Question

(y−1)dx−(x−y−1)dy=0

Answer 1

$(y-1)dx-(x-y-1)dy=0\\ M=y-1 \implies M_y=1\\ N=-(x-y-1) \implies N_x=-1\\ \text{Thus } M_y\neq N_x \implies \text{ It is not exact}\\ \text{To get the integrating factor;}\\ \frac{N_x-M_y}{M}=-\frac{2}{y-1}. \\ I.F=e^{\int -\frac{2}{y-1}dy}=e^{-2\ln (y-1)}=(y-1)^{-2}\\ \text{Multiply the DE by I.F}\\ (y-1)^{-1}dx-\frac{x-y-1}{(y-1)^2}dy=0\\ M_y=N_x=-\frac{1}{(y-1)^2}\\ F_x=(y-1)^{-1}dx\\ F=\int(y-1)^{-1}dx\\ F=x(y-1)^{-1}+\Phi(y)\\ \text{Differentiate w.r.t y}\\ F_y=-\frac{x}{(y-1)^2}+\Phi'(y)\\ -\frac{x-y-1}{(y-1)^2}=-\frac{x}{(y-1)^2}+\Phi'(y)\\ \Phi'(y)=-\frac{y+1}{(y-1)^2}\\ \Phi(y)=\frac{2}{y-1}-\ln(y-1)\\ \text{Hence}\\ F=x(y-1)^{-1}+\frac{2}{y-1}-\ln(y-1)$