The solution of the differential equation $\dfrac{dx}{x} + \dfrac{dy}{y} = 0$ is : (a) $\dfrac{1}{x} + \dfrac{1}{y} = C$ (b) $xy = C$ (c) $\log x \log y = C$ (d) $x + y = C$

> [!TLDR] > Integrating $\dfrac{dx}{x}+\dfrac{dy}{y}=0$ gives $\ln x+\ln y=k$, hence $xy=C$. > [!FORMULA] > $\int\dfrac{dt}{t}=\ln|t|+\text{const}$; a sum of exact differentials integrates term by term. 1. Integrate each term: $\int\dfrac{dx}{x}+\int\dfrac{dy}{y}=\int 0$. 2. This gives $\ln x+\ln y=k$ (a constant). …

The solution of the differential equation (dx)/(x) + (dy)/(y) = 0 is : (a) (1)/(x) + (1)/(y) = C (b) xy = C (c

Question

\dfrac{dx}{x} + \dfrac{dy}{y} = 0

(a)

\dfrac{1}{x} + \dfrac{1}{y} = C

(b)

xy = C

(c)

\log x \log y = C

(d)

x + y = C

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

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Concept understanding — Solution Verification

Solution Verification — The Intuition

When you solve an equation, you perform operations on both sides to isolate the variable. But here's the catch: some operations are reversible (like adding 5 or multiplying by 3) and some are not (like squaring both sides or multiplying by an expression that could be zero). When you use a non-reversible step, you might introduce extra "solutions" that don't actually satisfy the original equation. These are called extraneous solutions.

Solution verification is the simple act of plugging your final answers back into the original equation to check which ones actually work. It's your safety net against these fake solutions.

Watch out

Never check against a simplified version of the equation — only the original one. A simplified version might have already lost information (like a denominator that was cancelled), so it could still accept an extraneous solution.

The Precise Statement

Definition:

Given an equation $E(x) = 0$ and a set of candidate solutions $S = \{x_1, x_2, \dots, x_n\}$ obtained by algebraic manipulation, solution verification is the process of substituting each $x_i$ into the original equation $E(x) = 0$ and retaining only those $x_i$ for which the equality holds true.

Why it's necessary:

If at any step you performed an operation that is not bijective (one-to-one and onto) on the domain of the equation, the solution set of the transformed equation may be a superset of the solution set of the original equation. Common culprits:

Squaring both sides: $x = 2 \implies x^2 = 4$ , but $x^2 = 4$ also gives $x = -2$ , which is extraneous.
Multiplying by an expression that could be zero: $\frac{x}{x-1} = 2 \implies x = 2(x-1)$ loses the restriction $x \neq 1$ .
Taking logarithms or exponentials without domain checks.

A Worked Example

Solve: $\sqrt{x+6} = x$

Step 1 — Square both sides:

$(\sqrt{x+6})^2 = x^2 \implies x+6 = x^2 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$

So $x = 3$ or $x = -2$ .

Step 2 — Verify in the original equation:

For $x = 3$ : $\sqrt{3+6} = \sqrt{9} = 3$ , and RHS is $3$ . Works.
For $x = -2$ : $\sqrt{-2+6} = \sqrt{4} = 2$ , but RHS is $-2$ . $2 \neq -2$ . Extraneous.

Final answer: $x = 3$ only. …

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Integrating $\dfrac{dx}{x}+\dfrac{dy}{y}=0$ gives $\ln x+\ln y=k$ , hence $xy=C$ .

\int\dfrac{dt}{t}=\ln|t|+\text{const}

Integrate each term: $\int\dfrac{dx}{x}+\int\dfrac{dy}{y}=\int 0$ .
This gives $\ln x+\ln y=k$ (a constant). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/W1XZY/41 markMCQ
$y = mx$
›Reveal solutionSolution
$y=mx\Rightarrow \tfrac{dy}{dx}=m=\tfrac{y}{x}\Rightarrow x\tfrac{dy}{dx}-y=0$ ; the Reason states exactly this method, so it explains the Assertion.

$To form a differential equation from a family with one arbitrary constant: differentiate once and eliminate the constant.$

Start with the family: $y=mx$ , where $m$ is the arbitrary constant.

Differentiate with respect to $x$ : $\dfrac{dy}{dx}=m$ .

Eliminate $m$ : from $y=mx$ , $m=\dfrac{y}{x}$ ; substituting, $\dfrac{dy}{dx}=\dfrac{y}{x}$ , i.e. $x\dfrac{dy}{dx}-y=0$ . …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\dfrac{dx}{x} + \dfrac{dy}{y} = 0$
(a) $\dfrac{1}{x} + \dfrac{1}{y} = C$
(b) $xy = C$
(c) $\log x \log y = C$
(d) $x + y = C$
›Reveal solutionSolution
Integrating $\dfrac{dx}{x}+\dfrac{dy}{y}=0$ gives $\ln x+\ln y=k$ , hence $xy=C$ .

$\int\dfrac{dt}{t}=\ln|t|+\text{const}$

Integrate each term: $\int\dfrac{dx}{x}+\int\dfrac{dy}{y}=\int 0$ .

This gives $\ln x+\ln y=k$ (a constant). …
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Q.The solution of the differential equation dxx+dyy=0\dfrac{dx}{x} + \dfrac{dy}{y} = 0xdx​+ydy​=0 is :

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Q.The solution of the differential equation $\dfrac{dx}{x} + \dfrac{dy}{y} = 0$ is :

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