Question

The solution of $x^{2}dy-y^{2}dx-x{y}^2(x-y)dy=0$

• A. $\ln \left|\frac{x-y}{xy}\right|=\frac{y^2}{2}-c$
• B. $\ln \left|\frac{xy}{x-y}\right|=\frac{x^2}{2}+c$
• C. $\ln \left|\frac{x-y}{xy}\right|=\frac{x^2}{2}+c$
• D. $\ln \left|\frac{x-y}{xy}\right|=x+c$

Answer 1

The correct option is A $\ln \left|\frac{x-y}{xy}\right|=\frac{y^2}{2}-c$

$x^{2}dy-y^{2}dx=xy(x-y)\left(ydy\right)$

let $\frac{dy}{dx}=y^{\prime }$

$x^2{y}^{\prime }-y^2=xy(x-y)\left(y{y}^{\prime }\right)$

it appears that we need and subtract the terms to manage beause u can see on the RHS iyt is obtained by some $\ln$ function resulting with terms $xy,(x-y)$

$x^2{y}^{\prime }-y^2+xy{y}^{\prime }-xy{y}^{\prime }+xy-xy=xy{y}^{\prime }-xy-y^2+xy-xy{y}^{\prime }+x^2{y}^{\prime }$

$=xy(y^{\prime }-1)-(y-x)(y+x{y}^{\prime })$

$\int [\frac{xy}{x-y}\times \frac{xy(1-y^{\prime })-(x-y)(y+x{y}^{\prime })}{(xy{)}^2}]dx=\int y{y}^{\prime }dx$

$\ln |\frac{x-y}{xy}|=\frac{y^2}{2}-c$