What is a solution to the differential equation #dy/dx=(2x)/y#?

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Answer 1 · 2016-07-12T22:29:14

I found: #y=+-sqrt(2x+c)#

We can try separating and write:
#ydy=2xdx#
then integrate:
#intydy=int2xdx#
#y^2/2=x^2+c#
or:
#y^2=2x^2+c#
and:
#y=+-sqrt(2x^2+c)#

Answer 2 · 2016-07-13T00:08:09

#y = pm sqrt( 2x^2 + C)#

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

#y \ dy/dx=2x#

#int \ y \ dy/dx \ dx=int \ 2x \ dx#

#int \ y \ dy=int \ 2x \ dx#

#y^2/2=2 * x^2/2 + C#

#y^2/2 = x^2 + C#

#y^2 = 2x^2 + C#

#y = pm sqrt( 2x^2 + C)#