Solution of differential equation $x\,dy-y\,dx=0$ represents: (A) a rectangular hyperbola (B) parabola whose vertex is at origin (C) straight line passing through origin (D) a circle whose centre is at origin

> [!TLDR] > The differential equation $x\,dy - y\,dx = 0$ simplifies to $\frac{dy}{y} = \frac{dx}{x}$, which integrates to $\log|y| = \log|x| + C$, giving $y = kx$. This is a straight line through the origin, so the correct option is **(C)**. The heart of this problem is recognising that the equation $x\,dy - y\,dx = 0$ is a classic first-order differential equation that can be rearranged into a separable form. When you see $x\,dy$ and $y\,dx$ together, your instinct should be to separate the variables — bring all $y$ terms with $dy$ and all $x$ terms with $dx$. Let’s walk through it step by step. 1. **Rewrite the equation** Start with $x\,dy - y\,dx = 0$. Add $y\,dx$ to both sides: $$x\,dy = y\,dx$$ 2. **Separate the variables** Divide both sides by $x y$ (assuming $x \neq 0$, $y \neq 0$ for now): $$\frac{dy}{y} = \frac{dx}{x}$$ This is now a separable differential equation — each side depends only on one variable. 3. **Integrate both sides** $$\int \frac{dy}{y} = \int \frac{dx}{x}$$ The integrals are standard: $$\log|y| = \log|x| + C$$ where $C$ is the constant of integration. 4. **Simplify the result** Exponentiate both sides: $$|y| = e^{\log|x| + C} = e^C \cdot |x|$$ Let $k = \pm e^C$ (absorbing the absolute value), we get: $$y = kx$$ This is the equation of a straight line passing through the origin, with slope $k$. > [!WARNING] > A common mistake is to think the equation represents a circle or hyperbola because of the $x\,dy - y\,dx$ form, which appears in polar coordinate derivatives. But here, the variables separate cleanly — no squares or products of $x$ and $y$ remain after integration. …

Solution of differential equation xdy-ydx=0 represents: (A) a rectangular hyperbola (B) parabola whose vertex

NCERT Exemplar · Q61

x\,dy-y\,dx=0

Sikkim CbseMCQ· 1mImportance★★★★★

Appeared in past exams:KCET 2021· Set A-1· 1mexact

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Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

\frac{dy}{dx} = f(x) \cdot g(y)

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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The differential equation $x\,dy - y\,dx = 0$ simplifies to $\frac{dy}{y} = \frac{dx}{x}$ , which integrates to $\log|y| = \log|x| + C$ , giving $y = kx$ . This is a straight line through the origin, so the correct option is (C).

The heart of this problem is recognising that the equation $x\,dy - y\,dx = 0$ is a classic first-order differential equation that can be rearranged into a separable form. When you see $x\,dy$ and $y\,dx$ together, your instinct should be to separate the variables — bring all $y$ terms with $dy$ and all $x$ terms with $dx$ .

Let’s walk through it step by step.

Rewrite the equation Start with $x\,dy - y\,dx = 0$ . Add $y\,dx$ to both sides:

x\,dy = y\,dx

Separate the variables Divide both sides by $x y$ (assuming $x \neq 0$ , $y \neq 0$ for now):

\frac{dy}{y} = \frac{dx}{x}

This is now a separable differential equation — each side depends only on one variable.

Integrate both sides

\int \frac{dy}{y} = \int \frac{dx}{x}

The integrals are standard:

\log|y| = \log|x| + C

where $C$ is the constant of integration.

Simplify the result Exponentiate both sides:

|y| = e^{\log|x| + C} = e^C \cdot |x|

Let $k = \pm e^C$ (absorbing the absolute value), we get:

y = kx

This is the equation of a straight line passing through the origin, with slope $k$ .

Watch out

A common mistake is to think the equation represents a circle or hyperbola because of the $x\,dy - y\,dx$ form, which appears in polar coordinate derivatives. But here, the variables separate cleanly — no squares or products of $x$ and $y$ remain after integration. …

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Q.Solution of differential equation $x\,dy-y\,dx=0$ represents:
(A) a rectangular hyperbola
(B) parabola whose vertex is at origin
(C) straight line passing through origin
(D) a circle whose centre is at origin

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Method: Solving $x\,dy-y\,dx=0$ and Naming the Curve

Steps

Common Mistakes

Q.Solution of differential equation x dy−y dx=0x\,dy-y\,dx=0xdy−ydx=0 represents: (A) a rectangular hyperbola (B) parabola whose vertex is at origin (C) straight line passing through origin (D) a circle whose centre is at origin

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Q.Solution of differential equation $x\,dy-y\,dx=0$ represents:
(A) a rectangular hyperbola
(B) parabola whose vertex is at origin
(C) straight line passing through origin
(D) a circle whose centre is at origin

Unlock everything free for 14 days