Question

The differential equation dy/dx=√1-y2/y determines a family of circles with

• (A) Variable radius and fixed centre at (0,1)
• (B) Variable radius and fixed centere at (0,-1)
• (C) Fixed radius of 1 Unit and variable centre along the X-axis
• (D) Fixed radius of 1 Unit and variable centre along the X- axis

Answer 1

The Correct Option is D

The given differential equation is:

\[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{y} \]

Step 1: Rewrite the equation:
We can write the given equation in the form: \[ \frac{dy}{\sqrt{1 - y^2}} = \frac{dx}{y} \] This separation of variables allows us to integrate both sides with respect to \( y \) and \( x \), respectively.

Step 2: Integrate both sides:
Now, we integrate both sides: \[ \int \frac{dy}{\sqrt{1 - y^2}} = \int \frac{dx}{y} \] The integral on the left-hand side is the standard integral for the arcsine function, and the integral on the right-hand side is the natural logarithm. So we get: \[ \arcsin(y) = \ln|x| + C \] Where \( C \) is the constant of integration.

Step 3: Solve for \( y \):
Solving for \( y \), we get: \[ y = \sin(\ln|x| + C) \] This is the general solution of the differential equation.

Step 4: Interpretation of the solution:
We can observe that this solution represents a family of curves, which are circles centered on the x-axis. To demonstrate this, we rewrite the solution in a different form: \[ y = \sin(\ln|x| + C) \] Using the identity for sine: \[ \sin(A) = \frac{e^A - e^{-A}}{2} \] We get: \[ y = \frac{e^{\ln|x| + C} - e^{-\ln|x| - C}}{2} \] Simplifying the exponentials: \[ y = \frac{x - \frac{1}{x}}{2} \]

Step 5: Equation of a Circle:
This equation represents the equation of a circle centered at (0,0) with a radius of \( \frac{1}{2} \). Therefore, we can conclude that the solution describes a family of circles with a fixed radius.

Final Answer:
The correct option is (D) fixed radius of 1 unit and variable center along the x-axis.