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

The solution of the differential equation (dy)/(dx)=(1+y²)/(1+x²) is: (A) y=tan⁻¹x (B) y-x=k(1+xy) (C) x=tan⁻¹

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

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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 is a separable first-order ODE. By separating variables and integrating, we get $\tan^{-1}y = \tan^{-1}x + C$ , which simplifies to $y - x = k(1 + xy)$ . The correct option is (B).

The key insight here is that the equation is separable — the right-hand side is a product of a function of $x$ and a function of $y$ . That means we can rearrange it so that all $y$ terms are on one side and all $x$ terms on the other, then integrate each side independently. This is the most direct method for first-order ODEs of this form.

Let’s work through it step by step.

Separate the variables. The given equation is

\frac{dy}{dx} = \frac{1+y^2}{1+x^2}.

Multiply both sides by $dx$ and by $1+x^2$ , then divide by $1+y^2$ :

\frac{dy}{1+y^2} = \frac{dx}{1+x^2}.

Now the variables are isolated — $y$ on the left, $x$ on the right.

Integrate both sides. The integrals are standard:

\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}.

Each integral gives an inverse tangent:

\tan^{-1} y = \tan^{-1} x + C,

where $C$ is the constant of integration.

Tip

Remember: $\int \frac{du}{1+u^2} = \tan^{-1} u + \text{constant}$ . This is one of the most common integrals in differential equations — commit it to memory.

Rewrite the constant in a convenient form. Let $C = \tan^{-1} k$ , where $k$ is a new constant. This is allowed because $\tan^{-1}$ maps $\mathbb{R}$ to $(-\pi/2, \pi/2)$ , so any real $C$ can be expressed this way. Then:

\tan^{-1} y = \tan^{-1} x + \tan^{-1} k.

Apply the tangent addition formula. Take $\tan$ of both sides:

y = \tan(\tan^{-1} x + \tan^{-1} k).

Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ , we get:

$y = \frac{x + k}{1 - xk}.$ …

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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 20241 markMCQ
$\frac{dy}{dx} = 1 - x + y - xy$
›Reveal solutionSolution
The equation $\frac{dy}{dx} = 1 - x + y - xy$ can be factored and solved by separation of variables. The correct solution is $\log |1 + y| = x - \frac{x^2}{2} + c$ , which corresponds to option (A).

The key to solving this is noticing that the right-hand side can be grouped into factors, each depending on only one variable. That’s the signal to use Separation of Variables — a method where we rewrite the equation so that all $y$ terms are on one side and all $x$ terms on the other, then integrate both sides.

Let’s see why this works here.

Factor the right-hand side The given equation is:

$\frac{dy}{dx} = 1 - x + y - xy$
Group the terms cleverly:
$\frac{dy}{dx} = (1 - x) + y(1 - x)$
Factor out $(1 - x)$ :
$\frac{dy}{dx} = (1 - x)(1 + y)$
Now the derivative equals a product of a function of $x$ and a function of $y$ . That’s the perfect setup for separation.

Separate the variables Multiply both sides by $dx$ and divide by $(1 + y)$ (assuming $1 + y \neq 0$ for now):

$\frac{dy}{1 + y} = (1 - x)\,dx$
The variables are now isolated — $y$ on the left, $x$ on the right.

Integrate both sides

$\int \frac{dy}{1 + y} = \int (1 - x)\,dx$
The left integral is a standard logarithmic form:
$\log |1 + y| = x - \frac{x^2}{2} + C$
where $C$ is the constant of integration. …
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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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Q.The solution of the differential equation $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ is:
(A) $y=\tan^{-1}x$
(B) $y-x=k(1+xy)$
(C) $x=\tan^{-1}y$
(D) $\tan(xy)=k$

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: Separating $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ and Finishing with the Tangent Identity

Steps

Common Mistakes

Solving the differential equation

Checking the particular solution

Q.The solution of the differential equation dydx=1+y21+x2\frac{dy}{dx}=\frac{1+y^2}{1+x^2}dxdy​=1+x21+y2​ is: (A) y=tan⁡−1xy=\tan^{-1}xy=tan−1x (B) y−x=k(1+xy)y-x=k(1+xy)y−x=k(1+xy) (C) x=tan⁡−1yx=\tan^{-1}yx=tan−1y (D) tan⁡(xy)=k\tan(xy)=ktan(xy)=k

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Q.The solution of the differential equation $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ is:
(A) $y=\tan^{-1}x$
(B) $y-x=k(1+xy)$
(C) $x=\tan^{-1}y$
(D) $\tan(xy)=k$

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