IIT JEE Test Series — Practice smarter, perform stronger.

Differential Equation formulas

Master Differential Equation through 32 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Practise these formulas Know more about IIT JEE

Back to Formula Book

Differential Equation, every formula

32 formulas, typeset and free. Print it, or keep it open beside your practice.

Order of a DE
$order = order of the highest derivative present$
Degree of a DE
$degree = power of the highest-order derivative (rational, radical-free form)$
General solution
$contains as many arbitrary constants as the order$
Formation of a DE
$eliminate n arbitrary constants \Rightarrow DE of order n$
Variable separable
$\frac{d y}{d x} = \frac{f ( x )}{g ( y )} \Rightarrow \int g (y) d y = \int f (x) d x$
Q1MCQVariable separable
The solution of $\dfrac{dy}{dx}=\dfrac{x}{y}$ is:
- A
  $y^{2}-x^{2}=C$
- B
  $y^{2}+x^{2}=C$
- C
  $xy=C$
- D
  $y=Cx$
Reducible to separable
$\frac{d y}{d x} = f (a x + b y + c) : put z = a x + b y + c$
Q1MCQReducible
For $\dfrac{dy}{dx}=(x+y)^{2}$, the useful substitution is:
- A
  $z=x+y$
- B
  $z=xy$
- C
  $y=vx$
- D
  $z=x-y$
Homogeneous function
$f (k x, k y) = k^{n} f (x, y)$
degree n
Q1MCQHomogeneous function
$f(x,y)=x^{2}+xy$ is homogeneous of degree:
- A
  $2$
- B
  $1$
- C
  $3$
- D
  $0$
Homogeneous DE substitution
$\frac{d y}{d x} = F (\frac{y}{x}) : put y = v x, \frac{d y}{d x} = v + x \frac{d v}{d x}$
Q1MCQHomogeneous DE
For $\dfrac{dy}{dx}=\dfrac{x+y}{x}$, the substitution is:
- A
  $y=vx$
- B
  $z=x+y$
- C
  $x=vy$ only
- D
  $z=xy$
Homogeneous DE (x=vy)
$put x = v y, \frac{d x}{d y} = v + y \frac{d v}{d y}$
Q1MCQHomogeneous (x=vy)
When the RHS is a function of $x/y$, a convenient substitution is:
- A
  $x=vy$
- B
  $y=vx$
- C
  $z=x+y$
- D
  $z=xy$
Non-homogeneous (parallel lines)
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} : put a_{1} x + b_{1} y = t$
Q1MCQParallel-lines case
For $\dfrac{dy}{dx}=\dfrac{x+y+1}{x+y+2}$ (parallel numerator/denominator), put:
- A
  $t=x+y$
- B
  $y=vx$
- C
  $x=vy$
- D
  shift origin
Non-homogeneous (shift origin)
$put x = X + h, y = Y + k with a_{i} h + b_{i} k + c_{i} = 0$
Q1MCQShift origin
When $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$, the non-homogeneous DE is solved by:
- A
  shifting the origin $x=X+h,\ y=Y+k$
- B
  $t=a_1x+b_1y$
- C
  $y=vx$ directly
- D
  squaring
Linear DE in y
$\frac{d y}{d x} + P (x) y = Q (x)$
Q1MCQLinear DE
$\dfrac{dy}{dx}+2y=x$ is a linear DE with $P(x)=$:
- A
  $2$
- B
  $x$
- C
  $-2$
- D
  $2y$
Integrating factor (in y)
$I.F. = e^{\int P d x}$
Q1MCQIntegrating factor
The integrating factor of $\dfrac{dy}{dx}+2y=x$ is:
- A
  $e^{2x}$
- B
  $e^{x}$
- C
  $e^{x^{2}}$
- D
  $2x$
Linear DE solution
$y \cdot (I.F.) = \int Q \cdot (I.F.) d x + C$
Q1MCQLinear DE solution
The solution of a linear DE is:
- A
  $y\cdot\text{I.F.}=\int Q\cdot\text{I.F.}\,dx+C$
- B
  $y=\int Q\,dx$
- C
  $y\cdot\text{I.F.}=Q$
- D
  $y=\text{I.F.}\cdot Q$
Linear DE in x
$\frac{d x}{d y} + P (y) x = Q (y), I.F. = e^{\int P d y}$
Q1MCQLinear in x
The integrating factor of $\dfrac{dx}{dy}+\dfrac{x}{y}=y$ is:
- A
  $y$
- B
  $e^{y}$
- C
  $\dfrac1y$
- D
  $\ln y$
Bernoulli's equation
$\frac{d y}{d x} + P (x) y = Q (x) y^{n}$
divide by y^n, put z=y^{1-n}
Q1MCQBernoulli
$\dfrac{dy}{dx}+Py=Qy^{3}$ is a Bernoulli equation; divide by:
- A
  $y^{3}$
- B
  $y$
- C
  $y^{2}$
- D
  $x$
Bernoulli substitution
$z = y^{1 - n} \Rightarrow linear in z$
Q1MCQBernoulli substitution
