Differential Equations Formula Sheet — JEE Main Mathematics

Every key Differential Equations formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 6 sub-topics

Ordinary differential equations
Order and degree
Formation of differential equations
Solution of differential equations by method of separation of variables
Solution of homogeneous differential equations of first order and first degree
Solutions of linear differential equation

Order, Degree & Classification

Differential equation: An equation relating x, y and the derivatives

\frac{d y}{d x}, \frac{d ^{2} y}{d x ^{2}}, \dots

. The

order

is the order of the highest derivative present; the

degree

is the power of that highest-order derivative (when the equation is a polynomial in the derivatives).

Reading off order and degree

▸ $Order$ = highest derivative; $degree$ = its power (polynomial form only).
▸ $(\frac{d ^{2} y}{d x ^{2}})^{3} + \frac{d y}{d x} = 0$ : order 2, degree $3$ .
▸Clear radicals/fractions of derivatives first (e.g. square $y^{'} = y$ ).
▸If a derivative sits inside $sin, ln, e^{(\cdot)}$ , the $degree is undefined$ .

★ Remember · Order = number of arbitrary constants

A general solution carries exactly as many independent arbitrary constants as the order of the equation. So a 2nd-order DE has a two-parameter family of solutions (

y = A sin x + B cos x

), and a 1st-order DE a one-parameter family.

🎯 Exam · Linear differential equation

A DE is

linear

if y and all its derivatives appear to the first power with no products among them, i.e.

a_{0} (x) y^{(n)} + \dots + a_{n} (x) y = g (x)

. Anything like y\,y',

(y^{'})^{2}

sin y

makes it non-linear.

⚠️ Watch out · Make it a polynomial in derivatives first

Degree is only defined once the equation is free of radicals and fractional powers of the derivatives.

1 + (\frac{d y}{d x})^{2} = \frac{d ^{2} y}{d x ^{2}}

must be squared before you can read its degree (here

2

🚫 Examiner Trap · Examiner traps

(1) DEGREE is defined ONLY when the equation is polynomial in the derivatives —

sin (\frac{d y}{d x}) = x

or a

/fraction of a derivative has NO degree until you clear it. (2) Square/rationalise FIRST, then read order & degree (degree

=

power of the HIGHEST-order derivative). (3) Order

=

highest derivative present (independent of degree). (4) The general solution has as many arbitrary constants as the ORDER.

Formation of a Differential Equation

★ Remember · The procedure

Given a family with n independent arbitrary constants,

differentiate

times

and eliminate the constants between the original equation and its derivatives. The resulting constant-free equation is a DE of order n.

Worked ideas

▸ $y = A sin x + B cos x$ ( $2$ constants) $\Rightarrow$ differentiate twice $\Rightarrow$ $\frac{d ^{2} y}{d x ^{2}} + y = 0$ .
▸ $y = m x$ ( $1$ constant) $\Rightarrow$ $\frac{d y}{d x} = m = \frac{y}{x} \Rightarrow x \frac{d y}{d x} = y$ .
▸ $y = A e^{2 x} + B e^{3 x}$ ( $2$ constants) $\Rightarrow$ a 2nd-order linear DE.
▸Count the $independent$ constants first — that fixes the order.

🎯 Exam · Independent constants only

Combine constants that aren't truly independent before counting:

y = A e^{x + B} = (A e^{B}) e^{x} = C e^{x}

has only

one

arbitrary constant, so it gives a first-order DE, not second.

⚠️ Watch out · Eliminate, don't just differentiate

Differentiating is only half the job — you must then

eliminate

every arbitrary constant. A correct formed DE contains x,y and derivatives but

none

of the original constants.

🚫 Examiner Trap · Examiner traps

(1) To form a DE, count the INDEPENDENT arbitrary constants — that number is the ORDER, and you differentiate that many times. (2) Eliminate ALL constants; the final equation must be free of them. (3)

y = A sin x + B cos x

(2 constants)

\Rightarrow

order 2 (

y^{''} + y = 0

);

y = m x

(1 constant)

\Rightarrow

order 1. (4) Don't count a removable/duplicate constant twice.

Variable Separable Equations

Variable separable: If

\frac{d y}{d x} = f (x) g (y)

, the variables can be separated: gather all the y's (with dy) on one side and the x's (with dx) on the other, then integrate —

\int \frac{d y}{g ( y )} = \int f (x) d x + C

Recognising & handling

▸Anything that factors as (x-only) $\times$ (y-only) is separable.
▸ $\frac{d y}{d x} = e^{x + y} = e^{x} e^{y}$ separates as $e^{- y} d y = e^{x} d x$ .
▸ $\frac{d y}{d x} = g (a x + b y + c)$ : substitute $t = a x + b y + c$ to make it separable.
▸Apply the initial condition $y (x_{0}) = y_{0}$ at the end to find C.

🎯 Exam · The t = ax + by + c substitution

For

\frac{d y}{d x} = g (a x + b y + c)

, let

t = a x + b y + c

\frac{d t}{d x} = a + b \frac{d y}{d x}

. This converts it into a separable equation in t and x — a very common exam disguise.

