Question

Find an explicit solution of the given initial- value problem

square root of 1-y^2 dx - square root 1-x^2 dy = 0, y(0)= 1/2

y = _________________

Answer 1

As,

$$\sqrt{1-y^2}dx-\sqrt{1-x^2}dy=0$$

Divide both side of the above equation by $\sqrt{(1-y^2)(1-x^2)}$ ,thus

$$\dfrac{dx}{\sqrt{1-x^2}}-\dfrac{dy}{\sqrt{1-y^2}}=0\\ \implies \int\dfrac{dx}{\sqrt{1-x^2}}-\int\dfrac{dy}{\sqrt{1-y^2}}=c\\ \implies\sin^{-1}(x)-\sin^{-1}(y)=c$$

Now, $y(0)=1/2$ ,thus

$$\sin^{-1}(0)-\sin^{-1}(1/2)=c$$

Thus, particular solution is

$$\sin^{-1}(x)-\sin^{-1}(y)=-\frac{\pi}{6}$$

Thus,

$$y=\sin(\sin^{-1}(x)+\frac{\pi}{6})$$