Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 20

The given problem: ysqrt(1-x^2)y' -xsqrt(1-y^2)=0 is written in a form of first order "ordinary differential equation" or first order ODE.
To evaluate this, we can apply variable separable differential equation in which we express it in a form of f(y) dy= g(x) dx before using direct integration on each side.
To rearrange the problem, we move xsqrt(1-y^2) to the other to have an equation as:ysqrt(1-x^2)y' = xsqrt(1-y^2) .
Divide both sides by sqrt(1-y^2)sqrt(1-x^2) :
(ysqrt(1-x^2)y')/(sqrt(1-y^2)sqrt(1-x^2)) = (xsqrt(1-y^2))/(sqrt(1-y^2)sqrt(1-x^2))
(y*y')/sqrt(1-y^2)= x/sqrt(1-x^2)
Applying direct integration: int(y*y')/sqrt(1-y^2)= int x/sqrt(1-x^2)
Express y' as (dy)/(dx) : int(y*(dy)/(dx))/sqrt(1-y^2)= int x/sqrt(1-x^2)
Express in a form of f(y) dy= g(x) dx : int(y*dy)/sqrt(1-y^2)= int (x*dx)/sqrt(1-x^2)

To find the indefinite integral on both sides, we let:
u = 1-y^2 then du =-2y dy or (du)/(-2) =y dy
v = 1-x^2 then dv =-2x dx or (dv)/(-2) =x dx
The integral becomes:
int(y*dy)/sqrt(1-y^2)= int (x*dx)/sqrt(1-x^2)
int((du)/(-2))/sqrt(u)= int ((dv)/(-2))/sqrt(v)
Apply the basic integration property: int c*f(x) dx= c int f(x) dx .
(-1/2) int((du))/sqrt(u)= (-1/2) int (dv)/sqrt(v)
Apply the Law of Exponents: sqrt(x) = x^(1/2) and 1/x^n = x^(-n) .
Then, the integral becomes:
(-1/2) int((du))/u^(1/2)= (-1/2) int (dv)/v^(1/2)
(-1/2) int u^(-1/2)du= (-1/2) int v^(-1/2)dv
Applying Power Rule of integration: int x^ndx= x^(n+1)/(n+1)
(-1/2) int u^(-1/2)du= (-1/2) int v^(-1/2)dv
(-1/2) u^(-1/2+1)/(-1/2+1)= (-1/2) v^(-1/2+1)/(-1/2+1)+C
(-1/2) u^(1/2)/(1/2)= (-1/2) v^(1/2)/(1/2)+C
-u^(1/2)= - v^(1/2)+C
Note: (-1/2)/(1/2) = -1
In radical form: - sqrt( u)= -sqrt(v)+C
Plug-in u =1-y^2 and v=1-x^2 , we get the general solution of differential equation:
- sqrt( 1-y^2)= -sqrt(1-x^2)+C
Divide both sides by -1 , we get: sqrt( 1-y^2)= sqrt(1-x^2)+C .
Note:C/(-1) = C as arbitrary constant
For particular solution, we consider the initial condition y(0) =1 where x_0=0 and y_0=1 .
Plug-in the values, we get:
sqrt( 1-1^2)= sqrt(1-0^2)+C
sqrt(0)=sqrt(1)+C
0=1+C
C = 0-1
C =-1 .
Then plug-in C =-1 on the general solution: sqrt( 1-y^2)= sqrt(1-x^2)+C .
sqrt( 1-y^2)= sqrt(1-x^2)+(-1)

(sqrt(1-y^2))^2 =(sqrt(1-x^2) -1)^2
1-y^2= (1-x^2) -2sqrt(1-x^2) +1
Rearrange into:
y^2=-(1-x^2) +2sqrt(1-x^2)
y^2=-1+x^2 +2sqrt(1-x^2)
y^2=x^2+2sqrt(1-x^2)-1
Taking the square root on both sides:
y =sqrt(x^2+2sqrt(1-x^2) -1)