How do you find the particular solution to #ysqrt(1-x^2)y'-x(1+y^2)=0# that satisfies y(0)=sqrt3?

Given: #ysqrt(1 - x^2)y' - x(1 + y^2) = 0; y(0) = sqrt(3)#

Use the notation #dy/dx# for y':

#ysqrt(1 - x^2)dy/dx - x(1 + y^2) = 0#

Move the second term to the right side:

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

Multiply both sides of the equation by #dx/(sqrt(1 - x^2)(1 + y^2))#

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

Integrate both sides:

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

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

Multiply both sides by 2:

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

Use the exponential function:

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

Adding a constant in the exponent is the same as multiplying by a constant:

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

Subtract 1 from both sides:

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

Square root both sides:

#y = +-sqrt(Ce^(-2sqrt(1 - x^2)) - 1)" [1]"#

Use the point to solve for C, substitute 0 for x and #sqrt(3)# for y:

#sqrt(3) = +-sqrt(Ce^(-2sqrt(1 - 0^2)) - 1)#

The above equation can only be true, if we can drop the #+-# and make it only positive:

#sqrt(3) = sqrt(Ce^(-2sqrt(1 - 0^2)) - 1)#

Square both sides and solve for C:

#3 = Ce^-2 - 1#

#4 = Ce^-2#

#C = 4e^2#

Substitute #4e^2# for C and remove the #+-# in equation [1]:

#y = sqrt((4e^2)e^(-2sqrt(1 - x^2)) - 1)#