$\mathit{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{2}\mathbf{x}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{-}\mathbf{y}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{=}\mathbf{0}$when $\mathit{x}\mathbf{=}\sqrt{\mathbf{2}}$.

The given differential equation is $y\text{'}=\frac{2x{y}^2+x}{x^{2}y-y}$with boundaries y = 0 when$x=\sqrt{2}$.

A differential equation is a mathematical equation that connects one or more unknown functions with their derivatives. The study of differential equations primarily entails looking at their solutions (the set of functions that satisfy each equation) and characteristics.

Separate the variables in $y\text{'}=\frac{2x{y}^2+x}{x^2-y}$and keep y on one side and x on the other.

$$\begin{gathered} \frac{dy}{dx}=\frac{2x{y}^2+x}{x^{2}y-y} \\ \frac{dy}{dx}=\frac{x\left(2y^2+1\right)}{y\left(x^2-1\right)} \\ \frac{y}{2y^2+1}dy=\frac{x}{x^2-1}dx \end{gathered}$$

Integrate the above differential equation using the variable separable form.

Substitute,

$$\begin{gathered} 2y^2+1-t \\ ydy-\frac{1}{4}dt \\ {x}^2-1-u \\ sdx-\frac{1}{2}du \end{gathered}$$

Solve the separated form of the differential equation with an arbitrary constant C.

$$\begin{gathered} \frac{y}{2y^2+1}dy=\frac{x}{2^2-1}dx \\ \frac{1}{4}\int \frac{dt}{t}=\frac{1}{2}\int \frac{du}{u} \\ \frac{1}{4}Int=\frac{1}{2}Inu+c \\ {t}^4=u^{2}C \end{gathered}$$

Re-substitute $t=2y^2+1$and $u=x^2-1$in the above equation.

${\left(2y^2+1\right)}^4={\left(x^2-1\right)}^2+C$ …… (1)

Put the boundary conditions, y = 0 when $x=\sqrt{2}$, in (1) and find the value of C.

$$\begin{gathered} {\left(2{\left(0\right)}^2+1\right)}^4={\left({\left(\sqrt{2}\right)}^2-1\right)}^2+C \\ 1={\left(2-1\right)}^2+C \\ 1=1+C \\ C=0 \end{gathered}$$

Substitute the value of C in (1) and find the particular solution.

${\left(2y^2+1\right)}^4={\left(x^2-1\right)}^2$

Plot the slope field of $\left(1+y^2\right)dx+xydy=0$and some solution curves.

Therefore, for the differential equation $y\text{'}=\frac{2x{y}^2+x}{x^{2}y-y}$, the solution containing one arbitrary constant is ${\left(2y^2+1\right)}^4={\left(x^2-1\right)}^2+C$, and with boundary condition, y = 0 when $x=\sqrt{2}$. The value of the constant is C = 0 and the particular solution is ${\left(2y^2+1\right)}^4={\left(x^2-1\right)}^2$.The plot of a slope field and some solution curves are as follows.