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

## Expert Answers

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)

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