What is a solution to the differential equation #ydy/dx=e^x# with y(0)=4?

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Answer 1 · 2016-07-09T18:09:01

#implies y^2/2 = e^x + 7#

or if you like...
#implies y = pm sqrt{2(e^x + 7)}#

this is separable and has already been separated

so we integrate both side wrt x

#int ydy/dx \ dx=int e^x \ dx#

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

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

applying the IV

#4^2/2 = e^0 + C implies C = 7#

#implies y^2/2 = e^x + 7#