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Differential Equations hero

Differential Equations

~14 min read

In 30 seconds
  • What: An equation relating a function to its derivatives — NDA tests order, degree, variable-separable, homogeneous, and linear first-order types every year.
  • Why it matters: Differential equations appear in 3–5 questions per paper; they are direct-formula questions once you know the method, so they are high-yield marks.
  • Key fact: From NDA 2010 to NDA 2025, over 125 differential-equation questions appear across papers — order/degree and solution-type splits are roughly 50/50.

Differential equations are one of the most consistent scoring areas in NDA Mathematics. The paper tests a narrow set of skills — identifying order and degree, matching a given equation to a solution method, and solving straightforward first-order equations. There is no reason to lose marks here if you practise the standard question types.

This page covers every concept that has appeared in NDA papers from 2010-I through 2025-II, using the exact question types from those papers as examples. Work through each section in order; the worked examples in particular map directly onto what the NDA paper will ask you.

What This Topic Covers

The NDA syllabus for differential equations draws directly from NCERT Class 12, Chapter 9. The examinable content falls into four clean groups:

Four Groups the NDA Tests

  • Definitions — order, degree, ordinary vs partial, general vs particular solution.
  • Formation — deriving the differential equation for a family of curves by eliminating arbitrary constants.
  • Solutions — variable separable — separating \(f(x)\,dx\) from \(g(y)\,dy\) and integrating both sides.
  • Solutions — homogeneous and linear first-order — substitution \(y = vx\) for homogeneous; integrating factor (\(I.F.\)) for linear types.

The NDA does not ask for higher-order equations, Laplace transforms, or partial differential equations. Every question from 2010 to 2025 fits one of the groups above. Partial differential equations are explicitly excluded — only ordinary differential equations appear.

This topic connects directly to integration. If your integration is weak, review Indefinite and Definite Integration before working through the solution methods below.

Exam Pattern & Weightage

The table below uses question numbers from the NDA topicwise solved papers to show how many questions appeared per paper. Years with direction-based sets (e.g., Qs. 35–37 in 2014-I, Qs. 110–111 in 2023-I, Qs. 119–120 in 2024-II) count as separate items.

NDA Paper No. Focus Area
2010-II5Degree, solution, family of circles/parabolas
2011-I5Degree, variable separable, circle condition
2011-II2Curve through a point (sin x cos y type)
2012-I4Degree, linearity, order-degree pairs
2012-II5General solutions, separable equations
2013-I5Degree, order, general solutions
2013-II3Family of circles, y = 2cos x + 3sin x
2014-I7Direction set (Qs. 35–37), degree, particular solution
2014-II2Degree, solution of log(dy/dx) equation
2015-I5Variable separable, degree statements
2015-II4Degree, family of circles, order/degree of parabola
2016-I3Order, degree, unit-distance lines
2016-II3Degree/order, y(–3) value
2017-I4Formation, homogeneous, degree/order
2017-II4Circle condition for general solution, elimination of constants
2018-I5Variable separable, periodic solution, order/degree
2018-II6Linear, homogeneous, cos(y−x) substitution
2019-I4Degree, general solution, circle touching y-axis
2019-II4Degree, family curves, elimination
2020-I3Variable separable solutions
2021-I5Degree, order, general solution with given IC
2021-II4Ellipses (order 2), circles (degree 2), differential equation derivation
2022-I3Decay constant, degree statement, y = cx²
2022-II3Degree, parabola with axis along y, solution
2023-I3Direction set Qs. 110–111, particular solution
2023-II1Order and degree
2024-I4Intersection of solution curves, differential equation of curve
2024-II2Direction set Qs. 119–120, order/degree and solution
2025-I3Direction set Qs. 121–122 (slope = 4), degree
2025-II2Statements on solutions, general solution of (x+y)² dy/dx = k²
NDA Alert

Order/degree identification and variable-separable solutions together account for roughly half of all differential-equation marks across papers. Never skip these two sub-topics.

