$$\quad x e ^ { x ^ { 2 } + y } d x = y d y$$
$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x^{2}e^{x^{2}+y}-y^{2}=0\end{matrix}\right.$
$$x^{2}e^{x^{2}+y}d=ydy$$
$$x^{2}e^{x^{2}+y}d=y^{2}d$$
$$x^{2}e^{x^{2}+y}d-y^{2}d=0$$
$$dx^{2}e^{x^{2}+y}-dy^{2}=0$$
$$\left(x^{2}e^{x^{2}+y}-y^{2}\right)d=0$$
$$d=0$$
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