Find the general solution of the differential equation $(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0$.

Find the general solution of the differential equation (1+y²)+(x-e^tan⁻¹y)(dy)/(dx)=0.

NCERT Exemplar · Q17

(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0

Odisha ChseShort· 3mImportance★★★★★

Appeared in past exams:AP EAPCET 2026· Set eng-2026-05-15-AN· 1mexactMHT-CET 2025· Set pcm-2025-04-20-E· 2mexactCOMEDK 2021· Set 2021· 1mexactMHT-CET 2020· Set pcm-2020-10-19-E· 2mexact

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

y\cdot x = \int x\cdot x\,dx + C = \frac{x^3}{3} + C.

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This is a first-order linear differential equation in $x$ as a function of $y$ . By rewriting it as $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{\tan^{-1}y}}{1+y^2}$ and using the integrating factor $e^{\tan^{-1}y}$ , the general solution is $x\,e^{\tan^{-1}y} = \frac{1}{2}e^{2\tan^{-1}y} + C$ .

The given equation is:

(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0.

At first glance, this looks like a standard first-order differential equation in $y$ as a function of $x$ . But notice the term $e^{\tan^{-1}y}$ — that inverse tangent suggests that treating $y$ as the independent variable might be far cleaner. When you see a complicated function of $y$ alongside a simple $x$ , swapping roles often simplifies the algebra.

Why treat $x$ as a function of $y$ ?

If we try to write it as $\frac{dy}{dx} = -\frac{1+y^2}{x - e^{\tan^{-1}y}}$ , the right-hand side is messy — $x$ appears in the denominator, and the expression isn't linear in $y$ . But if we instead write $\frac{dx}{dy}$ , we get:

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

That is:

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

Now this is a first-order linear differential equation in $x$ with independent variable $y$ . That's a form we know exactly how to solve.

Step-by-step solution

1. Identify the standard linear form

We have:

\frac{dx}{dy} + P(y)\,x = Q(y),

where

P(y) = \frac{1}{1+y^2}, \quad Q(y) = \frac{e^{\tan^{-1}y}}{1+y^2}.

2. Find the integrating factor

The integrating factor $\mu(y)$ is given by:

\mu(y) = e^{\int P(y)\,dy} = e^{\int \frac{1}{1+y^2}\,dy}.

The integral $\int \frac{1}{1+y^2}\,dy = \tan^{-1}y + C$ . We only need one antiderivative, so:

\mu(y) = e^{\tan^{-1}y}.

Tip

The integrating factor $e^{\tan^{-1}y}$ is exactly the same exponential that appears in $Q(y)$ . This is no coincidence — it's a sign the problem was designed to cancel nicely.

3. Multiply through by the integrating factor

Multiply the entire differential equation by $e^{\tan^{-1}y}$ :

e^{\tan^{-1}y}\frac{dx}{dy} + \frac{e^{\tan^{-1}y}}{1+y^2}\,x = \frac{e^{2\tan^{-1}y}}{1+y^2}.

The left-hand side is now the derivative of $x \cdot e^{\tan^{-1}y}$ with respect to $y$ , because:

\frac{d}{dy}\left(x\,e^{\tan^{-1}y}\right) = \frac{dx}{dy}\,e^{\tan^{-1}y} + x\cdot e^{\tan^{-1}y}\cdot\frac{1}{1+y^2}.

So the equation becomes: …

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Q.Find the general solution of the differential equation $(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0$ .

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Why treat $x$ as a function of $y$ ?

Step-by-step solution

Method: Linear in $x$ with an inverse-tangent integrating factor

Steps

Common Mistakes

Q.Find the general solution of the differential equation (1+y2)+(x−etan⁡−1y)dydx=0(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0(1+y2)+(x−etan−1y)dxdy​=0.

Why treat xxx as a function of yyy?

Step-by-step solution

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Q.Find the general solution of the differential equation $(1+y^2)+(x-e^{\tan^{-1}y})\frac{dy}{dx}=0$ .

Why treat $x$ as a function of $y$ ?

Unlock everything free for 14 days