Core Concepts

Order and Degree

The order of a differential equation is the order of the highest-order derivative present. The degree is the power to which that highest-order derivative is raised, but only after the equation is expressed as a polynomial in its derivatives.

Order and Degree — Key Rule Order = highest derivative present (e.g., \(\tfrac{d^2y}{dx^2}\) → order 2). Degree = power of that derivative once the equation is free of radicals and fractions in derivatives. If the equation cannot be made into a polynomial in derivatives (e.g., contains \(\sin\!\left(\tfrac{dy}{dx}\right)\)), degree is NOT defined.
Rationalize First, Then Read Degree For $$\left[1 + \left(\tfrac{dy}{dx}\right)^2\right]^{3/2} = k\,\tfrac{d^2y}{dx^2}$$ square both sides to clear the 3/2 power: $$\left[1 + \left(\tfrac{dy}{dx}\right)^2\right]^3 = k^2\left(\tfrac{d^2y}{dx^2}\right)^2$$ Order = 2 (highest derivative is \(\tfrac{d^2y}{dx^2}\)); Degree = 2 (its power after rationalization). Ignore expanded powers of lower-order terms.

NDA 2014-I Q. 40 and NDA 2022-I Q. 107 both test the "degree not defined" rule for equations containing \(\cos\!\left(\tfrac{dy}{dx}\right)\) or similar transcendental functions. The order of such equations is still well-defined.

Trap

Degree is only defined when the DE can be made polynomial in its derivatives. Equations like \(\tfrac{d^2y}{dx^2} + \sin\!\left(\tfrac{dy}{dx}\right) = 0\) or \(\tfrac{d^2y}{dx^2} = e^{dy/dx}\) have a well-defined order (= 2) but degree is Not Defined. If "Not Defined" is an MCQ option for such an equation, mark it instantly.

NDA Alert

Degree of a differential equation is always a positive integer. The pair \((m, n) = (3, 2)\) is feasible; the pair \((2, 3/2)\) is NOT feasible as a degree — this was directly tested in NDA 2012-I.

General and Particular Solutions

A general solution contains as many arbitrary constants as the order of the differential equation. A particular solution has zero arbitrary constants — all constants are fixed using initial conditions.

Number of Arbitrary Constants General solution of an \(n\)th-order equation → \(n\) arbitrary constants. Particular solution → 0 arbitrary constants. (NDA 2014-I Q. 39 directly tests this: particular solution of a third-order equation has 0 arbitrary constants.)

Variable Separable Method

Use this when \(\dfrac{dy}{dx} = f(x)\cdot g(y)\). Rewrite as \(\dfrac{dy}{g(y)} = f(x)\,dx\), then integrate both sides.

Variable Separable — Steps 1. Rewrite \(\dfrac{dy}{dx} = g(x)\,h(y)\) as \(\dfrac{dy}{h(y)} = g(x)\,dx\). 2. Integrate both sides: $$\int \dfrac{dy}{h(y)} \;=\; \int g(x)\,dx \;+\; C$$ 3. Express \(y\) in terms of \(x\) if possible.
Variable Separable — Pattern Form Whenever \(\dfrac{dy}{dx} = f(x)\,g(y)\), the equation is separable: $$\int \dfrac{dy}{g(y)} \;=\; \int f(x)\,dx \;+\; C$$ Always pull the y-pieces with dy to one side and the x-pieces with dx to the other before integrating.

Example from NDA 2018-I Q. 70: \((1+2x)\,dy - (1-2y)\,dx = 0\) is solved by separating: \(\dfrac{dy}{1-2y} = \dfrac{dx}{1+2x}\), integrating both sides to get \(x - y - 2xy = c\).

