−x=y+1‌, which is linear. ‌ ‌∴‌‌‌‌‌ IF ‌=e∫−1dy=e−y ‌∴‌‌‌ Solution is ‌x⋅e−y=∫(y+1)e−ydy ‌⇒xe−y=∫(ye−y+e−y)dy+c ‌⇒xe−y=−ye−y+∫1⋅e−ydy+e−y⋅(−1)+c ‌⇒xe−y=−ye−y−e−y−e−y+c ‌⇒xe−y=−(y+2)e−y+c ‌⇒‌‌x=−(y+2)+cey