What is the solution to the differential equation: #(x^2+y+y^2)dx=xdy# ?

We have the following Differential Equation in differential form

# (x^2+y+y^2) dx = xdy #

Which we can re-arrange as follows:

# dy/dx = (x^2+y+y^2)/x #
# " " = x+y/x+y^2/x #
# " " = x+y/x+y(y/x) #

Try a substitution #v=y/x => y=xv#, then

# dy/dx = v+x(dv)/dx #

Substituting in the DE we get:

# v+x(dv)/dx = x+v+(xv)v #

# :. x(dv)/dx = x+xv^2 #

# :. (dv)/dx = 1+v^2 #

This is a separable DE, so we can seperate the variables to get:

# int \ 1/(1+v^2) \ dv = int \ dx #

These integrals are standard and so trivial to evaluate:

# arctan v = x + c #
# :. v = tan(x + c) #

And restoring the initial substitution we get:

# y/x = tan(x + c) #

# :. y = xtan(x + c) #

Making #y = lambda x# and

#dy = lambda dx + x dlambda# we have

#(x^2+y+y^2)dx -x dy -> (x^2+lambda x + x^2 lambda^2) dx - x(x dlambda + lambda dx)#

or

#x((x+lambda+x lambda^2)dx-(x dlambda+lambda dx))=0#

or

#x dx+lambda dx+x lambda^2dx - x dlambda-lambda dx = 0#

or

#x(1+lambda^2)dx-x dlambda = 0#

or

#(dlambda)/(1+lambda^2) = dx#

and after integration

#arctan(lambda) = x + C# or

#lambda = tan(x+C)=y/x# and finally

#y = x tan(x+C)#