What is a solution to the differential equation #y'=xe^(x^2-lny^2)#?

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Answer 1 · 2016-08-29T20:33:36

#y =root3 (3/2e^(x^2) + C)#

#y'=xe^(x^2-lny^2)#

we can separate it out first

#y'=xe^(x^2)e^(-lny^2)#

#y'=xe^(x^2)1/e^(lny^2)#

#y'=xe^(x^2)1/y^2#

#y^2 y'=xe^(x^2)#

#int y^2 y' dx=int xe^(x^2) dx#

#int y^2 dy=int xe^(x^2) dx#

.....spotting the pattern on the RHS

#y^3/3 =int d/dx (1/2e^(x^2) ) dx#

#y^3/3 =1/2e^(x^2) + C#

#y^3 =3/2e^(x^2) + C#

#y =root3 (3/2e^(x^2) + C)#