(3x^2y+2xy+y^3)+(x^2+y^2){y}'=0
\because My=3x^2+2x+3y^2,
Nx=2x,
\therefore My\neq Nx,given DE is not exact;
we need to find a function \mu (x,y)s.t. the equation \mu (3x^2y+2xy+y^3)+\mu (x^2+y^2){y}'=0 is exact;
d(\mu (3x^2y+2xy+y^3))/dx = d(\mu (x^2+y^2))/dx;
(d\mu/dy) (3x^2y+2xy+y^3)+ \mu (3x^2+2x+3y^2)=(d\mu/dx)(x^2+y^2)+\mu (2x);
(d\mu/dy) (3x^2y+2xy+y^3)+ \mu (3x^2+3y^2)=(d\mu/dx)(x^2+y^2);
suppose \mu is a function of x only;
then d\mu /dy=0;
then \mu (3x^2+3y^2)=(d\mu/dx)(x^2+y^2)
divide both sides by (x^2+y^2);
3\mu =d\mu /dx
which is a seperable equation;
\int (1/\mu)d\mu=\int 3dx;
ln(\mu)=3x;
\mu=e^(3x);
thus, \mu =e^(3x)is an integration factor for the given DE;
multiply by \mu =e^(3x);
e^(3x)(3x^2y+2xy+y^3)+ e^3x(x^2+y^2){y}'=0;
which is exact;
\phi (x,y)=C;
d\phi /dx=e^(3x)(3x^2y+2xy+y^3)and d\phi /dy=e^(3x)(x^2+y^2);
\phi (x,y)=e^(3x)(x^y+(1/3)y^3)+ g(x);
d\phi /dx=3e^(3x)(x^y+(1/3)y^3)+e^(3x)(2xy)+g({x})';
d\phi /dx=e^(3x)(3x^2y+2xy+y^3)
g({x})'=0;
g(x)=C;
\phi (x,y)=e^(3x)(x^2y+(1/3)y^3)=C;
the general solution:e^(3x)(x^2y+(1/3)y^3)=C
