The general solution of the differential equation $x \text{ dy} - (1 + x^2) \text{ dx} = \text{ dx}$ is : (A) $y = 2x + \frac{x^3}{3} + C$ (B) $y = 2 \log x + \frac{x^3}{3} + C$ (C) $y = \frac{x^2}{2} + C$ (D) $y = 2 \log x + \frac{x^2}{2} + C$

> [!TLDR] > The given equation simplifies to a separable first-order ODE. Integrating both sides yields $y = \log x + \frac{x^2}{2} + C$, which matches option **(D)** after adjusting for the constant factor. We start with the differential equation: $$ x \, dy - (1 + x^2) \, dx = dx $$ The first step is always to see what form we have. This is a first-order ODE, and the variables $x$ and $y$ are not yet separated. But notice: the term $dy$ appears only once, multiplied by $x$, and all $dx$ terms can be collected on the other side. That suggests we can rearrange it into a **separable** form — where all $y$ terms are on one side and all $x$ terms on the other. Let’s work through it step by step. 1. **Combine the $dx$ terms** Bring the $dx$ term from the right to the left: $$ x \, dy - (1 + x^2) \, dx - dx = 0 $$ Simplify the $dx$ coefficients: $$ x \, dy - (1 + x^2 + 1) \, dx = 0 $$ $$ x \, dy - (x^2 + 2) \, dx = 0 $$ 2. **Separate the variables** Move the $dx$ term to the other side: $$ x \, dy = (x^2 + 2) \, dx $$ Now divide both sides by $x$ (assuming $x \neq 0$; we handle the constant solution separately later): $$ dy = \frac{x^2 + 2}{x} \, dx $$ 3. **Simplify the integrand** Break the fraction: $$ \frac{x^2 + 2}{x} = x + \frac{2}{x} $$ So we have: $$ dy = \left( x + \frac{2}{x} \right) dx $$ 4. **Integrate both sides** The left side integrates directly to $y$. The right side is a sum of simple terms: $$ \int dy = \int \left( x + \frac{2}{x} \right) dx $$ $$ y = \frac{x^2}{2} + 2 \ln |x| + C $$ Since the options use $\log x$ (common notation for natural log in Indian exams), we write: …

The general solution of the differential equation x dy — CBSE Class 12 Mathematics 2023 PYQ

Question

x \text{ dy} - (1 + x^2) \text{ dx} = \text{ dx}

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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 given equation simplifies to a separable first-order ODE. Integrating both sides yields $y = \log x + \frac{x^2}{2} + C$ , which matches option (D) after adjusting for the constant factor.

We start with the differential equation:

x \, dy - (1 + x^2) \, dx = dx

The first step is always to see what form we have. This is a first-order ODE, and the variables $x$ and $y$ are not yet separated. But notice: the term $dy$ appears only once, multiplied by $x$ , and all $dx$ terms can be collected on the other side. That suggests we can rearrange it into a separable form — where all $y$ terms are on one side and all $x$ terms on the other.

Let’s work through it step by step.

Combine the $dx$ terms Bring the $dx$ term from the right to the left:

x \, dy - (1 + x^2) \, dx - dx = 0

Simplify the $dx$ coefficients:

x \, dy - (1 + x^2 + 1) \, dx = 0

x \, dy - (x^2 + 2) \, dx = 0

Separate the variables Move the $dx$ term to the other side:

x \, dy = (x^2 + 2) \, dx

Now divide both sides by $x$ (assuming $x \neq 0$ ; we handle the constant solution separately later):

dy = \frac{x^2 + 2}{x} \, dx

Simplify the integrand Break the fraction:

\frac{x^2 + 2}{x} = x + \frac{2}{x}

So we have:

dy = \left( x + \frac{2}{x} \right) dx

Integrate both sides The left side integrates directly to $y$ . The right side is a sum of simple terms:

\int dy = \int \left( x + \frac{2}{x} \right) dx

y = \frac{x^2}{2} + 2 \ln |x| + C

Since the options use $\log x$ (common notation for natural log in Indian exams), we write: …

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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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