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

NCERT Exemplar · Q76

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

Telangana TsbieMCQ· 1mImportance★★★★★

Appeared in past exams:MHT-CET 2023· Set pcm-2023-05-11-M· 2mexact

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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.

TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$x^2 dy - (xy - y^2) dx = 0$
›Reveal solutionSolution

The equation is homogeneous; the substitution $y = vx$ integrates to $y\log x = x + cy$ — option (C).

Rewrite the equation $x^2\,dy - (xy - y^2)\,dx = 0$ as
$\frac{dy}{dx} = \frac{xy - y^2}{x^2} = \frac{y}{x} - \left(\frac{y}{x}\right)^2.$
This is homogeneous. Put $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ :
$v + x\frac{dv}{dx} = v - v^2 \quad\Rightarrow\quad x\frac{dv}{dx} = -v^2.$
Separate variables:
$\frac{dv}{v^2} = -\frac{dx}{x} \quad\Rightarrow\quad -\frac{1}{v} = -\log x + c_1.$
Since $\dfrac{1}{v} = \dfrac{x}{y}$ : …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\left( x \sin \frac{y}{x} \right) dy = \left( y \sin \frac{y}{x} - x \right) dx$
›Reveal solutionSolution

Homogeneous DE; put $y=vx$ , separate variables, and integrate to get $\cos\!\big(\tfrac{y}{x}\big)=\log|x|+c$ .

Write in standard form.
$\left(x\sin\frac{y}{x}\right)dy=\left(y\sin\frac{y}{x}-x\right)dx \;\Rightarrow\; \frac{dy}{dx}=\frac{y\sin\frac{y}{x}-x}{x\sin\frac{y}{x}}.$
This is homogeneous (degree 0 in $x,y$ ).

Substitute $y=vx$ , so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ and $\dfrac{y}{x}=v$ :
$v+x\frac{dv}{dx}=\frac{vx\sin v-x}{x\sin v}=v-\frac{1}{\sin v}.$
Separate variables. …
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TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
$\dfrac{dy}{dx} = \dfrac{2y^2 + 1}{2y^3 - 4xy + y}$
›Reveal solutionSolution

Treat $x$ as a function of $y$ ; the equation becomes linear in $x$ , giving $4xy^2 + 2x = y^4 + y^2 + c$ — option (A).

Step-by-step solution

Given $\dfrac{dy}{dx} = \dfrac{2y^2 + 1}{2y^3 - 4xy + y}$ , invert to make $x$ the dependent variable:
$\frac{dx}{dy} = \frac{2y^3 - 4xy + y}{2y^2 + 1}.$
Separate the $x$ -term:
$\frac{dx}{dy} + \frac{4y}{2y^2+1}\,x = \frac{2y^3 + y}{2y^2+1} = \frac{y(2y^2+1)}{2y^2+1} = y.$
This is linear in $x$ . Integrating factor:
$\mu = e^{\int \frac{4y}{2y^2+1}\,dy} = e^{\log(2y^2+1)} = 2y^2 + 1.$
Then …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$\dfrac{dy}{dx} = \dfrac{xy + x - 2y - 2}{xy - 2x + y - 2}$
›Reveal solutionSolution

The equation separates into (y−2)/(y+1)dy = (x−2)/(x+1)dx, integrating to x − y + 3log|(y+1)/(x+1)| = c.

Numerator xy+x−2y−2 = (x−2)(y+1); denominator xy−2x+y−2 = (x+1)(y−2). Separating variables: ∫(y−2)/(y+1)dy = ∫(x−2)/(x+1)dx. Each integrand is 1 − 3/(·+1), giving y − 3ln|y+1| = x − 3ln|x+1| + c. Rearrangin …
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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

Step-by-step 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) $\tan(xy)=k$

Unlock everything free for 14 days