Homogeneous Differential Equations

\(\dfrac{dy}{dx} = F(x, y)\) is homogeneous if \(F(\lambda x, \lambda y) = \lambda^0\, F(x, y)\) — that is, \(F\) is a homogeneous function of degree zero. This happens exactly when \(\dfrac{dy}{dx} = g\!\left(\dfrac{y}{x}\right)\) for some function \(g\).

Homogeneous — Substitution Let \(y = vx\), so \(\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}\). Substitute, simplify to a separable equation in \(v\) and \(x\). Integrate, then replace \(v\) with \(\dfrac{y}{x}\).
Homogeneous — Why y = vx Works A DE is homogeneous when every term in numerator and denominator has the same total degree in x and y, e.g.\ \(\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}\) (every term is degree 2). Putting y = vx: $$v + x\,\dfrac{dv}{dx} \;=\; \dfrac{1 + v^2}{v}$$ which is now variable-separable in v and x.

If the equation is in \(\dfrac{dx}{dy}\) form, use \(x = vy\) instead. From NDA 2017-I Q. 69: \(y\,dx - (x + 2y^2)\,dy = 0\), rewritten as \(\dfrac{dx}{dy} - \dfrac{x}{y} = 2y\), is actually linear in \(x\) (treated as a function of \(y\)).

Linear First-Order Differential Equations

Standard form: \(\dfrac{dy}{dx} + P(x)\cdot y = Q(x)\). Multiply through by the integrating factor, then integrate the resulting exact derivative.

Linear First-Order — Integrating Factor $$\text{I.F.} = e^{\int P\,dx}$$ Multiply both sides by I.F.: $$\dfrac{d}{dx}\!\left[y \cdot \text{I.F.}\right] = Q \cdot \text{I.F.}$$ Solution: $$y \cdot \text{I.F.} \;=\; \int Q \cdot \text{I.F.}\,dx \;+\; C$$
Linear First-Order — KaTeX Form Standard form: $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$ Integrating factor: $$\text{I.F.} = e^{\int P(x)\,dx}$$ Direct solution: $$y \cdot \text{I.F.} \;=\; \int Q(x)\cdot\text{I.F.}\,dx \;+\; C$$
Trap

Always force the coefficient of \(\dfrac{dy}{dx}\) to 1 before reading off \(P(x)\) and computing the I.F. If you skip this normalisation step, your \(P\) is wrong and every subsequent integration is wrong. Multiply or divide the entire equation through first.

When \(x\) is the dependent variable: \(\dfrac{dx}{dy} + P_1(y)\cdot x = Q_1(y)\), use \(\text{I.F.} = e^{\int P_1\,dy}\) instead.

Circle Condition for \(\dfrac{dy}{dx} = \dfrac{ax+h}{by+k}\) The general solution represents a circle only when \(a = -b \neq 0\). (NDA 2017-II Q. 63 directly tests this fact.)

Formation of Differential Equations

To get the differential equation of a family of curves, differentiate as many times as there are arbitrary constants and then eliminate those constants.

Family of Parabolas — \(y^2 = 4ax\) Differentiate: \(2y\,y' = 4a \;\Rightarrow\; a = \tfrac{y y'}{2}\). Substitute back: \(y^2 = 4\cdot\tfrac{y y'}{2}\cdot x \;\Rightarrow\; y = 2x\,y'\). Order = 1, Degree = 1. (NDA 2010-II, 2015-II)
Family of Circles — \(x^2 + y^2 = r^2\) Differentiate: \(2x + 2y\,\tfrac{dy}{dx} = 0 \;\Rightarrow\; x\,dx + y\,dy = 0\). Order = 1, Degree = 1. (NDA 2010-II, 2013-II)
Parabolas Parallel to Y-axis — \(y = Ax^2 + Bx + C\) Three constants → three differentiations → \(\tfrac{d^3y}{dx^3} = 0\). Order = 3, Degree = 1. (NDA 2011-I)
NDA Alert

For \(y = A[\sin(x+C) + \cos(x+C)]\), two constants but they combine — actual order = 1 (effectively one constant). The result is \(y'' + y = 0\). NDA 2017-II Q. 65 tested this exact collapse.

