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Homogeneous Equations

Mediummathematics

A curve satisfies the homogeneous differential equation \frac{dy}{dx} = \frac{x+y}{x}, and we wish to express its general solution using a suitable substitution for the ratio y over x.

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About This Question

Subject
mathematics
Chapter
differential equations
Topic
homogeneous equations
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillhomogeneous equationsubstitution y equals vxseparable reductiongeneral solution

Solution

Correct Answer:

A homogeneous differential equation is one where the right-hand side can be written purely as a function of the ratio y/x, which signals the substitution y = vx so that \frac{dy}{dx} = v + x\frac{dv}{dx}. The power of this substitution is that it converts a homogeneous equation into a separable one in v and x. Rewriting the right side, \frac{x+y}{x} = 1 + \frac{y}{x} = 1 + v. Substituting gives v + x\frac{dv}{dx} = 1 + v, so the v terms cancel and x\frac{dv}{dx} = 1. Separating, dv = \frac{dx}{x}, and integrating gives v = \ln|x| + C. Restoring v = y/x produces \frac{y}{x} = \ln|x| + C, hence y = x\ln|x| + Cx. Option y = x^2 + C ignores the homogeneous structure entirely. Option y = Ce^x is the solution of a constant-coefficient equation, not this one. Option y = \ln|x| + C forgets to multiply back by x when undoing the substitution. This is the textbook JEE Advanced homogeneous-equation procedure. As a final plausibility check, differentiating y = x\ln|x| + Cx gives y' = \ln|x| + 1 + C, and \frac{x+y}{x} = 1 + \ln|x| + C, which match exactly.

This medium difficulty mathematics question is from the chapter differential equations, covering the topic of homogeneous equations. It appeared in the 2025 exam.

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