`ysqrt(1-x^2)y' - xsqrt(1-y^2) = 0 , y(0) = 1` Find the particular solution that satisfies the initial condition

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)`