Trap

To get the DE of a family of curves, you must differentiate the equation as many times as there are arbitrary constants and then eliminate the constants. Order of the DE = number of independent arbitrary constants (not just the number of letters that look like constants — count independent ones only).

Worked Examples

Example 1 — Order and Degree (NDA 2014-I Q. 40)

Find the degree of $$\left(\dfrac{dy}{dx}\right)^2 + 1 = \left[x\,\dfrac{d^2y}{dx^2}\right]^2$$ (after clearing radicals on the original form).

  • Highest-order derivative present: \(\dfrac{d^2y}{dx^2}\).
  • Square both sides to clear any fractional powers. The equation becomes polynomial in \(\dfrac{d^2y}{dx^2}\).
  • The power to which \(\dfrac{d^2y}{dx^2}\) is raised after clearing radicals is 2.
  • Answer: Degree = 2. (From PYQ solution, NDA M-470 answer key option (c).)

Example 2 — Variable Separable: \(x\,dy - y\,dx = 0\) (NDA 2018-I, 2017-I)

Solve \(x\,dy - y\,dx = 0\).

  • Rewrite: \(x\,dy = y\,dx\).
  • Separate variables: $$\dfrac{dy}{y} \;=\; \dfrac{dx}{x}$$
  • Integrate both sides: $$\log|y| \;=\; \log|x| + \log|c|$$
  • \(\log\left|\dfrac{y}{x}\right| = \log|c|\), so \(y = cx\).
  • Answer: \(y = cx\) — a family of straight lines through the origin. (NDA 2018-I Q. 67, 2012-I Q. 15.)

Example 3 — Variable Separable with Solution Curve (NDA 2018-I Q. 70)

Solve \((1 + 2x)\,dy - (1 - 2y)\,dx = 0\).

  • Separate: $$\dfrac{dy}{1-2y} \;=\; \dfrac{dx}{1+2x}$$
  • Integrate left side: \(-\tfrac{1}{2}\log|1-2y|\).
  • Integrate right side: \(\tfrac{1}{2}\log|1+2x|\).
  • Combine: $$\log|1-2y| + \log|1+2x| = \log c \;\;\Longrightarrow\;\; (1+2x)(1-2y) = c$$
  • Expand: \(1 + 2x - 2y - 4xy = c \;\Rightarrow\; x - y - 2xy = c\) (absorbing constant).
  • Answer: \(x - y - 2xy = c\). (Option (a) in NDA 2018-I.)

Example 4 — Linear First-Order: \(\dfrac{dy}{dx} + 2y = 1\) with \(y(0) = 0\) (NDA 2014-I Q. 33)

Find the particular solution of \(\dfrac{dy}{dx} + 2y = 1\) satisfying \(y(0) = 0\).

  • \(P = 2,\; Q = 1\). $$\text{I.F.} = e^{\int 2\,dx} = e^{2x}$$
  • Multiply: $$\dfrac{d}{dx}\!\left[y\,e^{2x}\right] = e^{2x}$$
  • Integrate: $$y\,e^{2x} \;=\; \tfrac{1}{2}\,e^{2x} + C$$
  • Apply \(y(0) = 0\): \(0 = \tfrac{1}{2} + C \;\Rightarrow\; C = -\tfrac{1}{2}\).
  • So $$y\,e^{2x} = \tfrac{1}{2}\,e^{2x} - \tfrac{1}{2} \;\;\Longrightarrow\;\; y = \dfrac{1 - e^{-2x}}{2}$$
  • Answer: \(y = \dfrac{1 - e^{-2x}}{2}\). (Option (a) in NDA 2014-I.)

Example 5 — Formation: Differential Equation of Family of Circles Touching Y-axis at Origin (NDA 2019-II Q. 82)

Find the differential equation of circles touching the y-axis at the origin.