For $\dfrac{dy}{dx}+Py=Qy^{3}$, the substitution is:
- A
  $z=y^{-2}$
- B
  $z=y^{2}$
- C
  $z=y^{3}$
- D
  $z=\ln y$
Orthogonal trajectory rule
$replace \frac{d y}{d x} \to - \frac{d x}{d y}$
Q1MCQOrthogonal trajectory
To find the orthogonal trajectory, replace $\dfrac{dy}{dx}$ by:
- A
  $-\dfrac{dx}{dy}$
- B
  $\dfrac{dx}{dy}$
- C
  $-\dfrac{dy}{dx}$
- D
  $0$
Exact differential xdy+ydx
$x d y + y d x = d (x y)$
Q1MCQd(xy)
$x\,dy+y\,dx$ equals:
- A
  $d(xy)$
- B
  $d\!\left(\dfrac{x}{y}\right)$
- C
  $d(x+y)$
- D
  $d(x^{2}+y^{2})$
Exact differential (ydx-xdy)/y²
$\frac{y d x - x d y}{y ^{2}} = d (\frac{x}{y})$
Q1MCQd(x/y)
$\dfrac{y\,dx-x\,dy}{y^{2}}$ equals:
- A
  $d\!\left(\dfrac{x}{y}\right)$
- B
  $d\!\left(\dfrac{y}{x}\right)$
- C
  $d(xy)$
- D
  $d\!\left(\ln\dfrac{x}{y}\right)$
Exact differential (xdy-ydx)/x²
$\frac{x d y - y d x}{x ^{2}} = d (\frac{y}{x})$
Q1MCQd(y/x)
$\dfrac{x\,dy-y\,dx}{x^{2}}$ equals:
- A
  $d\!\left(\dfrac{y}{x}\right)$
- B
  $d\!\left(\dfrac{x}{y}\right)$
- C
  $d(xy)$
- D
  $d(x^{2}+y^{2})$
d(x/y)
$d (\frac{x}{y}) = \frac{2 x y d x - x ^{2} d y}{y ^{2}} (for x^{2} / y)$
as in the table
Q1MCQtan⁻¹(y/x)
$\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}$ equals:
- A
  $d\!\left(\tan^{-1}\dfrac{y}{x}\right)$
- B
  $d\!\left(\dfrac{y}{x}\right)$
- C
  $d\big(\ln xy\big)$
- D
  $d(x^{2}+y^{2})$
tan⁻¹(y/x) differential
$\frac{x d y - y d x}{x ^{2} + y ^{2}} = d (tan^{- 1} \frac{y}{x})$
Q1MCQd(ln xy)
$\dfrac{x\,dy+y\,dx}{xy}$ equals:
- A
  $d\big(\ln xy\big)$
- B
  $d(xy)$
- C
  $d\!\left(\dfrac{x}{y}\right)$
- D
  $d\!\left(\ln\dfrac{y}{x}\right)$
log(xy) differential
$\frac{x d y + y d x}{x y} = d (ln x y)$
Q1MCQd(ln y/x)
$\dfrac{x\,dy-y\,dx}{xy}$ equals:
- A
  $d\!\left(\ln\dfrac{y}{x}\right)$
- B
  $d\big(\ln xy\big)$
- C
  $d\!\left(\dfrac{y}{x}\right)$
- D
  $d(xy)$
log(y/x) differential
$\frac{x d y - y d x}{x y} = d (ln \frac{y}{x})$
Q1MCQd(x²+y²)
$2(x\,dx+y\,dy)$ equals:
- A
  $d(x^{2}+y^{2})$
- B
  $d(xy)$
- C
  $d(x^{2}-y^{2})$
- D
  $d(x+y)^{2}$
d(x²+y²)
$2 (x d x + y d y) = d (x^{2} + y^{2})$
Q1MCQexp roots DE
The DE whose solution is $y=A e^{2x}+B e^{3x}$ is:
- A
  $y_2-5y_1+6y=0$
- B
  $y_2+5y_1+6y=0$
- C
  $y_2-6y_1+5y=0$
- D
  $y_2-y_1-6y=0$
y=A e^{αx}+B e^{βx} ⇒ DE
$y_{2} - (α + β) y_{1} + α β y = 0$
Q1MCQcomplex roots DE
The DE whose solution is $y=e^{x}(A\cos x+B\sin x)$ is:
- A
  $y_2-2y_1+2y=0$
- B
  $y_2+2y_1+2y=0$
- C
  $y_2-2y_1+y=0$
- D
  $y_2+2y=0$
y=e^{αx}(A cosβx+B sinβx) ⇒ DE
$y_{2} - 2 α y_{1} + (α^{2} + β^{2}) y = 0$
Q1MCQx-intercept of tangent
The x-intercept of the tangent at $(x,y)$ is:
- A
  $x-\dfrac{y}{\,dy/dx\,}$
- B
  $y-x\dfrac{dy}{dx}$
- C
  $x+\dfrac{y}{\,dy/dx\,}$
- D
  $\dfrac{dy}{dx}$
Tangent intercept on x-axis
$x - \frac{y}{d y / d x}$
Q1MCQy-intercept of tangent
The y-intercept of the tangent at $(x,y)$ is:
- A
  $y-x\dfrac{dy}{dx}$
- B
  $x-\dfrac{y}{\,dy/dx\,}$
- C
  $y+x\dfrac{dy}{dx}$
- D
  $\dfrac{dy}{dx}$
Tangent intercept on y-axis
$y - x \frac{d y}{d x}$
Q1MCQSimple separable
The solution of $\dfrac{dy}{dx}=e^{x}$ is:
- A
  $y=e^{x}+C$
- B
  $y=e^{x}$
- C
  $y=xe^{x}+C$
- D
  $y=\ln x+C$
dy/dx = e^{x} solution
$y = e^{x} + C$
simple separable
Q1MCQOrthogonal to circles
The orthogonal trajectories of the family of concentric circles $x^{2}+y^{2}=a^{2}$ are:
- A
  straight lines $y=mx$ through the origin
- B
  parabolas
- C
  other circles
- D
  ellipses
Family y=mx orthogonal to circles
$y = m x ⊥ x^{2} + y^{2} = a^{2}$

