What is a general solution to the differential equation #1+y^2-y'sqrt(1-x^2)=0#?

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Answer 1 · 2016-07-20T16:40:25

#y = tan ( sin^(-1) x + C)#

#1+y^2-y'sqrt(1-x^2)=0#

first we separate it
#1+y^2=y'sqrt(1-x^2)#

#1/(1+y^2) y'=1/sqrt(1-x^2)#

then integrate

#int 1/(1+y^2) y' \ dx= int 1/sqrt(1-x^2) \ dx#

#int 1/(1+y^2) \ dy= int 1/sqrt(1-x^2) \ dx#

these are standard integrals, done using tan and sin subs, which means....

#tan^(-1) y = sin^(-1) x + C#

#y = tan ( sin^(-1) x + C)#