When a differential equation of the form<\/p>\n
P(x,y) +Q(x,y)y’ = 0<\/p>\n
results from the implicit differentiation of an original equation F(x,y)=c, the equation<\/p>\n
P(x,y) +Q(x,y)y’ = 0<\/p>\n
is said to be an exact differential equation.<\/p>\n
The way to tell is that, for an exact differential equation,<\/p>\n
Py<\/sub>(x,y) = Qx<\/sub>(x,y)<\/p>\n Example: Solve the differential equation<\/strong><\/p>\n siny +1 +xy’cosy +2y’ = 0<\/p>\n First, we render it to P(x,y) +Q(x,y)y’ = 0:<\/p>\n siny + 1 +(xcosy +2)y’ = 0<\/p>\n P(x,y) = siny+1; Q(x,y) = xcosy + 2<\/p>\n Now we take the “other derivative” of each one:<\/p>\n Py<\/sub>(x,y) = cosy<\/p>\n Qx<\/sub>(x,y) = cosy<\/p>\n Py<\/sub>(x,y) = Qx<\/sub>(x,y): the equation is exact. Therefore,<\/p>\n siny +1 +xy’cosy +2y’ = 0<\/p>\n is the implicit derivative of some equation F(x,y) = c, which we need to find. Furthermore,<\/p>\n siny + 1 = Fx<\/sub><\/p>\n We integrate siny + 1 with respect to x:<\/p>\n ∫siny +1 = xsiny + x + g(y)<\/p>\n where g(y) is a function only of y that was lost in the original derivative by x.<\/p>\n g'(y) should be recognizable in Q(x,y): specifically, it’s 2⇒ g(y)=2y<\/p>\n Therefore, the solution F(x)=c is<\/p>\n xsiny +x +2y=c<\/p>\n Note that its implicit derivative is<\/p>\n siny + 1 + x(cosy)y’ +2y’ = 0<\/p>\n which matches the original differential equation posed.<\/p>\n Source:<\/p>\n Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems<\/u>. Toronto: John Wiley & Sons, 1986.<\/p>\n