Answer to Question #316640 in Differential Equations for Tobias Felix

Question

Integrating factors found by inspection

y(x^2 + y^2 − 1)dx + x(x^2 + y^2 + 1)dy = 0 when y(1) = 0

Answer 1

$y(x^2 + y^2 − 1)dx + x(x^2 + y^2 + 1)dy = 0\\ (x^2y + y^3 − y)dx + (x^3 + xy^2 + x)dy = 0\\ x^2ydx + y^3dx − ydx + x^3dy + xy^2dy + xdy = 0\\ x^2(ydx+xdy)+y^2(ydx+xdy)+xdy-ydx=0\\ \left[(ydx+xdy)(x^2+y^2)+xdy-ydx =0\right]\frac{1}{x^2+y^2}\\ ydx+xdy+ \frac{xdy-ydx}{x^2+y^2}=0\\ d(xy)+d(\arctan (\tfrac{y}{x}))=0\\$

integrating through, we have:

$xy+ \arctan (\tfrac{y}{x})=C$

When y(1) = 0,

$(1)(0) + \arctan (\tfrac{0}{1}) =C\\ 0+0=C \implies C=0\\[6mm] \therefore xy+ \arctan (\tfrac{y}{x})=0$