Question

Slove [1+xy]xdy+[1-xy]ydx=0

Answer 1

We have(1+xy)ydx+(1−xy)xdy=0⇒ydx+xy2dx+xdy−x2ydy=0⇒ydx+xdy+xy2dx−x2ydy=0By product rule we know that:ydx+xdy=d(xy)⇒d(xy)+xy2dx−x2ydy=0Dividing out by x2y2, we getd(xy)x2y2+dxx−dyy=0⇒∫d(xy)(xy)2+∫dxx−∫dyy=0On integrating we get,−1xy+lnx−lny=C [Note that as ∫d(t)t=−1t2, similarly ∫d(xy)(xy)2=−1xy]We know that log(m)−log(n)=log(mn)⇒−1xy+ln(xy)=C⇒ln(xy)−1xy=C (Answer)We have(1+xy)ydx+(1−xy)xdy=0⇒ydx+xy2dx+xdy−x2ydy=0⇒ydx+xdy+xy2dx−x2ydy=0By product rule we know that:ydx+xdy=d(xy)⇒d(xy)+xy2dx−x2ydy=0Dividing out by x2y2, we getd(xy)x2y2+dxx−dyy=0⇒∫d(xy)(xy)2+∫dxx−∫dyy=0On integrating we get,−1xy+lnx−lny=C [Note that as ∫d(t)t=−1t2, similarly ∫d(xy)(xy)2=−1xy]We know that log(m)−log(n)=log(mn)⇒−1xy+ln(xy)=C⇒ln(xy)−1xy=C (Answer)