  • General equation: \((x - a)^2 + y^2 = a^2 \;\Rightarrow\; x^2 - 2ax + y^2 = 0\).
  • From this: \(2a = \dfrac{x^2 + y^2}{x} \;\Rightarrow\; a = \dfrac{x^2 + y^2}{2x}\).
  • Differentiate \(x^2 + y^2 - 2ax = 0\) with respect to \(x\): $$2x + 2y\,\dfrac{dy}{dx} - 2a = 0$$
  • Substitute \(a\): $$2x + 2y\,\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{x}$$
  • Multiply by \(x\): $$2x^2 + 2xy\,\dfrac{dy}{dx} = x^2 + y^2$$
  • Rearrange: $$x^2 - y^2 + 2xy\,\dfrac{dy}{dx} = 0$$ (Option (c) in NDA 2019-II.)

Example 6 — Variable Separable with Trig (NDA-style)

Solve \(e^x \tan y\, dx + (1 - e^x)\sec^2 y\, dy = 0\).

  • Separate variables: \(\dfrac{\sec^2 y}{\tan y}\, dy = \dfrac{e^x}{e^x - 1}\, dx\) (after dividing by \((1 - e^x)\tan y\) and flipping the sign).
  • Integrate the LHS: \(\int \dfrac{\sec^2 y}{\tan y}\, dy = \log|\tan y|\).
  • Integrate the RHS: \(\int \dfrac{e^x}{e^x - 1}\, dx = \log|e^x - 1|\).
  • Combine: \(\log|\tan y| = \log|e^x - 1| + \log c \;\Rightarrow\; \tan y = c\,(e^x - 1)\).
  • Equivalently, \(\sin y = c\,(1 - e^x)\) (after absorbing signs into the constant). Matches the option in NDA 2012-II Q. 25.

Example 7 — Linear DE Using Integrating Factor

Solve \(\dfrac{dy}{dx} + \dfrac{y}{x} = x^2\).

  • Compare with \(\dfrac{dy}{dx} + P y = Q\): here \(P = \dfrac{1}{x}\), \(Q = x^2\).
  • Integrating factor: \(\text{I.F.} = e^{\int (1/x)\, dx} = e^{\log x} = x\).
  • Direct solution: \(y \cdot x = \int x^2 \cdot x\, dx + C = \int x^3\, dx + C\).
  • So \(xy = \dfrac{x^4}{4} + C\), i.e.\ \(y = \dfrac{x^3}{4} + \dfrac{C}{x}\).

Example 8 — Homogeneous DE via y = vx

Solve \(\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}\).

  • Every term is total-degree 2 in x and y → homogeneous. Put y = vx, so \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\).
  • Substitute: \(v + x\dfrac{dv}{dx} = \dfrac{x^2 + v^2 x^2}{x \cdot vx} = \dfrac{1 + v^2}{v}\).
  • Isolate: \(x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v} - v = \dfrac{1}{v}\).
  • Separate: \(v\, dv = \dfrac{dx}{x}\). Integrate: \(\dfrac{v^2}{2} = \log|x| + C\).
  • Back-substitute v = y/x: \(\dfrac{y^2}{2x^2} = \log|x| + C\), i.e.\ \(y^2 = 2x^2\log|x| + C' x^2\).

Test Yourself on Differential Equations

NDA mock tests include 3–5 differential-equation questions per paper. Practice full-length mocks to build speed on order/degree and solution-type identification.

Start Free Mock Test

Exam Shortcuts (Pro-Tips)

Differential equations are formula-driven once you spot the type. The five hacks below turn a 90-second solve into a 15-second one, and each is drawn from a recurring NDA pattern.

Shortcut 1 — Read Order First, Then Degree

Order = order of the highest derivative; degree = power of that derivative only, after the equation has been rationalised (no radicals, no fractional exponents on any derivative). Powers of lower-order derivatives that show up after expansion do not affect the degree.

