Formulation of differential equations

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2017, 2025)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~10 min
• Difficulty mix: medium 1, easy 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

Formulation questions test whether you can connect geometry (a parametric family of curves) to calculus (its governing ODE) — a skill that reappears implicitly in every method question. Each UPSC appearance has been worth 10 marks and reduces to one clear algorithm: count the parameters, differentiate that many times, eliminate them. The two exam instances — all circles (3 parameters, third-order) and all ellipses with axes along coordinate axes (2 parameters, second-order) — are clean enough that a prepared student can complete them in under eight minutes and collect all marks.

Minimum Theory

Eliminating parameters. A family of plane curves defined by $F(x,y,c_1,c_2,\ldots,c_n)=0$ with $n$ arbitrary constants yields an ODE of order $n$ after successively differentiating $n$ times and eliminating $c_1,\ldots,c_n$. The result is a relation among $x$, $y$, $y'$, $y''$, $\ldots$, $y^{(n)}$ that holds for every curve in the family.

General circle. The family $x^2+y^2+2gx+2fy+c=0$ has three parameters ($g,f,c$). Each differentiation eliminates one: after the first, $c$ disappears; after the second, $g$; after equating two expressions for $f$ using the second and third derivatives, $f$ disappears too, leaving a third-order ODE in $x,y,y',y'',y'''$.

Family of ellipses with axes along coordinate axes. The family $x^2/A+y^2/B=1$ has two parameters ($A,B$). Differentiating twice and eliminating both gives a second-order ODE.

Question Archetypes

form-ode-from-family (2 question(s); 2017, 2025)

Recognition Cues

• “Find the differential equation representing all [circles / ellipses / parabolas / …].”
• “Form the differential equation of the family …” where the family has a stated number of parameters.
• Geometric families: circles (3 params), ellipses with axis constraints (2 params), parabolas with vertex at origin (1 param).
• The answer order equals the number of arbitrary constants to be eliminated.

Solution Template

1. Write the general equation of the family, identifying every free parameter.
2. Count parameters $n$; the result will be an order-$n$ ODE.
3. Differentiate successively $n$ times, obtaining equations (1), (2), …, ($n$).
4. Eliminate the parameters from the system of $n+1$ equations (original + $n$ derivatives):
   • Express each parameter from one equation; substitute into another.
   • Or use cross-multiplication to cancel parameters from two equations.
5. Verify: the final ODE contains no arbitrary constants; a specific member of the family satisfies it.

Worked Example

2017 Paper 1, 2017-P1-Q5a (10 marks)

Find the differential equation representing all the circles in the $x$-$y$ plane.

The general circle has equation

$x^2 + y^2 + 2gx + 2fy + c = 0,$

with three arbitrary constants $g,f,c$. Expect a third-order ODE.

Step 1 — Differentiate once. Treating $y=y(x)$, $y'=dy/dx$:

$2x + 2yy' + 2g + 2fy' = 0 \qquad\Longrightarrow\qquad x + yy' + g + fy' = 0.\quad(1)$

Here $c$ has already vanished. So after one differentiation, only $g$ and $f$ remain.

Step 2 — Differentiate again.

$1 + y'^2 + yy'' + fy'' = 0.\quad(2)$

Solve for $f$:

$f = -\frac{1 + y'^2 + yy''}{y''}.$

Step 3 — Differentiate once more.

$3y'y'' + yy''' + fy''' = 0.\quad(3)$

Solve for $f$:

$f = -\frac{3y'y'' + yy'''}{y'''}.$

Step 4 — Equate the two expressions for $f$ and simplify. Cross-multiply:

$(1 + y'^2 + yy'')y''' = (3y'y'' + yy''')y''.$

Expand both sides:

$(1+y'^2)y''' + yy''y''' = 3y'y''^2 + yy''y'''.$

The $yy''y'''$ terms cancel:

$(1+y'^2)y''' = 3y'y''^2.$

$\boxed{\;(1+y'^2)\,y''' = 3\,y'\,(y'')^2.\;}$

All three constants $g,f,c$ are gone. Verify: the circle $(x-1)^2+(y-2)^2=25$ gives $y'=(1-x)/(y-2)$, and one can check this satisfies the ODE identically.

2025 Paper 1, 2025-P1-Q5b (10 marks)

Form the differential equation of all ellipses whose axes coincide with the coordinate axes.

The family is

$\frac{x^2}{A} + \frac{y^2}{B} = 1,$

with two arbitrary constants $A,B$. Expect a second-order ODE.

Step 1 — Differentiate once.

$\frac{2x}{A} + \frac{2yy'}{B} = 0 \qquad\Longrightarrow\qquad \frac{x}{A} + \frac{yy'}{B} = 0.\quad(1)$

Step 2 — Differentiate again.

$\frac{1}{A} + \frac{(y')^2 + yy''}{B} = 0.\quad(2)$

Step 3 — Eliminate $A$ and $B$. From (1): $\dfrac{1}{A} = -\dfrac{yy'}{Bx}$. Substitute into (2):

$-\frac{yy'}{Bx} + \frac{(y')^2 + yy''}{B} = 0.$

Multiply through by $Bx$:

$-yy' + x\bigl[(y')^2 + yy''\bigr] = 0.$

$\boxed{\;xyy'' + x(y')^2 - yy' = 0.\;}$

Verify with $A=4$, $B=9$: $y=3\sqrt{1-x^2/4}$ at $x=1$ gives $y=3\sqrt{3}/2$, $y'=-3/(4\sqrt{3}/2)$, $y''=\ldots$, and substitution yields residual $0$. ✓

Common Traps

• For circles: stopping at the second derivative still carries $f$ — you need a third differentiation. The order of the ODE equals the number of parameters, not the degree.
• For circles: the $yy''y'''$ cross-terms on both sides of the cross-multiplication cancel; recognising this simplification is essential. If you expand carelessly you may not see the cancellation.
• For ellipses: the centre need not be at the origin when axes only “coincide with coordinate axes” — but the standard UPSC interpretation (as in 2025) uses the form $x^2/A+y^2/B=1$ centred at the origin. If a question specifies “axes along but not necessarily through origin,” add a $(h,k)$ shift (four parameters, fourth-order).
• Missing the elimination step: differentiating is not enough; you must actively eliminate every constant to produce a pure ODE.

Marks-Aware Writing

10-mark questions: State the general equation of the family and identify the number of parameters (1–2 marks). Carry out each differentiation step clearly, labelling equations (1), (2), (3) (2–3 marks each). Write the elimination step explicitly and box the final ODE. For circles, the cross-multiplication and cancellation of $yy''y'''$ are the two critical steps the examiner checks; for ellipses, the substitution of $1/A$ from equation (1) into equation (2) is the pivot.

Practice Set

• 2024-P1-Q7a (15 m) — — Hint: state Picard–Lindelöf theorem; then test Lipschitz condition for $f(x,y)=2\sqrt y$ near $y=0$ (fails: $\partial f/\partial y\to\infty$) to explain non-uniqueness.
• 2017-P1-Q6a-ii (8 m) — — Hint: model as $N'=kN$; use $N(1)=2N_0$ to find $k=\ln 2$; evaluate $N(4)=16N_0$.