Risolvi dx+xdy=e^-ytext{logy}dy

$$d x + x d y = e ^ { - y } \text { logy } d y$$

Trova d (soluzione complessa)

$$\left\{\begin{matrix}d=0\text{, }&y\neq 0\\d\in \mathrm{C}\text{, }&x=\frac{y\log(y)}{\left(y+1\right)e^{y}}\text{ and }y\neq -1\text{ and }y\neq 0\end{matrix}\right,$$

Trova x (soluzione complessa)

$$\left\{\begin{matrix}x=\frac{y\log(y)}{\left(y+1\right)e^{y}}\text{, }&y\neq -1\text{ and }y\neq 0\\x\in \mathrm{C}\text{, }&d=0\text{ and }y\neq 0\end{matrix}\right,$$

Trova d

$$\left\{\begin{matrix}d=0\text{, }&y>0\\d\in \mathrm{R}\text{, }&x=\frac{y\log(y)}{\left(y+1\right)e^{y}}\text{ and }y>0\end{matrix}\right,$$

Trova x

$$\left\{\begin{matrix}x=\frac{y\log(y)}{\left(y+1\right)e^{y}}\text{, }&y>0\\x\in \mathrm{R}\text{, }&d=0\text{ and }y>0\end{matrix}\right,$$