How do you integrate #ydx = 2(x+y)dy#?

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Answer 1 · 2015-10-20T21:44:39

#x=-2y+Cy^2#

#ydx = 2(x+y)dy#

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

#dx/dy - (2(x+y))/y = 0#

#x' - 2/y*x -2 = 0#

Let #x=u*v => x'=u'v+v'u#

#u'v+v'u-2/yuv-2 = 0#

#v(u'-2/yu)+v'u-2 = 0#

#u'-2/yu= 0#

#(du)/dy =2/yu#

#(du)/u = 2(dy)/y#

#lnu = 2lny#

#lnu = lny^2#

#u=y^2#

#v(u'-2/yu)+v'u-2 = 0 => v'u-2=0#

#v'(y^2)-2=0#

#(dv)/dy=2 1/y^2#

#dv=2 (dy)/y^2#

#v= -2 1/y + C#

#x=y^2(-2/y+C)#

#x=-2y+Cy^2#