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

# Solve dy/dx = y/x + tan(y/x)

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

- Asked in: 2072 Kartik; 2069 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-dy-dx-y-over-x-tan
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation dy/dx = y/x + tan(y/x).

## Short answer

Homogeneous. Put y=vx to get cot v dv = dx/x. Integrate: ln|sin v| = ln|x|+ln c, so sin v = cx. Solution: sin(y/x) = cx.

## Answer

The right side is a function of y/x, so the equation is homogeneous.

#### Substitution
Put y = vx, so dy/dx = v + x dv/dx. The equation becomes

v + x dv/dx = v + tan v
x dv/dx = tan v

#### Separate and integrate
cot v dv = dx/x

ln|sin v| = ln|x| + ln c

sin v = cx

**sin(y/x) = cx**, where c is an arbitrary constant.

(Check: y = x sin⁻¹(cx) gives dy/dx = y/x + cx/√(1 − c²x²) = y/x + tan(y/x) ✓.)

## Check yourself

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

1. Why is dy/dx = y/x + tan(y/x) homogeneous?
   A. It is linear in y
   B. It is exact
   C. The right side depends only on y/x
   D. It has constant coefficients
   Answer: C. A function of y/x alone is the signature of a homogeneous equation. (Key point: Type)

2. With y = vx, the equation becomes:
   A. x dv/dx = tan v
   B. v dx = x dv
   C. x dv/dx = v + tan v
   D. dv/dx = tan v
   Answer: A. v + x dv/dx = v + tan v, so the v terms cancel. (Key point: Substitution)

3. Separating variables gives:
   A. cot v dv = dx/x
   B. cot v dv = x dx
   C. tan v dv = dx/x
   D. sec v dv = dx/x
   Answer: A. x dv/dx = tan v gives dv/tan v = dx/x. (Key point: Separation)

4. The solution is:
   A. cos(y/x) = cx
   B. sin(y/x) = c/x
   C. tan(y/x) = cx
   D. sin(y/x) = cx
   Answer: D. Integrating cot v gives ln|sin v| = ln|x| + ln c. (Key point: Final solution)

## Close variants set in other exams

- [Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-y2-x-minus-e-atan-y) — 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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