What is a particular solution to the differential equation #dy/dx=x/y# with #y(1)=2#?

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Answer 1 · 2016-07-31T11:51:54

#y = sqrt (x^2 + 3)#

this is separable

#y' = x/y#

#y \ y' = x#

#int \ y \ y' \ dx = int \ x \ dx#

#int \d/dx(y^2 / 2) \ dx = int \ x \ dx#

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

#y^2 = x^2 + C#

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

applying IV: #y(1) = 2#

#2 = pm sqrt (1 + C)#

suggest the + root solution. so

#2 = sqrt (1 + C) implies C = 3#

#y = sqrt (x^2 + 3)#