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

The solution of the differential equation (dx)/(x) + (d — CBSE Class 12 Mathematics 2023 PYQ

Question

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

(a)

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

(b)

\log x - \log y = C

(c)

xy = C

(d)

x + y = C

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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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This differential equation is solved by separating variables and integrating. The sum of the integrals of $\frac{1}{x}$ and $\frac{1}{y}$ leads to $\log|x| + \log|y| = C'$ , which simplifies to the product $xy = C$ .

The problem asks us to find the solution to the given differential equation. A differential equation relates a function to its derivatives. Our goal is to find the original function or the relationship between the variables that satisfies the equation.

The given equation, $\frac{dx}{x} + \frac{dy}{y} = 0$ , is a first-order differential equation. The key observation here is that the terms involving $x$ are entirely separate from the terms involving $y$ . This type of equation is known as a "variables separable" differential equation. The fundamental idea behind solving such an equation is that if the sum of two exact differentials is zero, then the sum of their integrals must be a constant. Integration allows us to "undo" the differentiation and find the original relationship between $x$ and $y$ .

Here's how we solve it step-by-step:

Recognize the variables separable form. The given differential equation is already in a form where all terms involving $x$ are grouped with $dx$ , and all terms involving $y$ are grouped with $dy$ :

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

This is the simplest form of a variables separable equation, as no rearrangement is needed to separate the variables.

2. Integrate both sides of the equation.

Since the variables are separated, we can integrate each term independently. Remember that the integral of $0$ is an arbitrary constant.

\int \frac{dx}{x} + \int \frac{dy}{y} = \int 0 \, dx

The standard integral for $\frac{1}{u}$ with respect to $u$ is $\log|u|$. Applying this rule to both terms:

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

Here, $C'$ is the constant of integration. It's essential to include this constant because the derivative of any constant is zero, meaning there are infinitely many functions whose derivative is zero. This constant represents the family of all possible solutions.

> [!WARNING]
> Forgetting the constant of integration is a common mistake in differential equations. It leads to an incomplete general solution.

3. Simplify the expression using logarithm properties.

We use the logarithm property that states the sum of logarithms is the logarithm of the product: $\log a + \log b = \log(ab)$ .

Applying this to our equation:

$\log(|x||y|) = C'$ …

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

Showing the 12 most recent of 44 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$
›Reveal solutionSolution
This is a separable first-order ODE. Separate variables, integrate, and simplify to get $\sqrt{y} - \sqrt{x} = C$ , which is option (C).

The key idea: whenever you see $\frac{dy}{dx}$ expressed as a ratio of functions of $y$ and $x$ alone, you can "separate" them — move all $y$ terms to one side and all $x$ terms to the other — then integrate each side independently. That's exactly what we have here: $\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$ .

Notice that $\sqrt{y}$ depends only on $y$ , and $\sqrt{x}$ only on $x$ . So the equation is already in separable form — we just need to rearrange it properly.

Separate the variables. Multiply both sides by $dx$ and divide by $\sqrt{y}$ :

$\frac{dy}{\sqrt{y}} = \frac{dx}{\sqrt{x}}$
This is valid as long as $x > 0$ and $y > 0$ (so the square roots are defined and non-zero).

Integrate both sides. Each side is a standard power integral:

$\int y^{-1/2}\, dy = \int x^{-1/2}\, dx$ $\frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C_1$
which simplifies to:
$2\sqrt{y} = 2\sqrt{x} + C_1$

Simplify the constant. Divide through by 2:

$\sqrt{y} = \sqrt{x} + \frac{C_1}{2}$
Let $C = \frac{C_1}{2}$ (just renaming the arbitrary constant). Then:
$\sqrt{y} - \sqrt{x} = C$
Watch out

A common mistake is to forget the constant of integration or to combine the two integration constants incorrectly. When you integrate both sides, you get a constant on each side — but they can be merged into a single constant. Also, don't lose the factor of 2 from the power rule: $\int u^{-1/2}\, du = 2u^{1/2}$ , not $u^{1/2}$ . …
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CBSE 2026Set 65/3/11 markMCQ
$\frac{dy}{dx} = e^{x+y}$
›Reveal solutionSolution
Separate variables in $\frac{dy}{dx} = e^{x+y}$ to find the general solution $e^x + e^{-y} = C$ ; the particular solution $e^x + e^{-y} = -2$ is impossible because the left side is always positive while the right is negative.

The differential equation $\frac{dy}{dx} = e^{x+y}$ is separable. The key insight is to rewrite the exponential sum in the exponent as a product: $e^{x+y} = e^x \cdot e^y$ . This lets us collect all $x$ -terms with $dx$ and all $y$ -terms with $dy$ .

Solving the differential equation

Separate the variables.

$\frac{dy}{dx} = e^x \cdot e^y$
Rearranging:
$\frac{dy}{e^y} = e^x \, dx$
or equivalently,
$e^{-y} \, dy = e^x \, dx$

Integrate both sides.

$\int e^{-y} \, dy = \int e^x \, dx$
The left side gives $-e^{-y}$ and the right gives $e^x$ :
$-e^{-y} = e^x + C_1$
where $C_1$ is an arbitrary constant.

Rearrange to standard form. Multiply through by $-1$ :

$e^{-y} = -e^x - C_1$
or equivalently,
$e^x + e^{-y} = -C_1$
Renaming $-C_1$ as $C$ (still an arbitrary constant):
$e^x + e^{-y} = C$
This is the general solution, so Reason (R) is true.

