Question 293 marksShort AnswerSection C

(a) Find the general solution of the following differential equation: x2dydx=x2+xy+y2x^2 \frac{dy}{dx} = x^2 + xy + y^2 OR (b) Find the particular solution of the differential equation xydydx=(x+2)(y+2)xy \frac{dy}{dx} = (x + 2)(y + 2), given that y(1)=−1y(1) = -1.

Differential Equations
Differential Equations — Homogeneous and Variable Separable Forms

Official answer

Option (a): Dividing by x2x^2 gives dydx=1+y/x+(y/x)2\frac{dy}{dx} = 1 + y/x + (y/x)^2. Substituting y=vxy=vx reduces this to the separable form dv1+v2=dxx\frac{dv}{1+v^2} = \frac{dx}{x}, integrating to tan⁡−1v=ln⁡∣x∣+C\tan^{-1}v = \ln|x| + C. The general solution is tan⁡−1(y/x)=ln⁡∣x∣+C\tan^{-1}(y/x) = \ln|x| + C.


Option (b): Separating variables gives [1−2y+2]dy=[1+2x]dx\left[1 - \frac{2}{y+2}\right]dy = \left[1 + \frac{2}{x}\right]dx. Integrating: y−2ln⁡∣y+2∣=x+2ln⁡∣x∣+Cy - 2\ln|y+2| = x + 2\ln|x| + C. Using y(1)=−1y(1) = -1: −1−0=1+0+C  ⟹  C=−2-1 - 0 = 1 + 0 + C \implies C = -2. The particular solution is y−x+2=2ln⁡∣x(y+2)∣y - x + 2 = 2\ln|x(y+2)|.

Answer keywords:homogeneous differential equationvariable separablegeneral solutionparticular solutioninitial conditionsubstitution y=vx

Marking scheme

  1. 1.1 mark: correct identification of equation type and substitution/separation setup (y=vxy=vx for a; splitting fractions for b).
  2. 2.1 mark: correct integration to reach tan⁡−1v=ln⁡∣x∣+C\tan^{-1}v = \ln|x| + C (a) or y−2ln⁡∣y+2∣=x+2ln⁡∣x∣+Cy - 2\ln|y+2| = x + 2\ln|x| + C (b).
  3. 3.1 mark: correct final general solution in terms of y/xy/x (a), or correct evaluation of C using the initial condition and final particular solution (b).

Hint

For (a), divide by x2x^2 and substitute y=vxy=vx (homogeneous equation). For (b), separate variables using yy+2=1−2y+2\frac{y}{y+2} = 1 - \frac{2}{y+2} and x+2x=1+2x\frac{x+2}{x} = 1 + \frac{2}{x}.

Quick oral answer

Part (a) uses the substitution y=vxy=vx to get tan⁡−1(y/x)=ln⁡∣x∣+C\tan^{-1}(y/x) = \ln|x| + C; part (b) separates variables to get y−2ln⁡∣y+2∣=x+2ln⁡∣x∣+Cy - 2\ln|y+2| = x + 2\ln|x| + C, and using y(1)=−1y(1) = -1 gives C=−2C = -2.

Common mistakes

  1. 1.Forgetting the product rule term x(dv/dx)x(dv/dx) when substituting y=vxy=vx in part (a).
  2. 2.Integrating yy+2\frac{y}{y+2} or x+2x\frac{x+2}{x} directly without splitting into a constant plus a proper fraction in part (b), leading to an incorrect logarithmic term.
  3. 3.Applying the initial condition y(1)=−1y(1) = -1 before simplifying the general solution fully, causing an incorrect value of the constant C.

Analysis and explanation

This OR question tests the two most common first-order differential equation techniques: the homogeneous substitution method and the variable separable method.


Concept: Part (a)'s equation is homogeneous because every term (x2,xy,y2)(x^2, xy, y^2) has the same total degree; dividing by x2x^2 expresses it purely in terms of v=y/xv=y/x, converting it into a separable equation. Part (b) is directly separable once each side is split into a constant plus a proper fraction (1−2y+21 - \frac{2}{y+2} and 1+2x1 + \frac{2}{x}), since yy+2\frac{y}{y+2} and x+2x\frac{x+2}{x} are not directly integrable as written.


Exam trap: In part (a), a frequent error is forgetting the product rule when substituting dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}. In part (b), students often try to integrate yy+2\frac{y}{y+2} directly without splitting it first, giving an incorrect log form, or apply the initial condition before fully simplifying.


Real-world link: Homogeneous and variable-separable differential equations model population growth, radioactive decay, and rate-based mixing problems where the rate of change depends proportionally on the current state.

Interesting facts

Homogeneous differential equations of this type were systematically studied by Johann Bernoulli in the late 17th century as part of the early development of calculus-based modelling.

Variable-separable equations of the form in part (b) are the mathematical basis of the logistic growth model used in population biology and epidemiology.

Question details

Q29 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Differential Equations
Topic
Differential Equations — Homogeneous and Variable Separable Forms
Bloom's level
apply
Syllabus topic
Unit 4: Differential Equations — Homogeneous Differential Equations and Variable Separable Form
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you recognise a homogeneous differential equation?

A differential equation dydx=f(x,y)\frac{dy}{dx} = f(x,y) is homogeneous if f(x,y) can be written as a function of y/xy/x alone, i.e. every term has the same total degree in x and y, as with x2,xy,y2x^2, xy, y^2 in part (a).

Why split yy+2\frac{y}{y+2} into 1−2y+21 - \frac{2}{y+2} in part (b)?

Direct integration of yy+2\frac{y}{y+2} is not a standard form; rewriting it as 1 minus a proper fraction converts it into two integrable standard terms, yy and ln⁡∣y+2∣\ln|y+2|.