What is a particular solution to the differential equation #dy/dx=2xsqrt(1-y^2)# and y(0)=0?

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Answer 1 · 2016-09-07T21:34:06

# y = sin x^2 #

this is separable

#dy/dx=2xsqrt(1-y^2)#

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

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

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

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

# y = sin ( x^2 + C) #

using the IV, #y(0) = 0#

# 0 = sin ( 0 + C) implies C = 0#

meaning

# y = sin x^2 #