Checking the particular solution

Now examine the proposed particular solution $e^x + e^{-y} = -2$ .

For this to be valid, we need $C = -2$ in the general solution. But notice:

$e^x > 0$ for all real $x$ …
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CBSE 2026Set CX1 markMCQ
$\dfrac{dy}{dx}=e^{x+y}$
(a) $e^{-y}=e^{x}+c$
(b) $e^{x}+e^{-y}=c$
(c) $e^{-x}-e^{-y}=c$
(d) $e^{-x}+e^{-y}=c$
›Reveal solutionSolution

The equation is variable-separable; separating and integrating gives $e^{x}+e^{-y}=c$ — option (b).

Why separate? Since $e^{x+y}=e^{x}\,e^{y}$ , the right side factors into an $x$ -part and a $y$ -part, so the variables separate cleanly.

$\dfrac{dy}{dx}=e^{x}\,e^{y}\;\Rightarrow\;e^{-y}\,dy=e^{x}\,dx$

…

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CBSE 2026Set A1 markMCQ
$\frac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = k$
(b) $e^x + e^y = k$
(c) $e^{-x} + e^y = k$
(d) $e^{-x} + e^{-y} = k$
›Reveal solutionSolution

Separate variables in $\tfrac{dy}{dx}=e^{x+y}$ : $\int e^{-y}dy=\int e^x dx\Rightarrow e^x+e^{-y}=k$ .

Write $\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$ and separate:

$e^{-y}\,dy=e^x\,dx$ .

…

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CBSE 2026Set A1 markMCQ
$x\frac{dy}{dx} = \cot y$
(a) $x\cos y = k$
(b) $x\tan y = k$
(c) $x\sec y = k$
(d) $x\sin y = k$
›Reveal solutionSolution

Separate variables: $\int\tan y\,dy=\int\tfrac{dx}{x}\Rightarrow \log|\sec y|=\log|x|+c\Rightarrow x\cos y=k$ .

From $x\dfrac{dy}{dx}=\cot y$ , separate:

$\dfrac{dy}{\cot y}=\dfrac{dx}{x}\Rightarrow \tan y\,dy=\dfrac{dx}{x}$ .

…

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CBSE 2026Set ANNUAL1 mark
$\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$
›Reveal solutionSolution

Separate variables and integrate both sides using the standard integral of $1/(1+t^2)$ .

$\dfrac{dy}{1+y^2} = \dfrac{dx}{1+x^2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = c$
(b) $e^x + e^y = c$
(c) $e^{-x} + e^y = c$
(d) $e^{-x} + e^{-y} = c$
›Reveal solutionSolution

Separate variables using $e^{x+y}=e^x\cdot e^y$ and integrate both sides.

$\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$

Separating variables: $e^{-y}\,dy=e^x\,dx$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the general solution of the differential equation \frac{dy}{dx} = (1 + x^2)(1 + y^2).$
›Reveal solutionSolution
$\tan^{-1}y=x+\dfrac{x^3}{3}+c$ .

Concept. A separable differential equation $\dfrac{dy}{dx}=g(x)h(y)$ is solved by collecting all $y$ -terms on one side and all $x$ -terms on the other, then integrating both sides.

Steps.

$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$ .

Separate: $\dfrac{dy}{1+y^2}=(1+x^2)\,dx$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$(x^{2}+1)\dfrac{dy}{dx}=1,\ y(1)=\dfrac{\pi}{2}$
(a) $y=\tan^{-1}x+\dfrac{\pi}{3}$
(b) $y=\tan^{-1}x$
(c) $y=\tan^{-1}x+\dfrac{\pi}{6}$
(d) $y=\tan^{-1}x+\dfrac{\pi}{4}$
›Reveal solutionSolution

Integrating gives $y=\tan^{-1}x+C$ ; the condition fixes $C=\tfrac{\pi}{4}$ .

Step 1: $(x^{2}+1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x^{2}+1}$ .

Step 2: Integrating, $y=\tan^{-1}x+C$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{y\,dx - x\,dy}{y} = 0$
(a) $y = cx$
(b) $x = cy^2$
(c) $xy = c$
(d) $y = cx^2$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{y\,dx-x\,dy}{y}=0 \implies y\,dx = x\,dy \implies \dfrac{dx}{x}=\dfrac{dy}{y}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^y = C$
(b) $e^{-x} + e^y = C$
(c) $e^x + e^{-y} = C$
(d) $e^{-x} + e^{-y} = C$
›Reveal solutionSolution

Separate variables ( $e^{-y}dy = e^x dx$ ) and integrate both sides.

$\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$

Separating variables:

$e^{-y}\,dy = e^x\,dx$

Integrating both sides: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The general solution of the differential equation dy/dx = e^(x+y) is$
(a) eˣ+eʸ=c
(b) eˣ+e⁻ʸ=c
(c) e⁻ˣ+eʸ=c
(d) e⁻ˣ+e⁻ʸ=c
›Reveal solutionSolution

Separate variables using $e^{x+y}=e^x\cdot e^y$ , then integrate both sides.

Given $\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$ . Separate variables:

$e^{-y}\,dy = e^x\,dx$

Integrate both sides:

$\int e^{-y}\,dy = \int e^x\,dx \;\Rightarrow\; -e^{-y} = e^x + C_1$

…

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Q.The solution of the differential equation dxx+dyy=0\frac{dx}{x} + \frac{dy}{y} = 0xdx​+ydy​=0 is :

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

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