

$\Rightarrow \frac{dt}{dx}=2x{t}^3-xt$

$\Rightarrow \frac{dt}{dx}+xt=2x{t}^3$

$\Rightarrow \frac{1}{t^3}\frac{dt}{dx}+\frac{x}{t^2}=2x ...\left(i\right)$

Putting, $\frac{1}{t^2}=u$

$\Rightarrow \frac{-2}{t^3}\frac{dt}{dx}=\frac{du}{dx}$

$\Rightarrow \frac{1}{t^3}\frac{dt}{dx}=\frac{-1}{2}\frac{du}{dx}$

Putting this value in equation $\left(i\right)$

$\Rightarrow \frac{-1}{2}\frac{du}{dx}+xu=2x$

$\Rightarrow \frac{du}{dx}-2xu=-4x$

Integrating factor, $IF=e^{\int -2xdx}$

$\Rightarrow IF=e^{-x^2}$

So, solution of the given differential equation is,

$u\times e^{-x^2}=\int e^{-x^2}\left(-4x\right)dx$

$\Rightarrow \frac{e^{-x^2}}{t^2}=\int e^{-x^2}\left(-4x\right)dx$

Putting, $-x^2=z$

$\Rightarrow -2xdx=dz$

$\Rightarrow \frac{e^{-x^2}}{{\left(x+y\right)}^2}=\int 2e^{z}dz$

$\Rightarrow \frac{e^{-x^2}}{{\left(x+y\right)}^2}=2e^z+C$

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

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

It is given that, at $x=0$, $y=1$.

$\Rightarrow 1=2+C$

$\Rightarrow C=-1$

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

Putting, $x=\frac{1}{\sqrt{2}}$

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

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