⚠️ Watch out · Don't lose constant solutions

Dividing by g(y) silently assumes

g (y) \neq = 0

. If

g (y_{0}) = 0

, then

y = y_{0}

is also a solution — check these

separately

, since they can be the very answer asked for.

🚫 Examiner Trap · Examiner traps

(1) Get ALL y (with dy) on one side and ALL x (with dx) on the other BEFORE integrating — half-separated is the common slip. (2) Don't LOSE constant solutions: if

g (y_{0}) = 0

then

y = y_{0}

is also a solution. (3)

\frac{d y}{d x} = g (a x + b y + c)

is made separable by

t = a x + b y + c

. (4) Add

+ C

and apply any initial condition at the END to get the particular solution.

Homogeneous Differential Equations

Homogeneous equation: A first-order DE is homogeneous if it can be written

\frac{d y}{d x} = F (\frac{y}{x})

— equivalently

\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}

with P,Q homogeneous of the

same degree

. The substitution

y = vx

reduces it to a separable equation.

★ Remember · Put y = vx

With

y = vx

\frac{d y}{d x} = v + x \frac{d v}{d x}

, so

v + x \frac{d v}{d x} = F (v)

— now separable:

\frac{d v}{F ( v ) - v} = \frac{d x}{x}

. Integrate, then substitute

v = \frac{y}{x}

back at the end.

How to spot one

▸Every term has the same total degree: $f (λ x, λ y) = λ^{n} f (x, y)$ .
▸Both numerator and denominator of $\frac{d y}{d x}$ are homogeneous of equal degree.
▸If it appears as $\frac{d x}{d y} = G (\frac{x}{y})$ instead, put $x = v y$ .
▸After substituting it is $always$ variable-separable in the new variable.

⚠️ Watch out · Substitute v back at the end

A common slip is to leave the answer in terms of v. After integrating, replace

v = \frac{y}{x}

everywhere so the solution is in x and y. Keep the constant C throughout.

🎯 Exam · Non-homogeneous-in-x,y but reducible

Equations like

\frac{d y}{d x} = \frac{a _{1} x + b _{1} y + c _{1}}{a _{2} x + b _{2} y + c _{2}}

are made homogeneous by shifting

x \to X + h, y \to Y + k

to kill the constants (choose h,k as the intersection point of the two lines).

🚫 Examiner Trap · Examiner traps

(1) Homogeneous means

\frac{d y}{d x} = F (\frac{y}{x})

(every term same total degree) — substitute

y = vx

, so

\frac{d y}{d x} = v + x \frac{d v}{d x}

(DON'T forget the

x \frac{d v}{d x}

term). (2) After substituting it ALWAYS becomes variable-separable in v,x. (3) If it's

\frac{d x}{d y} = G (\frac{x}{y})

instead, put

x = v y

. (4) Remember to put

v = \frac{y}{x}

back at the end.

Linear Equations & the Integrating Factor

Linear first-order equation: An equation of the form

\frac{d y}{d x} + P (x) y = Q (x)

, where P,Q are functions of x alone. It is solved with the

integrating factor

IF = e^{\int P d x}

, which turns the left side into an exact derivative.

★ Remember · The solution formula

IF = e^{\int P d x}

, and the solution is

y \cdot IF = \int Q \cdot IF d x + C

. (Multiplying by the IF makes the left side

\frac{d}{d x} (y \cdot IF)

, which is why it works.)

The three steps

▸Write it in standard form to read off $P$ and $Q$ .
▸Compute $IF = e^{\int P d x}$ (no constant needed here).
▸Apply $y \cdot IF = \int Q \cdot IF d x + C$ .
▸If y is awkward but x is linear, swap roles: $\frac{d x}{d y} + P (y) x = Q (y)$ .

🎯 Exam · The dx/dy variant

When

\frac{d y}{d x}

is not linear in y but the equation is linear in x, treat x as the dependent variable:

\frac{d x}{d y} + P (y) x = Q (y)

with

IF = e^{\int P d y}

and

x \cdot IF = \int Q \cdot IF d y + C

⚠️ Watch out · P and Q must be single-variable

For the IF method, P and Q must depend on x

only

(or y only in the variant). If a stray y remains on the coefficient side, the equation isn't yet in linear form — rearrange or try Bernoulli.

🚫 Examiner Trap · Examiner traps

(1) Write in STANDARD form

\frac{d y}{d x} + P (x) y = Q (x)

first; P,Q must be functions of x ALONE (no y). (2) Integrating factor

IF = e^{\int P d x}

— no

+ C

inside this exponent. (3) Solution:

y \cdot IF = \int Q \cdot IF d x + C

. (4) If y isn't linear but x is, swap roles:

\frac{d x}{d y} + P (y) x = Q (y)

with

IF = e^{\int P d y}

Reducible Forms — Bernoulli & Exact

★ Remember · Bernoulli equation

\frac{d y}{d x} + P y = Q y^{n}

: divide by

y^{n}

and substitute

v = y^{1 - n}

. This gives

\frac{d v}{d x} + (1 - n) P v = (1 - n) Q

, a

linear

equation in v — finish with the integrating factor.

