$\left(1+y^2\right)\mathit{d}\mathit{x}\mathbf{+}\mathit{x}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{=}\mathbf{0}$when $\mathit{x}\mathbf{=}\mathbf{5}$.

The given differential equation is $\left(1+y^2\right)dx+xydy=0$with boundaries $y=0$when$x=5$.

A differential equation is a mathematical equation that connects one or more unknown functions with their derivatives.

Separate the variables in $\left(1+y^2\right)dx+xydy=0$and keep y on one side and x on the other.

$$\begin{gathered} \left(1+y^2\right)dx+xydy=0 \\ \left(1+y^2\right)dx=-xydy \\ -\frac{1}{x}dx=\frac{y}{1+y^2}dy \\ \frac{y}{1+y^{2}dy}=\frac{1}{x}dx \end{gathered}$$

Integrate the above differential equation using a variable separable form.

Substitute

$$\begin{gathered} 1+y^2=t \\ 2ydy=dt \\ ydy=\frac{1}{2}dy \end{gathered}$$

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

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

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

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

Put the boundary conditions, $y=0$, when $x=5$, in (1) and find the value of C.

$$\begin{gathered} \left(5\right){\left(1+{\left(1\right)}^2\right)}^2=2 \\ C=5\times 4 \\ C=20 \end{gathered}$$

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

$x\left(1+y^2\right)=20$

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

Therefore, for the differential equation role="math" localid="1655103818090" $\left(1+y^2\right)dx+xydy=0$, the solution containing one arbitrary constant is $x{\left(1+y^2\right)}^4=C$, and with boundary condition, y = 0. When x = 5, the value of constant is C = 20. The particular solution is $x{\left(1+y^2\right)}^2=20$. The plot of a slope field and some solution curves are as follows.