Question:

The particular solution of the differential equation $(1+y^{2})dx-xy \, dy=0$ at $x=1$, $y=0$, represents

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Logic Tip: Equations of the form $x^2 + y^2 = a^2$ represent circles, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ represent ellipses, and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ represent hyperbolas. Since the signs on the $x^2$ and $y^2$ terms are opposite, it must be a hyperbola.
Updated On: Apr 28, 2026
  • circle
  • pair of straight lines
  • hyperbola
  • ellipse
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The Correct Option is C

Solution and Explanation

Concept:
This is a first-order, first-degree differential equation that can be solved using the variable separable method. Once the general solution is obtained by integrating both sides, we use the initial conditions to find the constant of integration and identify the resulting conic section.
Step 1: Separate the variables x and y.
Given the differential equation: $$(1+y^2)dx - xy \, dy = 0$$ $$(1+y^2)dx = xy \, dy$$ Divide both sides by $x(1+y^2)$ to group $x$'s with $dx$ and $y$'s with $dy$: $$\frac{1}{x} dx = \frac{y}{1+y^2} dy$$
Step 2: Integrate both sides.
$$\int \frac{1}{x} dx = \int \frac{y}{1+y^2} dy$$ For the left side, the integral is simply $\log|x|$. For the right side, let $u = 1+y^2$, then $du = 2y \, dy$, so $y \, dy = \frac{du}{2}$. The integral becomes $\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log|u| = \frac{1}{2}\log(1+y^2)$. $$\log|x| = \frac{1}{2}\log(1+y^2) + c$$
Step 3: Apply initial conditions to find the particular solution.
We are given $x=1$ when $y=0$. Substitute these into the general solution to find $c$: $$\log(1) = \frac{1}{2}\log(1+0^2) + c$$ $$0 = \frac{1}{2}\log(1) + c$$ $$0 = 0 + c \implies c = 0$$ Substitute $c=0$ back into the equation: $$\log x = \frac{1}{2}\log(1+y^2)$$
Step 4: Simplify the equation to identify the conic section.
Multiply by 2 to remove the fraction: $$2\log x = \log(1+y^2)$$ Using the logarithm power rule $\log(a^b) = b\log a$: $$\log(x^2) = \log(1+y^2)$$ Remove the logarithms from both sides: $$x^2 = 1 + y^2$$ $$x^2 - y^2 = 1$$ This equation matches the standard form of a rectangular hyperbola.
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