Given DE
#(2xy-3)dx+(x^2+4y)dy=0#
Comparing above equation with the standard form of DE #Mdx+Ndy=0# we get
#M=2xy-3\implies \frac{\partial M}{\partial y}=2x# &
#N=x^2+4y\implies \frac{\partial N}{\partial x}=2x#
Since, # \frac{\partial M}{\partial y}= \frac{\partial N}{\partial 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#
#\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#