Two-Step Order/Degree Check Step 1 — Spot the highest derivative → that is the Order.
Step 2 — Rationalise the equation, then read the exponent on that highest derivative → that is the Degree.
If a derivative is locked inside \(\sin\), \(\cos\), \(\log\), or \(e^{(\cdot)}\), degree is Not Defined.

Shortcut 2 — Variable-Separable Pattern Recognition

The instant you can write \(\dfrac{dy}{dx} = f(x)\,g(y)\) — a pure product of an x-only function and a y-only function — the DE is separable. No substitution needed; just split and integrate.

Separable Pattern $$\dfrac{dy}{dx} = f(x)\,g(y) \;\;\Longrightarrow\;\; \int \dfrac{dy}{g(y)} = \int f(x)\,dx + C$$

Shortcut 3 — Linear DE Integrating Factor

For \(\dfrac{dy}{dx} + P(x)\,y = Q(x)\), the I.F. is always \(e^{\int P\,dx}\). Once you have it, jump straight to the boxed solution — do not re-derive the product rule in the exam.

Linear DE — Memorise These Standard I.F.s \(\dfrac{dy}{dx} + \dfrac{1}{x}\,y = \ldots\) → \(\text{I.F.} = x\)
\(\dfrac{dy}{dx} - \dfrac{1}{x}\,y = \ldots\) → \(\text{I.F.} = \dfrac{1}{x}\)
\(\dfrac{dy}{dx} + (\cot x)\,y = \ldots\) → \(\text{I.F.} = \sin x\)
\(\dfrac{dy}{dx} + (\tan x)\,y = \ldots\) → \(\text{I.F.} = \sec x\)

Examiners often ask only for the I.F. — recognising one of the four standard forms above saves the full integration.

Shortcut 4 — Homogeneous DE Substitution y = vx

If every term in numerator and denominator of \(\dfrac{dy}{dx}\) has the same total degree in \(x\) and \(y\), the equation is homogeneous. Substitute \(y = vx\) (so \(\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}\)) — it will collapse into a separable equation in \(v\) and \(x\).

Homogeneous Substitution Box Recognise: \(\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)\).
Put y = vx → \(v + x\,\dfrac{dv}{dx} = F(v)\).
Separate, integrate, then replace v = y/x.

Shortcut 5 — Syllabus Boundary: Skip Bernoulli

Bernoulli's equation \(\dfrac{dy}{dx} + P y = Q y^n\) is outside the NDA syllabus. If a question looks like Bernoulli, it is almost certainly either a misprint or actually reducible to a standard linear/separable form by inspection. Do not waste prep time on Bernoulli reductions.

Common Question Patterns

Every differential-equation question in NDA from 2010 to 2025 falls into one of seven recognisable patterns. Learn to spot the pattern before you start calculating.

Pattern 1 — Order and Degree Identification

  • You are given an equation; find order and/or degree.
  • Watch for equations with \(\sin\!\left(\tfrac{dy}{dx}\right)\) or \(e^{dy/dx}\) — degree is not defined.
  • Watch for fractional-power derivatives — clear radicals first, then read degree.
  • Appeared in: 2010-II Q. 1, 2011-I Q. 9, 2012-I Q. 16, 2013-I Q. 24, 2014-I Q. 38, 2015-II Q. 47, 2017-II Q. 63, 2018-I Q. 71, 2019-II Q. 84, 2020-I Q. 94, 2021-I Q. 98, 2022-I Q. 102, 2023-I Q. 113, 2024-I Q. 117, 2025-II Q. 123.

Pattern 2 — What Does the Equation Represent?

  • Solve the equation, then identify the curve family: lines, circles, parabolas, ellipses.
  • \(\dfrac{dy}{dx} = \dfrac{y}{x} \;\Rightarrow\; y = cx\) (family of lines through origin).
  • \(y\,\dfrac{dy}{dx} + x = k \;\Rightarrow\; x^2 + y^2 = 2kx\) (family of circles, centre on x-axis).
  • \(x\,\dfrac{dy}{dx} - 2y = 0 \;\Rightarrow\; y = cx^2\) (family of parabolas).

