The given differential equation is
"2xy\\ d x+(1-x^2+2y)dy=0" .............(1)
Let "x^2=t" .
On differentiation, "2x\\ dx=dt"
Substituting these in Eq.(1) we get,
"ydt+(1-t+2y)dy=0" .This is the linear form of Eq.(1).
"ydt+(1+2y)dy-tdy=0\\\\\nydt-tdy+(1+2y)dy=0"
Dividing by "y^2" ,
"\\frac{ydt-tdy}{y^2}+\\frac{1+2y}{y^2}dy=0"
"d(t\/y)+\\left(\\frac{1}{y^2}+\\frac{2}{y}\\right)dy=0" since "d(t\/y)=\\frac{ydt-tdy}{y^2}"
Integrating,
"\\int d(t\/y)+\\int \\left(\\frac{1}{y^2}+\\frac{2}{y}\\right)dy=C" , where "C" is the constant of integration.
"\\frac{t}{y}+\\left(-\\frac{1}{y} +2\\ln y\\right)=C"
Substituting "t=x^2,"
"\\frac{x^2}{y}-\\frac{1}{y}+2\\ln y=C"
This is the solution of the differential equation (1).
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