(-xysinx +2ycosx)dx +(2xcosx)dy=0
There is no need for integrating factor, since it is a separable ODE :
2dy/y = ((xsinx -2cosx)/xcosx)dx
2 ln(y) = -ln(cos(x)) -2 ln(x) +c
y² = C/(x²cos(x))
I start with the equation:
(-xysinx +2ycosx)dx +(2xcosx)dy with the integrating factor given u(x,y)=xy
after multiplying by u I get:
(-x^2 * y^2 * sinx +2xy^2 * cosx) +(2x^2 * ycosx)y'=0
then taking partial derivatives for M and N
Msuby= -2x^2*y*sinx +4xycosx
Nsubx= -4xysinx
I get Msuby does not equal Nsubx
Do I have to find another integrating factor for this equation or is there some calculus/algebra rule I'm not seeing where Msuby does equal Nsubx?