Homogeneous Equations and Reduction

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2020)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

This atom appeared once (2020) in Section A. UPSC tests whether you can recognise a homogeneous ODE (or one reducible to homogeneous form) and execute the standard substitution cleanly. Most of the marks are awarded for the substitution and subsequent separation and integration — errors at any one step cascade, so accuracy at each stage matters.

Minimum Theory

Homogeneous ODEs

A first-order ODE $\dfrac{dy}{dx} = f(x,y)$ is homogeneous of degree zero if $f(tx,ty) = f(x,y)$ for all $t \neq 0$, i.e., $f$ depends only on $y/x$.

Standard form: $\dfrac{dy}{dx} = g\!\left(\dfrac{y}{x}\right)$.

Substitution: Set $v = \dfrac{y}{x}$, so $y = vx$ and

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

The ODE becomes:

$v + x\frac{dv}{dx} = g(v) \implies x\frac{dv}{dx} = g(v) - v$

This is separable:

$\int \frac{dv}{g(v)-v} = \int \frac{dx}{x} = \ln|x| + C$

Equations reducible to homogeneous form

The ODE $\dfrac{dy}{dx} = \dfrac{ax+by+c}{dx+ey+f}$ with $c^2 + f^2 \neq 0$ is not immediately homogeneous.

Case 1: $ae \neq bd$ (lines intersect).

Find the intersection $(h,k)$ of $ax+by+c = 0$ and $dx+ey+f = 0$:

$ah + bk + c = 0, \quad dh + ek + f = 0$

Substitute $X = x - h$, $Y = y - k$. Then $dY/dX = (aX+bY)/(dX+eY)$, which is homogeneous.

Case 2: $ae = bd$, i.e., $a/d = b/e = k$ (lines parallel).

Let $z = ax + by$. Then $dz/dx = a + b\,dy/dx$, reducing the ODE to separable form in $z$ and $x$.

Solution checklist

1. Check whether the ODE is homogeneous as written.
2. If not, identify whether lines are parallel or intersecting.
3. Apply the appropriate substitution.
4. Separate variables and integrate both sides.
5. Back-substitute $v = y/x$ (or the translation) to express the solution in original variables.
6. Apply initial condition if given.

Question Archetypes

homogeneous-substitution (1 question; 2020)

Recognition Cues

• The right-hand side of $dy/dx$ is a ratio of two expressions that are both degree 1 in $x$ and $y$.
• Setting $v = y/x$ makes the right-hand side a function of $v$ alone.
• May be disguised as $M\,dx + N\,dy = 0$ where $M$ and $N$ are both homogeneous of the same degree.
• The word “homogeneous” may not appear — check by the degree test.

Solution Template

1. Write $dy/dx = g(v)$ where $v = y/x$.
2. Substitute $y = vx$, $dy/dx = v + xdv/dx$.
3. Rearrange to $x\,dv/dx = g(v) - v$; separate: $\dfrac{dv}{g(v)-v} = \dfrac{dx}{x}$.
4. Integrate both sides; partial fractions if needed.
5. Replace $v = y/x$; simplify; apply initial condition.

Worked Example

2020 Paper 1, 2020-P1-Q2b (10 marks)

Solve the differential equation: $\frac{dy}{dx} = \frac{y - x}{y + x}$

Step 1. Check homogeneity.

Numerator $y - x$ and denominator $y + x$ are both degree 1 in $(x,y)$. Dividing by $x$:

$\frac{dy}{dx} = \frac{(y/x) - 1}{(y/x) + 1} = \frac{v-1}{v+1}$

This is a function of $v = y/x$ alone. The ODE is homogeneous.

Step 2. Substitute $v = y/x$.

$y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Substituting:

$v + x\frac{dv}{dx} = \frac{v-1}{v+1}$

$x\frac{dv}{dx} = \frac{v-1}{v+1} - v = \frac{v-1 - v(v+1)}{v+1} = \frac{v - 1 - v^2 - v}{v+1} = \frac{-v^2 - 1}{v+1}$

Step 3. Separate variables.

$x\frac{dv}{dx} = \frac{-(v^2+1)}{v+1}$

$\frac{v+1}{v^2+1}\,dv = -\frac{dx}{x}$

Integrate both sides:

$\int \frac{v+1}{v^2+1}\,dv = -\int \frac{dx}{x}$

Split the left side:

$\int \frac{v}{v^2+1}\,dv + \int \frac{1}{v^2+1}\,dv = -\ln|x| + C_1$

$\frac{1}{2}\ln(v^2+1) + \arctan(v) = -\ln|x| + C_1$

Step 4. Back-substitute $v = y/x$.

$\frac{1}{2}\ln\!\left(\frac{y^2}{x^2}+1\right) + \arctan\!\left(\frac{y}{x}\right) = -\ln|x| + C_1$

$\frac{1}{2}\ln\!\left(\frac{x^2+y^2}{x^2}\right) + \arctan\!\left(\frac{y}{x}\right) = -\ln|x| + C_1$

$\frac{1}{2}\ln(x^2+y^2) - \ln|x| + \arctan\!\left(\frac{y}{x}\right) = -\ln|x| + C_1$

$\frac{1}{2}\ln(x^2+y^2) + \arctan\!\left(\frac{y}{x}\right) = C_1$

Writing $C_1 = \ln C$ (absorbing the constant):

$\boxed{\ln\!\sqrt{x^2+y^2} + \arctan\!\left(\frac{y}{x}\right) = C}$

Common Traps

• Forgetting to subtract $v$ when forming $x\,dv/dx = g(v)-v$: writing $x\,dv/dx = g(v)$ directly.
• Sign error when combining the fraction: $v - (v^2+v)/(v+1)$ requires careful algebra over a common denominator.
• Splitting $\int(v+1)/(v^2+1)\,dv$ incorrectly; the $v/(v^2+1)$ part integrates to $\tfrac{1}{2}\ln(v^2+1)$, not $\ln(v^2+1)$.
• Omitting the absolute value in $\ln|x|$ or $\ln|v^2+1|$, losing marks for rigor.

Marks-Aware Writing

This is a 10-mark computation. Examiners award:

1. Homogeneity check / identification (1 mark).
2. Correct substitution $v = y/x$, $dy/dx = v + xdv/dx$ (2 marks).
3. Separation of variables with correct algebraic simplification of $g(v)-v$ (3 marks).
4. Integration of both sides, splitting the left-side integral correctly (2 marks).
5. Back-substitution and correct final form (2 marks).

Show every algebraic step from substitution to separation; partial fractions or integral splits must be written out explicitly.

Practice Set

Only one historical question on this atom (shown above).