In problem, solve the equation.$(x+x{y}^2)dx+e^{x^2}ydy=0$

A first-order ordinary differential equation $\frac{dy}{dx}=f(x,y)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions g(x)that is a function of x alone and h(y) that is a function of y alone.

A separable differential equation can be expressed as $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$. By separating the variables, the equation follows$\frac{dy}{h\left(y\right)}=g\left(x\right)dx$ . Then, on direct integration of both sides, the solution of the differential equation is determined.

The given equation is

$(x+x{y}^2)dx+e^{x^2}ydy=0.....\left(1\right)$

Re-write equation (1) as follows:

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

After separating the variables, equation (2) can be written as

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

Integrate both sides of equation (3). It results,

$$\begin{gathered} \int \frac{y}{1+y^2}dy=\int -x{e}^{x^2}dx \\ \frac{1}{2}\int \frac{2y}{1+y^2}dy=\frac{1}{2}\int -2x{e}^{x^2}dx \\ \frac{1}{2}\int \frac{d(1+y^2)}{1+y^2}=\frac{1}{2}\int 2e^{x^2}d(-x^2) \\ \frac{1}{2}\ln (1+y^2)=\frac{1}{2}{e}^{-x^2}+\ln k[\ln k=Integratingcons\tan t] \\ \ln (1+y^2)=e^{-x^2}+\ln C[\ln C=2\ln k=cons\tan t] \\ \ln \left(\frac{1+y^2}{C}\right)=e^{x^2} \\ 1+y^2=C{e}^{x^2} \\ y=\pm \sqrt{C{e}^{x^2}-1} \\ y=\pm \sqrt{Cexp\left[exp\right(x^2\left)\right]-1} \end{gathered}$$

Therefore, the solution of the given equation is $y=\pm \sqrt{Cexp\left[exp\right(x^2\left)\right]-1}$.