Question Number 103931 by bemath last updated on 18/Jul/20

(y^2 +2) dx = (xy+2y+y^3 ) dy

Answered by bobhans last updated on 18/Jul/20

(dy/dx) = ((y^2 +2)/(y(x+2+y^2 )))  set : q = x+2+y^2  ⇒(dq/dx) = 1+2y (dy/dx)  2y (dy/dx) = (dq/dx)−1⇒(dy/dx) = (1/(2y)) (dq/dx)−(1/(2y))  ⇔ (1/(2y)) (dq/dx)−(1/(2y)) = ((q−x)/(qy))   (1/2)((dq/dx)−1) =1−(x/q)   (dq/dx) −1 = 2−((2x)/q) ⇒(dq/dx) = 3−((2x)/q)  (dq/dx) + ((2x)/q) = 3

Answered by bramlex last updated on 18/Jul/20

(∂M/∂y) = 2y ; (∂N/∂x) = −y ⇒non exact   (∂N/∂x)−(∂M/∂y) = −3y is a function only in y .   integrating factor u(y)=e^(∫(((−3y)/(y^2 +2))) dy)   u(y)= e^(−(3/2)∫ ((d(y^2 +2))/(y^2 +2))) = (1/((y^2 +2)^(3/2) ))  ⇒(((y^2 +2)/(√((y^2 +2)^3 )))) dx − (((xy+2y+y^3 )/(√((y^2 +2)^3 )))) dy=0  (1/(√(y^2 +2))) dx − ((x+2y+y^3 )/(√(y^2 +2))) dy=0   { (((∂f/∂x) = M(x,y)=(1/(√(y^2 +2))))),(((∂f/∂y)=N(x,y)=((x+2y+y^3 )/(√((y^2 +2)^3 ))))) :}  F(x,y)=∫M(x,y)dx+g(y)  g′(y) =((−y)/(√(y^2 +2)))  g(y) = −(1/2)∫ ((d(y^2 +2))/((y^2 +2)^(1/2) ))   g(y) = (√(y^2 +2))  F(x,y)= ∫ (dx/(√(y^2 +2))) + (√(y^2 +2))  F(x,y) = (x/(√(y^2 +2))) +(√(y^2 +2))   F(x,y) = ((x+y^2 +2)/(√(y^2 +2))) .★