Given DE
(2xy-3)dx+(x^2+4y)dy=0(2xy−3)dx+(x2+4y)dy=0
Comparing above equation with the standard form of DE Mdx+Ndy=0Mdx+Ndy=0 we get
M=2xy-3\implies \frac{\partial M}{\partial y}=2xM=2xy−3⇒∂M∂y=2x &
N=x^2+4y\implies \frac{\partial N}{\partial x}=2xN=x2+4y⇒∂N∂x=2x
Since, \frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}∂M∂y=∂N∂x hence the given DE is an exact DE whose solution is given as
int_{y=const} Mdx+\int_{\text{free of x}}Ndy =C∫y=constMdx+∫free of xNdy=C
\int (2xy-3)\ dx+\int (x^2+4y)\ dy=C
2y\int x\ dx-3\int dx+4\int y\ dy=C
x^2y-3x+2y^2=C
Now, applying initial condition, y(2)=5 by setting x=2 & y=5 in above solution we get
2^2(5)-3(2)+2(5)^2=C
C=64
x^2y-3x+2y^2=64
x^2y+2y^2-3x-64=0