Question

Consider a curve $f(x, y) = 0$, satisfying $xdx = (\frac{x^2}{y}-y^3)dy, y \neq 0$, such that $f(0, 1) = 0$. Then which of the following option(s) is/are correct?

• A. Area bounded by $f(x, y) = 0$ is $\frac{4}{3}$ square units
• B. The curve $f(x, y) = 0$ is is symmetrical about both x and y axis
• C. $f(\frac{1}{2}, \frac{1}{\sqrt{2}}) = 0$
• D. $f(\frac{1}{\sqrt{2}}, \frac{1}{2}) = 0$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The given differential equation is $xdx = (\frac{x^2}{y}-y^3)dy$. Rewriting it as $xy \, dx + (y^4 - x^2) \, dy = 0$. This is not an exact equation. An integrating factor $\mu(y) = y^{-3}$ is found, making the equation $xy^{-2} \, dx + (y - x^2y^{-3}) \, dy = 0$. This equation is exact. Integrating $xy^{-2}$ with respect to $x$ gives $f(x, y) = \frac{x^2}{2y^2} + g(y)$. Differentiating with respect to $y$ and comparing with $N'(x, y) = y - x^2y^{-3}$ yields $g'(y) = y$, so $g(y) = \frac{y^2}{2}$. Thus, $f(x, y) = \frac{x^2}{2y^2} + \frac{y^2}{2} + C$. Using the condition $f(0, 1) = 0$, we get $C = -\frac{1}{2}$. The curve equation is $\frac{x^2}{2y^2} + \frac{y^2}{2} - \frac{1}{2} = 0$, which simplifies to $x^2 = y^2(1 - y^2)$.

Option A: The area bounded by $x^2 = y^2(1 - y^2)$ is calculated by integrating $2x$ with respect to $y$ from $y=-1$ to $y=1$. The area of one loop (e.g., for $y \in [0, 1]$) is $2 \int_{0}^{1} y\sqrt{1 - y^2} \, dy = \frac{2}{3}$. Due to symmetry, the total area is $\frac{4}{3}$ square units.

Option B: The equation $x^2 = y^2(1 - y^2)$ is symmetric with respect to the y-axis (replacing $x$ with $-x$) and the x-axis (replacing $y$ with $-y$).

Option C: For $f(\frac{1}{2}, \frac{1}{\sqrt{2}}) = 0$, we check if $(\frac{1}{2}, \frac{1}{\sqrt{2}})$ satisfies $x^2 = y^2(1 - y^2)$. $(\frac{1}{2})^2 = \frac{1}{4}$ and $(\frac{1}{\sqrt{2}})^2(1 - (\frac{1}{\sqrt{2}})^2) = \frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{4}$. The point lies on the curve.

Option D: For $f(\frac{1}{\sqrt{2}}, \frac{1}{2}) = 0$, we check if $(\frac{1}{\sqrt{2}}, \frac{1}{2})$ satisfies $x^2 = y^2(1 - y^2)$. $(\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$ and $(\frac{1}{2})^2(1 - (\frac{1}{2})^2) = \frac{1}{4}(1 - \frac{1}{4}) = \frac{3}{16}$. Since $\frac{1}{2} \neq \frac{3}{16}$, the point does not lie on the curve.