What is a solution to the differential equation #dy/dx=e^x/y# with y(0)=1?

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Answer 1 · 2016-06-28T17:52:25

#y = sqrt{2 e^x - 1}#

#dy/dx=e^x/y#

separate

#y \ dy/dx = e^x #

#int \ y \ dy =int e^x \ dx#

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

using the IV
#1^2/2 = 1 + C \implies C = - 1/2 #

#y^2 = 2 e^x - 1# or #y = sqrt{2 e^x - 1}#