Como determinar la solución de Ec Diferencial

Diosa Lara!

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Derivaremos implícitamente la respuesta para calcular y', y'' y luego veremos si se cumple la ecuación

$$\begin{align}&y=ln(xy)\\&\\&y' = \frac{y+xy'}{xy}\\&\\&xyy'=y+xy'\\&\\&xyy'-xy'=y\\&\\&y' = \frac{y}{xy-x}=\frac{y}{x(y-1)}\\&\\&\text{ahora derivamos de nuevo}\\&\\&y''=\frac{xy'(y-1)-y(y-1+xy')}{x^2(y-1)^2}=\\&\\&\frac{xyy'-xy'- y^2+y-xyy'}{x^2(y-1)^2}=\\&\\&\frac{y-xy'-y^2}{x^2(y-1)^2}\\&\\&\text{sustituimos el valor de y'}\\&\\&y'' = \frac{y-x \frac{y}{x(y-1)}-y^2}{x^2(y-1)^2}=\\&\\&\frac{xy(y-1)-xy-xy^2(y-1)}{x^3(y-1)^3}=\\&\\&\frac{xy(y-1-1-y^2+y)}{x^3(y-1)^3}=\\&\\&\frac{y(-y^2+2y-2)}{x^2(y-1)^3}\\&\\&\text{Y la ecuación diferencial retocada es}\\&\\&x(y-1)y''+y'(xy'+y-2)=0\\&\\&\text{sustituyendo}\\&\\&\frac{y(-y^2+2y-2)}{x(y-1)^2}+\frac{y}{x(y-1)}\left(\frac{xy}{x(y-1)}+y-2  \right)=\\&\\&\frac{y(-y^2+2y-2)}{x(y-1)^2}+\frac{y}{x(y-1)}\left(\frac{xy+xy(y-1)-2x(y-1)}{x(y-1)}  \right)=\\&\\&\frac{y(-y^2+2y-2)}{x(y-1)^2}+\frac{y}{x(y-1)}\left(\frac{ xy^2-2xy+2x}{x(y-1)}  \right)=\\&\\&\frac{y(-y^2+2y-2)}{x(y-1)^2}+\frac{y}{x(y-1)}\left(\frac{ y^2-2y+2}{y-1}  \right)=\\&\\&\frac{y(-y^2+2y-2)}{x(y-1)^2}+\frac{ y(y^2-2y+2)}{x(y-1)^2}=\\&\\&\frac{y(-y^2+2y-2+y^2-2y+2)}{x(y-1)^2}=\\&\\&\frac{y·0}{x(y-1)^2}=0\end{align}$$

Luego es verdad.