What is the solution for (1-y^2)^{1/2}dx-(1-x^2)^{1/2}dy=0 ?

The solution for (1-y^2)^{1/2}dx-(1-x^2)^{1/2}dy=0 is y=sin(arcsin(x)+c_{1})

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(1-y^2)^{1/2}dx-(1-x^2)^{1/2}dy=0

(1 - y^{2})^{\frac{1}{2}} d x - (1 - x^{2})^{\frac{1}{2}} d y = 0

y = sin (arcsin (x) + c_{1})

Solution steps

(1 - y^{2})^{\frac{1}{2}} d x - (1 - x^{2})^{\frac{1}{2}} d y = 0

Solve separable ODE:

y = sin (arcsin (x) + c_{1})

y = sin (arcsin (x) + c_{1})

\frac{d y}{d x} = y sin (3 x) - 3 y e^{- 3 x}, y (0) = 1

y^{^{'}} + y tan (x) = 0, y (0) = a cos (x)

x^{2} (x d y + y d x) = 8 x d y - 8 y d x

3 y^{^{'}} = - 9 y

y^{^{'}} \frac{2}{x} y = 4 x

What is the solution for (1-y^2)^{1/2}dx-(1-x^2)^{1/2}dy=0 ?
The solution for (1-y^2)^{1/2}dx-(1-x^2)^{1/2}dy=0 is y=sin(arcsin(x)+c_{1})