More JEE Advanced Mathematics formula sheets

Every chapter sheet is typeset, free and open without a sign-in.

Other ways to revise this chapter

Master this chapter with similar other learning materials.

Preparing students for India’s top institutes

Our students are currently into top technological and medical institutes of India.

IIT Bombay
IIT Delhi
IIT Madras
IIT Kanpur
IIT Kharagpur
IIT Roorkee
IIT Guwahati
IIT BHU Varanasi
AIIMS Delhi
NIT Tiruchirappalli
NIT Rourkela
IIT Bombay
IIT Delhi
IIT Madras
IIT Kanpur
IIT Kharagpur
IIT Roorkee
IIT Guwahati
IIT BHU Varanasi
AIIMS Delhi
NIT Tiruchirappalli
NIT Rourkela

Join QuestPix, Today!

Get notified first, with exam & curriculum updates, course & test series launch offers, motivation & success stories and free learning resources recommended by toppers.

Differential Equation formulas

Differential Equation, every formula

Order of a DE

Degree of a DE

General solution

Formation of a DE

Variable separable

Reducible to separable

Homogeneous function

Homogeneous DE substitution

Homogeneous DE (x=vy)

Non-homogeneous (parallel lines)

Non-homogeneous (shift origin)

Linear DE in y

Integrating factor (in y)

Linear DE solution

Linear DE in x

Bernoulli's equation

Bernoulli substitution

Orthogonal trajectory rule

Exact differential xdy+ydx

Exact differential (ydx-xdy)/y²

Exact differential (xdy-ydx)/x²

d(x/y)

tan⁻¹(y/x) differential

log(xy) differential

log(y/x) differential

d(x²+y²)

y=A e^{αx}+B e^{βx} ⇒ DE

y=e^{αx}(A cosβx+B sinβx) ⇒ DE

Tangent intercept on x-axis

Tangent intercept on y-axis

dy/dx = e^{x} solution

Family y=mx orthogonal to circles

More JEE Advanced Mathematics formula sheets

Other ways to revise this chapter

Preparing students for India’s top institutes

Join QuestPix, Today!