Pattern 3 — General Solution of Separable Equation

  • \(e^x \tan y\,dx + (1 - e^x)\sec^2 y\,dy = 0 \;\Rightarrow\; \sin y = c(1 - e^x)\). (NDA 2012-II Q. 25.)
  • \(x^2\,dy + y^2\,dx = 0 \;\Rightarrow\; \dfrac{1}{x} + \dfrac{1}{y} = c\) or \(xy = c\) in some forms.
  • Key: write the separated integrals carefully, combine log terms using \(\log(mn) = \log m + \log n\).

Pattern 4 — Particular Solution / Curve Through a Point

  • Solve the general solution, then substitute the given point to find \(C\).
  • NDA 2011-II Q. 14: $$\sin x \cos y\,dx + \cos x \sin y\,dy = 0$$ through \(\left(0, \tfrac{\pi}{4}\right)\) gives $$\cos x \cos y = \tfrac{1}{2}.$$
  • NDA 2025-I Q. 121–122: slope = 4 at all points, passes through origin → \(y = 4x\) (straight line), area under curve from \(x = 0\) to \(x = 4\) is 32 sq. units.

Pattern 5 — Formation (Elimination of Constants)

  • Number of differentiations = number of arbitrary constants.
  • \(y = A\sin(\lambda x + \alpha)\): two constants \(A, \alpha\) → two differentiations → \(\dfrac{d^2y}{dx^2} + \lambda^2 y = 0\). (NDA 2012-II Q. 20.)
  • \(y = a e^x + b e^{-x} \;\Rightarrow\; \dfrac{d^2y}{dx^2} - y = 0\). (NDA 2021-I Q. 100.)

Pattern 6 — Linear First-Order with Integrating Factor

  • Identify \(P\) and \(Q\), compute \(\text{I.F.} = e^{\int P\,dx}\), write solution as \(y\cdot\text{I.F.} = \int Q\cdot\text{I.F.}\,dx + C\).
  • NDA 2024-II Q. 119–120: \(y\,dx + (x - y^3)\,dy = 0\) → rearranges to \(\dfrac{dx}{dy} - \dfrac{x}{y} = y^2\) → \(\text{I.F.} = \dfrac{1}{y}\) → solution: \(4xy - y^4 = c\).

Pattern 7 — Application (Radioactive Decay / Growth)

  • \(\dfrac{dP}{dt} = -kP \;\Rightarrow\; P = P_0\,e^{-kt}\). Half-life condition: \(\tfrac{P_0}{2} = P_0\,e^{-100k} \;\Rightarrow\; k = \dfrac{\ln 2}{100}\).
  • NDA 2022-I Q. 103: half of substance decays in 100 years → decay constant \(k = \dfrac{\ln 2}{100}\).

How NDA Tests This Topic

NDA rarely combines two methods in one question. A question is either "find order/degree" OR "find general solution" — not both simultaneously. Read the question carefully to avoid wasting time solving when you only need to identify order and degree, or vice versa.

Preparation Strategy

Differential equations are a reliable source of marks if approached systematically. Here is a four-week plan built around the patterns identified above.

Week 1 — Definitions and Formation

  • Memorise: order = highest derivative; degree = power of that derivative after clearing radicals; degree undefined if not polynomial in derivatives.
  • Practice 20 order/degree identification questions from the PYQ list above — aim for under 30 seconds each.
  • Work through formation questions for standard families: circles, parabolas (\(y^2 = 4ax\) and \(x^2 = 4ay\)), ellipses, straight lines.

