NDA Maths · Differential Equations

Forming an ODE from a Family of Curves

To form the differential equation of a family of curves, differentiate enough times to eliminate every arbitrary constant — n constants need n differentiations and produce an order-n equation.

Why this matters

12 PYQs running the reverse of solving: you are given the answer (a family of curves) and must find the equation. The recipe never changes — differentiate, eliminate the constants — so these are dependable marks once the routine is automatic.

Concept 1 of 2

Eliminating arbitrary constants

Intuition

A family of curves with arbitrary constants is the general solution of some ODE. To recover that ODE, differentiate the family — each differentiation gives a new equation — until you have enough equations to eliminate every constant. With n constants, differentiate n times.

Definition

The elimination recipe:

  • Count the arbitrary constants — that is the order of the ODE you will get.
  • Differentiate the family that many times.
  • Eliminate the constants between the original equation and its derivatives; the constant-free relation is the ODE.
  • Examples: parabolas x2=4ayx^2=4ay (one constant) → xdydx=2yx\,\dfrac{dy}{dx}=2y; y=ex(acosx+bsinx)y=e^x(a\cos x+b\sin x) (two constants) → y2y+2y=0y''-2y'+2y=0.
y = c·x²(one curve per c)eliminate c → x·y′ = 2y

Worked example

Form the differential equation of the family y=cx2y = cx^2 (c arbitrary).
  1. One constant cc → differentiate once: dydx=2cx\dfrac{dy}{dx} = 2cx.
  2. From the original, c=yx2c = \dfrac{y}{x^2}. Substitute: dydx=2yx2x=2yx\dfrac{dy}{dx} = 2\cdot\dfrac{y}{x^2}\cdot x = \dfrac{2y}{x}.
  3. So xdydx=2yx\dfrac{dy}{dx} = 2y.
Answer:xdydx2y=0x\dfrac{dy}{dx} - 2y = 0.
Practice this conceptself-check · 3 quick reps

Try it yourself

Form the differential equation of the family of parabolas x2=4ayx^2 = 4ay with vertex at the origin.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    How many times to differentiate a 2-constant family?
  2. 2.
    ODE of y=ABxy = A - \frac{B}{x} (two constants)?
  3. 3.
    ODE of y=ex(acosx+bsinx)y = e^x(a\cos x + b\sin x)?

From the bank · past-year question

Example 1Differential EquationsMODERATE
What is the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis?

[Q99 · Sep · 2022]

Differentiate as many times as there are constants

A one-constant family needs one differentiation (order 1); a two-constant family like y2=4a(xb)y^2=4a(x-b) needs two (order 2, giving yy+(y)2=0yy''+(y')^2=0). Differentiating too few times leaves a constant stranded in the answer.

Concept 2 of 2

Matching an ODE to its general solution

Intuition

Sometimes you are handed both a candidate ODE and a family, and must check they correspond — or decide what condition makes a solution a particular shape (a circle, say). Either differentiate the family to confirm it fits the ODE, or solve the ODE and compare.

Definition

Two directions, one idea:

  • Family → ODE: differentiate and eliminate constants (as above), then compare with the given option.
  • ODE → family: integrate the separable ODE and read off the curve type.
  • A solved family is a circle only when the x2x^2 and y2y^2 coefficients are equal — e.g. dydx=ax+hby+k\dfrac{dy}{dx}=\dfrac{ax+h}{by+k} integrates to a circle exactly when a=ba=-b.

Worked example

For what relation between a and b does dydx=axby\dfrac{dy}{dx} = \dfrac{ax}{by} have circular solutions?
  1. Separate: bydy=axdxby\,dy = ax\,dx, integrate: b2y2=a2x2+C\dfrac{b}{2}y^2 = \dfrac{a}{2}x^2 + C.
  2. Rearrange: a2x2b2y2+C=0\dfrac{a}{2}x^2 - \dfrac{b}{2}y^2 + C = 0.
  3. A circle needs equal coefficients on x2x^2 and y2y^2: a2=b2\dfrac{a}{2} = \dfrac{b}{2} with opposite signs after moving terms, i.e. a=ba = -b.
Answer:a=ba = -b (and nonzero).
Practice this conceptself-check · 3 quick reps

Try it yourself

The general solution of dydx=ax+hby+k\dfrac{dy}{dx} = \dfrac{ax+h}{by+k} is a circle only when?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    To confirm a family solves a given ODE, you?
  2. 2.
    Integrated x2y2=cx^2 - y^2 = c — circle or hyperbola?
  3. 3.
    Equal, same-sign x2x^2 and y2y^2 coefficients give a?

From the bank · past-year question

Example 2Differential EquationsMODERATE
The general solution of dydx=ax+hby+k\dfrac{dy}{dx} = \dfrac{ax + h}{by + k} represents a circle only when

[Q74 · Sep · 2017]

A circle needs equal squared-term coefficients

After integrating, a2x2b2y2\frac{a}{2}x^2 - \frac{b}{2}y^2 is a circle only if those coefficients match in magnitude (giving a=ba=-b); otherwise it is an ellipse or hyperbola. Don't assume any separable solution is a circle.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Watch out for (2)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Differential EquationsHARD
The differential equation of the system of circles touching the y-axis at the origin is

[Q89 · Apr · 2019]

Example 2Differential EquationsEASY
Which one of the following differential equations has the general solution y=aex+bexy=ae^x+be^{-x}?

[Q98 · Apr · 2021]

Example 3Differential EquationsHARD
The differential equation representing the curve y=ex(acosx+bsinx)y=e^x(a\cos x+b\sin x), where aa and bb are arbitrary constants, is

[Q72 · Apr · 2024]

Example 4Differential EquationsEASY
The differential equation of the family of curves y=pcos(ax)+qsin(ax)y=p\cos(ax)+q\sin(ax), where p, q are arbitrary constants, is

[Q87 · Sep · 2018]

Example 5Differential EquationsMODERATE
What is the differential equation of all parabolas of the type y2=4a(xb)y^2=4a(x-b)?

[Q70 · Sep · 2023]

Drill every past-year question on this subtopic

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