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

1 Answer
Oct 20, 2015

x=-2y+Cy^2

Explanation:

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