🎯 Exam · Exact equations

M d x + N d y = 0

exact

iff

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. Then the solution is

u (x, y) = C

, where

u = \int M d x +

(the terms of N not containing x, integrated in y).

Quick recognition

▸A power of y multiplying the right side ( $Q y^{n}$ , $n \neq = 0, 1$ ) $\Rightarrow$ Bernoulli.
▸ $n = 0$ is already linear; $n = 1$ is separable — so Bernoulli is the in-between case.
▸ $M d x + N d y = 0$ with $M_{y} = N_{x}$ $\Rightarrow$ exact (no substitution needed).
▸If not exact, an integrating factor (in $x$ or $y$ only) can sometimes fix it.

⚠️ Watch out · Don't forget the (1 − n) factors

When you substitute

v = y^{1 - n}

, the derivative brings a factor

(1 - n)

, so the linear equation is

\frac{d v}{d x} + (1 - n) P v = (1 - n) Q

— carry the

(1 - n)

through, including its sign when

n > 1

🚫 Examiner Trap · Examiner traps

(1) Bernoulli

\frac{d y}{d x} + P y = Q y^{n}

: divide by

y^{n}

and substitute

v = y^{1 - n}

to LINEARISE — then use the IF method. (2)

\frac{d v}{d x} = (1 - n) y^{- n} \frac{d y}{d x}

, so the linear form is

\frac{d v}{d x} + (1 - n) P v = (1 - n) Q

(don't drop the

(1 - n)

). (3) Exact

M d x + N d y = 0

IFF

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

. (4) Solution

u (x, y) = C

via

\int M d x +

(terms of N free of x).

Applications & the Solution-Method Map

General solution as a family of curves vs a particular solution, a map for choosing the solution method, and modelling applications like growth/decay and Newton's cooling — General/particular solutions, the method map, and modelling.

Which method? — match the form

Form of the equation	Method
$f (x) d x = g (y) d y$	variable separable
$\frac{d y}{d x} = F (\frac{y}{x})$	homogeneous (put $y = vx$ )
$\frac{d y}{d x} + P y = Q$	linear (integrating factor)
$\frac{d y}{d x} + P y = Q y^{n}$	Bernoulli ( $v = y^{1 - n}$ )
$M d x + N d y = 0, M_{y} = N_{x}$	exact

★ Remember · Which method?

Separates into

f (x) d x = g (y) d y

\Rightarrow

separable

;

\frac{d y}{d x} = F (y / x)

\Rightarrow

homogeneous

(

y = vx

);

\frac{d y}{d x} + P y = Q

\Rightarrow

linear

(IF);

\frac{d y}{d x} + P y = Q y^{n}

\Rightarrow

Bernoulli

(

v = y^{1 - n}

Modelling

▸Growth/decay: $\frac{d y}{d t} = k y \Rightarrow y = y_{0} e^{k t}$ (population, radioactivity, interest).
▸Newton's cooling: $\frac{d T}{d t} = - k (T - T_{s})$ — temperature heads toward $T_{s}$ .
▸Geometry: a prescribed slope or tangent/normal condition becomes a first-order DE.
▸Translate words $\to$ $\frac{d y}{d x}$ , solve, then apply the condition for the particular solution.

🎯 Exam · General vs particular

The

general

solution is a family of curves (one for each value of the arbitrary constant). A

particular

solution is the single member of that family passing through the given initial point.

⚠️ Watch out · Apply conditions after integrating

Find the general solution (with C)

first

, then substitute the initial/boundary condition to evaluate C. Plugging in too early — before the constant of integration appears — loses the particular solution.

🚫 Examiner Trap · Examiner traps

(1) Method map: separable

\to

separate;

F (y / x) \to

homogeneous (

y = vx

);

\frac{d y}{d x} + P y = Q \to

linear (IF);

+ P y = Q y^{n} \to

Bernoulli. (2) Growth/decay

\frac{d y}{d t} = k y \Rightarrow y = y_{0} e^{k t}

(population/radioactivity/interest). (3) Newton cooling

\frac{d T}{d t} = - k (T - T_{s})

— note the MINUS and

T_{s}

. (4) 'Slope

=

…' or tangent/normal conditions translate into a first-order DE — turn the words into

\frac{d y}{d x}

first.

More JEE Main Mathematics formula sheets

Frequently Asked Questions

What are the most important Differential Equations formulas for JEE Main?

This Differential Equations formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Ordinary differential equations, Order and degree, Formation of differential equations, Solution of differential equations by method of separation of variables, Solution of homogeneous differential equations of first order and first degree — each shown with the key result and, where useful, a worked example.

Is this Differential Equations formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Differential Equations formulas?

Blurt the Differential Equations formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheets · JEE Main previous-year papers · most important chapters.