---
title: "Solve (1+x²)dy/dx+y=e^(tan⁻¹x)"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-1-plus-x2-dy-dx-etan"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-1-plus-x2-dy-dx-etan
site: BCT Engineering
---

# Solve (1+x²)dy/dx+y=e^(tan⁻¹x)

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2067 Asar
- Video: [(1 + x2) dy/dx + y = e^(tan-1 x) @EAG](https://www.youtube.com/watch?v=BCxMJHVeBgw) — Elite Academy Guntur
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-1-plus-x2-dy-dx-etan
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation (1 + x²) dy/dx + y = e^(tan⁻¹x).

## Short answer

Rewrite as linear in y with integrating factor e^{tan⁻¹x}. Multiply, substitute t=tan⁻¹x, and integrate to obtain y = (1/2)e^{tan⁻¹x} + c e^{-tan⁻¹x}.

## Answer

(1 + x²) dy/dx + y = e^(tan⁻¹x)

#### Standard linear form
dy/dx + y/(1 + x²) = e^(tan⁻¹x)/(1 + x²)

P = 1/(1 + x²), Q = e^(tan⁻¹x)/(1 + x²).

#### Integrating factor
IF = e^(∫dx/(1 + x²)) = e^(tan⁻¹x)

#### Solution
y e^(tan⁻¹x) = ∫ [e^(tan⁻¹x)/(1 + x²)] e^(tan⁻¹x) dx

Put t = tan⁻¹x, dt = dx/(1 + x²):

∫ e^(2t) dt = e^(2t)/2 + c

y e^(tan⁻¹x) = ½ e^(2 tan⁻¹x) + c

**y = ½ e^(tan⁻¹x) + c e^(−tan⁻¹x)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The linear form has Q equal to:
   A. e^(tan^-1 x)/(1 + x^2)
   B. e^(tan^-1 x)
   C. (1 + x^2)e^(tan^-1 x)
   D. 1/(1 + x^2)
   Answer: A. Divide by (1 + x^2). (Key point: Standard form)

2. The integrating factor is:
   A. 1 + x^2
   B. tan^-1 x
   C. e^x
   D. e^(tan^-1 x)
   Answer: D. Integral of dx/(1+x^2) = tan^-1 x. (Key point: Integrating factor)

3. With t = tan^-1 x, the integral reduces to:
   A. integral of dt/t
   B. integral of e^t dt
   C. integral of e^(2t) dt
   D. integral of t e^t dt
   Answer: C. IF × Q dx = e^(2t) dt. (Key point: Substitution)

4. The solution is:
   A. y = (1/2)e^(tan^-1 x) + c e^(-tan^-1 x)
   B. y = (1/2)e^(2 tan^-1 x) + c
   C. y = (1/2)e^(-tan^-1 x) + c e^(tan^-1 x)
   D. y = e^(tan^-1 x) + c
   Answer: A. Divide y e^t = e^(2t)/2 + c by e^t. (Key point: Final solution)

## Close variants set in other exams

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