\((2xy-9x^2)dx+(2y+x^2+1)dy=0\)

Comparing with M dx + N dy = 0 we get

\(M= 2xy-9x^2\), \(N= 2y+x^2+1\)

\((dM)/dy= 2x-0= 2x\)

\((dN)/dx= 0+2x+0= 2x\)

\((dM)/dy= (dN)/dx\)

int Mdx+int (terms of N not containing x) dy=C

\(=> \int(2xy-9x^2)dx+\int(2y+1)dy=C\)

\(=> (2y x^2/2)−9 x^3/3+2 y^2/2+y=C\)

\(\int x^2y−3x^3+y^2+y=C\)

\(\int y^2+(x^2+1)y−3x^3=C\) (1)

Given y(0) = −3

substitute x=0 and y= -3 in (1) we get

\((−3)^2+(0^2+1)(−3)−3(0^3)=C\)

9−3=C int C=6

substitute C=6 in (1) we get

\(y^2+(x^2+1)y−3x^3=6\)

\(\int y^2+(x^2+1)y−3x^3−6=0\)

\(2xy−9x^2+(2y+x^2+1)dy/dx=0\)

\(y(0)= −3 is y^2+(x^2+1)y\)

\(−3x^3−6=0\)