Let \(y=y(x)\) be the solution of the differential equation \(\frac{ d y}{ d x}=2 x(x+y)^3-x(x+y)-1, y(0)=1\). Then,\({\…

Given \( \frac{dy}{dx}=2x(x+y)^3-x(x+y)-1 \).

Put \(u=x+y\).

Then \( \frac{du}{dx}=1+\frac{dy}{dx} \).

So \( \frac{du}{dx}=1+2xu^3-xu-1 \).

Thus \( \frac{du}{dx}=2xu^3-xu=xu(2u^2-1) \).

Now put \(z=u^2\), then \( \frac{dz}{dx}=2u\frac{du}{dx} \).

So \( \frac{dz}{dx}=2x u^2(2u^2-1)=2xz(2z-1) \).

Put \(v=\frac{1}{z}\), then \( \frac{dv}{dx}=-\frac{1}{z^2}\frac{dz}{dx} \).

Hence \( \frac{dv}{dx}=-\frac{2xz(2z-1)}{z^2}=-2x\left(2-\frac{1}{z}\right) \).

So \( \frac{dv}{dx}=2xv-4x \).

Therefore \( \frac{dv}{dx}-2xv=-4x \).

Here integrating factor \(=\mathrm{e}^{\int -2x\,dx}=\mathrm{e}^{-x^2}\).

Multiplying by \( \mathrm{e}^{-x^2} \), we get

\( \frac{d}{dx}\left(v\mathrm{e}^{-x^2}\right)=-4x\mathrm{e}^{-x^2} \).

Integrating, \( v\mathrm{e}^{-x^2}=2\mathrm{e}^{-x^2}+C \).

Thus \(v=2+C\mathrm{e}^{x^2}\).

Since \(y(0)=1\), we get \(u(0)=0+y(0)=1\).

So \(z(0)=1\) and \(v(0)=1\).

Therefore \(1=2+C\), hence \(C=-1\).

So \(v=2-\mathrm{e}^{x^2}\).

Since \(v=\frac{1}{z}\) and \(z=u^2\), we get \(u^2=\frac{1}{2-\mathrm{e}^{x^2}}\).

Now \(u=x+y\).

Therefore \( \left(x+y(x)\right)^2=\frac{1}{2-\mathrm{e}^{x^2}} \).

At \(x=\frac{1}{\sqrt{2}}\), \(x^2=\frac{1}{2}\).

Hence \( \left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2=\frac{1}{2-\mathrm{e}^{\frac{1}{2}}}=\frac{1}{2-\sqrt{\mathrm{e}}} \).