Question

Let \(y=y(x)\) be a solution of the differential equation,
\(\sqrt{1-x^2}\,\frac{dy}{dx}+\sqrt{1-y^2}=0\), \(|x|<1\).
If \(y\left(\frac12\right)=\frac{\sqrt3}{2}\), then \(y\left(-\frac{1}{\sqrt2}\right)\) is equal to :

• A. \(-\frac{\sqrt3}{2}\)
• B. None of those
• C. \(\frac{1}{\sqrt2}\)
• D. \(-\frac{1}{\sqrt2}\)

Hint:

Separate the variables to get \(\sin^{-1}x+\sin^{-1}y=C\), find C from the initial condition and note the relation reduces to \(y=\sqrt{1-x^2}\).

Answer 1

The Correct Option is C

Separate variables: \(\frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}\).
Integrate: \(\sin^{-1}y=-\sin^{-1}x+C\), i.e. \(\sin^{-1}x+\sin^{-1}y=C\).
Using \(x=\frac12,\ y=\frac{\sqrt3}{2}\): \(C=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}\).
So \(\sin^{-1}y=\frac{\pi}{2}-\sin^{-1}x=\cos^{-1}x\), giving \(y=\sin(\cos^{-1}x)=\sqrt{1-x^2}\).
At \(x=-\frac{1}{\sqrt2}\): \(y=\sqrt{1-\frac12}=\frac{1}{\sqrt2}\).
Answer: (3) \(\frac{1}{\sqrt2}\)