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

Q: Why can't we just use energy conservation?

Energy conservation is a physical principle, but here we are dealing with a mathematical constraint defined by a differential equation. We aren't modeling a particle's motion under a potential; we are finding the locus of points that satisfy a specific slope condition. The differential equation itself is the constraint.

Q: Wait, why do we reject the negative root here?

Actually, we don't reject it! When we square both sides to remove the radical, the negative sign is absorbed. The equation $(x+C)^2 + y^2 = 1$ covers both the positive and negative branches of the original square root, effectively describing the entire circle.

Q: How do I know the center is on the x-axis?

Look at the standard form of a circle: $(x-h)^2 + (y-k)^2 = r^2$. Our result is $(x+C)^2 + (y-0)^2 = 1^2$. Comparing this, we see $k=0$. Since the $y$-coordinate of the center is $0$, the center must lie on the $x$-axis.

Question

The differential equation $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

Options:

(A) variable radii and a fixed centre at (0, 1)

(B) variable radii and a fixed centre at (0, -1)

(C) fixed radius 1 and variable centres along the x-axis

(D) fixed radius 1 and variable centres along the y-axis

Sigma Percentile · Accepted Answer

### The Geometry of Slopes

Welcome, fellow traveler of the mathematical landscape. Today, we are going to peel back the layers of a seemingly simple differential equation and reveal the elegant geometric truth hidden within.

We are looking at the equation:

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

At first glance, it might look like just another problem to solve, but I want you to see it as a story of constraints and curves.

### Phase 1: The Separation

Look closely at the structure. The right-hand side is purely a function of $y$. There is no $x$ to be found!

This is a beautiful moment in calculus—the moment of separation. We can isolate the variables, moving all the $y$ terms to one side and the $x$ terms to the other.

By cross-multiplying, we transform our equation into:

$$ \frac{y}{\sqrt{1-y^2}} \, dy = dx $$

This is the pivotal step. We have successfully decoupled the variables, allowing us to treat the $y$ side and the $x$ side as independent entities waiting to be integrated.

### Phase 2: The Substitution Dance

Now, we face the integral:

$$ \int \frac{y}{\sqrt{1-y^2}} \, dy = \int dx $$

The right side is trivial—it is just $x + C$. But the left side requires a bit of finesse.

Let us use substitution. If we set $t = 1 - y^2$, then the derivative is $dt = -2y \, dy$. This means $y \, dy = -\frac{1}{2} \, dt$.

Suddenly, the complexity collapses. The integral becomes:

$$ -\frac{1}{2} \int t^{-\frac{1}{2}} \, dt $$

Applying the power rule, we get:

$$ -\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}} = -\sqrt{t} $$

Substituting back, we find our left side is simply $-\sqrt{1-y^2}$.

### Phase 3: The Geometric Reveal

We are left with the equation:

$$ -\sqrt{1-y^2} = x + C $$

To see the shape, we need to remove that radical. Squaring both sides gives us:

$$ 1 - y^2 = (x + C)^2 $$

Rearranging this, we get:

$$ (x + C)^2 + y^2 = 1 $$

Does this look familiar? It is the standard equation of a circle, $(x-h)^2 + (y-k)^2 = r^2$.

Here, $r^2 = 1$, so the radius is fixed at $1$. The center is at $(-C, 0)$.

Because $C$ is an arbitrary constant, the center can be anywhere along the $x$-axis. We have discovered a **family of circles with a fixed radius of 1**, all sliding along the $x$-axis.

Isn't that beautiful? You have just derived a geometric family from a simple slope condition.

Animated Solution for Mathematics - Differential Equations: The differential equation $\frac{d y}{d x} = \frac{1 - y ^{2}}{y}$ determines a family of circles with

Select Answer:

Visualized Solution