Week 2 — Variable Separable Solutions

  • Drill separation, integration, and combining log terms. The most common error is sign mistakes in log combination.
  • Targets: \(x\,dy - y\,dx = 0\), \((1+2x)\,dy = (1-2y)\,dx\), \(e^x \tan y\,dx + (1-e^x)\sec^2 y\,dy = 0\), \(\sin x \cos y\,dx + \cos x \sin y\,dy = 0\).
  • After every solution, verify by checking if the answer satisfies the original equation.

Week 3 — Homogeneous and Linear First-Order

  • Homogeneous: check if \(\dfrac{dy}{dx}\) can be written as \(g\!\left(\dfrac{y}{x}\right)\), substitute \(y = vx\), separate, integrate, back-substitute.
  • Linear: write in standard form \(\dfrac{dy}{dx} + Py = Q\), compute I.F., write solution. Practice five questions per day on linear types.
  • Review: circle condition (\(a = -b \neq 0\)), and the form \(\dfrac{dx}{dy} + P_1\,x = Q_1\) for equations linear in \(x\).

Week 4 — Full Mock Integration

  • Do timed NDA-style mock tests. Differential-equation questions should take no more than 90 seconds each.
  • Review integration techniques if you are slow on the integration step of any method.
  • Check your work on derivatives — formation questions require confident differentiation.

For overall NDA Maths structure, see the NDA Maths subject index. Differential equations link closely to Limits, Continuity and Differentiability through the concept of derivative, and to Integration through the solution technique.

Frequently Asked Questions

How many differential-equation questions appear in one NDA paper?

Typically 3 to 5 questions per paper, though some papers (like 2014-I and 2018-II) have gone up to 6 or 7. Over the 2010–2025 window the average is close to 4 per paper.

Is the degree always defined for a differential equation?

No. Degree is defined only when the equation is a polynomial in its derivatives. If the equation contains \(\sin\!\left(\tfrac{dy}{dx}\right)\), \(e^{dy/dx}\), or any transcendental function of a derivative, the degree is not defined. Order is still defined in those cases — it equals the order of the highest derivative present.

What is the difference between general and particular solutions?

The general solution of an nth-order equation contains exactly n arbitrary constants. A particular solution is obtained from the general solution by assigning specific values to all arbitrary constants using initial or boundary conditions, leaving zero free constants.

How do I decide which method to use — separable, homogeneous, or linear?

Check in this order. (1) Can you write \(\dfrac{dy}{dx}\) as \(f(x)\cdot g(y)\)? Use variable separable. (2) Can you write \(\dfrac{dy}{dx}\) as \(g\!\left(\dfrac{y}{x}\right)\)? Use homogeneous with \(y = vx\). (3) Is the equation of the form \(\dfrac{dy}{dx} + P(x)y = Q(x)\) with \(P\) and \(Q\) not involving \(y\) in a nonlinear way? Use linear first-order with I.F. If none of the above, look for substitutions suggested by the equation structure.

What does dy/dx = y/x represent geometrically?

Separating and integrating gives \(\log y = \log x + \log c\), so \(y = cx\). This is a family of straight lines through the origin. NDA 2010-II Q. 4 and NDA 2013-I Q. 26 both ask this — the answer is "family of straight lines through the origin."

When does dy/dx = (ax+h)/(by+k) represent a family of circles?

Integration gives \(\tfrac{b}{2}y^2 + ky = \tfrac{a}{2}x^2 + hx + C\). For this to represent circles (equal coefficients of \(x^2\) and \(y^2\)), we need \(a = -b\) with \(b \neq 0\) (equivalently \(a = -b \neq 0\)). NDA 2017-II Q. 63 tests this condition directly.

How is the integrating factor computed for a linear equation?

For \(\dfrac{dy}{dx} + P(x)y = Q(x)\), the integrating factor is \(\text{I.F.} = e^{\int P(x)\,dx}\). After multiplying both sides by I.F., the left side becomes \(\dfrac{d}{dx}\!\left[y \cdot \text{I.F.}\right]\), which you integrate directly. The answer is \(y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C\).