What is a solution to the differential equation #x+2ysqrt(x^2+1)dy/dx# with y(0)=1?

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Answer 1 · 2016-08-01T22:44:14

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

IF its actually #x+2ysqrt(x^2+1)dy/dx = 0#

then we can say that

#2ysqrt(x^2+1)dy/dx = -x#

So #2ydy/dx = -x/sqrt(x^2+1) #

Integrating both sides

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

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

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

#implies y^2= - sqrt(x^2+1) + C#

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

#y= pm sqrt ( C - sqrt(x^2+1) )#

applying the IV: #y(0) = 1#

#1= pm sqrt ( C - sqrt(1) )#

so we can disregard the -ve sign

#1= sqrt ( C - 1 ) implies C = 2#

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