---
title: "Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-1-plus-y2-x-minus-e-atan-y"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-1-plus-y2-x-minus-e-atan-y
site: BCT Engineering
---

# Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0

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

- Asked in: 2073 Shrawan; 2070 Chaitra (asked 2 times)
- Video: [Solve: (1+y^2)dx=(tan^-1y-x)dy](https://www.youtube.com/watch?v=NitVyEw4Rrs) — RK Eduworld (Classic)
- 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-y2-x-minus-e-atan-y
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation (1+y²) + (x − e^(tan⁻¹y)) dy/dx = 0.

## Short answer

Rearrange to dx/dy + x/(1+y²) = e^(tan⁻¹y)/(1+y²), linear in x with IF e^(tan⁻¹y). Solving yields x e^(tan⁻¹y) = (1/2)e^(2 tan⁻¹y) + C. Thus x = ½ e^(tan⁻¹y) + C e^(−tan⁻¹y).

## Answer

The equation is not linear in y, but it is linear in x taken as the dependent variable.

#### Rewrite as a linear equation in x
Multiply the equation by dx/dy (using (dy/dx)(dx/dy) = 1):

(1 + y²) dx/dy + x − e^(tan⁻¹y) = 0, so

dx/dy + x/(1 + y²) = e^(tan⁻¹y)/(1 + y²)

This is linear in x with P = 1/(1 + y²) and Q = e^(tan⁻¹y)/(1 + y²).

#### Integrating factor
IF = e^(∫dy/(1+y²)) = e^(tan⁻¹y)

#### Solution
x e^(tan⁻¹y) = ∫ [e^(tan⁻¹y)/(1 + y²)] e^(tan⁻¹y) dy

Put t = tan⁻¹y: ∫ e^(2t) dt = e^(2t)/2 + C.

x e^(tan⁻¹y) = e^(2 tan⁻¹y)/2 + C

**x = ½ e^(tan⁻¹y) + C e^(−tan⁻¹y)**

Check: put x = ½e^t + Ce^(−t), t = tan⁻¹y: dx/dy = (½e^t − Ce^(−t))/(1+y²) and dx/dy + x/(1+y²) = e^t/(1+y²) ✓.

## Check yourself

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

1. How is the equation made linear?
   A. Multiply by an exponential in x
   B. Multiply by dx/dy and treat x as dependent
   C. Substitute y = vx
   D. Divide by y^2 (Bernoulli)
   Answer: B. Using (dy/dx)(dx/dy) = 1 gives an equation linear in x. (Key point: Choice of dependent variable)

2. The linear form is:
   A. dx/dy + x/(1+y^2) = e^(tan^-1 y)/(1+y^2)
   B. dx/dy + (1+y^2)x = e^(tan^-1 y)
   C. dx/dy + x = e^(tan^-1 y)
   D. dx/dy - x/(1+y^2) = e^(tan^-1 y)
   Answer: A. Dividing (1+y^2)dx/dy + x - e^(tan^-1 y) = 0 by (1+y^2) gives this. (Key point: Standard form)

3. With t = tan^-1 y the integral becomes e^(2t) dt, which equals:
   A. e^(2t) + C
   B. 2e^(2t) + C
   C. e^(2t)/2 + C
   D. e^(t)/2 + C
   Answer: C. The integral of e^(2t) is e^(2t)/2. (Key point: Right-hand integral)

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

## Close variants set in other exams

- [Solve dy/dx = y/x + tan(y/x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-dy-dx-y-over-x-tan) — Mathematics I, chapter 4, asked 6×
- [Solve dy/dx − 2y tan x = y² tan x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-bernoulli-2y-tan-x) — Mathematics I, chapter 4, asked 6×
- [Solve (1+x²)dy/dx+y=e^(tan⁻¹x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-x2-dy-dx-etan) — Mathematics I, chapter 4, asked 6×

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