MathsISC Class 12 · Chapter 9

Differential Equations ISC Class 12: Notes, Methods and Solved Questions

Updated 14 October 2026 · Fact-checked

Differential Equations is Chapter 9 of ISC Class 12 Mathematics. It covers order and degree, general and particular solutions, forming an equation by eliminating arbitrary constants, and solving first-order, first-degree equations: variables separable (and reducible forms), homogeneous equations, and linear equations of the forms dy/dx + Py = Q and dx/dy + Px = Q.

What Differential Equations is about

A differential equation relates a function to its derivatives. This chapter is part of Calculus (35 of 80 marks) and usually appears as a long question asking for a particular solution, which examiners say many candidates fail to complete.

Order and degree: the order is that of the highest derivative; the degree is the power of that highest derivative once the equation is free of radicals and fractions in the derivatives (it is defined only when the equation is a polynomial in the derivatives). Formation: a family of curves with n arbitrary constants gives a differential equation of order n, found by differentiating n times and eliminating the constants. A general solution contains as many arbitrary constants as the order; a particular solution uses given conditions to fix them.

ISC solves only first-order, first-degree equations (second-order equations are excluded):

1. Variables separable: write f(y)dy = g(x)dx and integrate both sides; equations such as $\frac{d y}{d x} = (x + y)^{2}$ are reducible to this form by a substitution v = x + y.
2. Homogeneous: $\frac{d y}{d x} = F (\frac{y}{x})$ ; substitute y = vx, separate and integrate (or x = vy for $\frac{d x}{d y} = F (\frac{x}{y})$ ).
3. Linear: $\frac{d y}{d x} + P y = Q$ with integrating factor $I F = e^{\int P d x}$ , so $y \cdot I F = \int Q \cdot I F d x + C$ ; similarly $\frac{d x}{d y} + P x = Q$ when the equation is linear in x.

The skill that decides marks is identifying the type before starting; the second is simplifying the integrating factor (for example $e^{l o g x} = x$ ); the third is finishing with the particular solution.

Weightage: Differential Equations is part of the unit Calculus, which carries 35 of the 80 marks in the ISC Mathematics, shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Order and degree: The order is the order of the highest derivative present. The degree is the highest power of that highest-order derivative, after the equation has been made free of radicals and fractional powers in the derivatives. If the equation involves derivatives inside functions like sin(y′) or $e^{y^{'}}$ , the degree is not defined. Example: $(\frac{d ^{2} y}{d x ^{2}})^{3} + (\frac{d y}{d x})^{4} + y = 0$ has order 2 and degree 3.; Worked example
Find the order and degree of $\frac{d ^{2} y}{d x ^{2}} = 1 + (\frac{d y}{d x})^{2}$ .
Try it first, then show the solutionSolution
Highest derivative: second order.
Square to remove the radical: $(\frac{d ^{2} y}{d x ^{2}})^{2} = 1 + (\frac{d y}{d x})^{2}$ ; the power of y″ is 2.
Answer: Order 2, degree 2
Forming a differential equation: Given a family of curves with n arbitrary constants, differentiate n times; together with the original equation you have n + 1 equations, from which the n constants are eliminated. The result is a differential equation of order n. Keep the algebra tidy: often the constants can be isolated from the first derivative and substituted.; Worked example
Form the differential equation of the family $y = A e^{2 x} + B e^{- x}$ .
Try it first, then show the solutionSolution
$y^{'} = 2 A e^{2 x} - B e^{- x}$ , $y^{''} = 4 A e^{2 x} + B e^{- x}$ .
$y^{''} - y^{'} = 2 A e^{2 x} + 2 B e^{- x} = 2 y$ .
Answer: $\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0$
General and particular solutions; verification: A solution is a function that satisfies the equation. The general solution of an nth-order equation contains n independent arbitrary constants; a particular solution is obtained by giving them values, usually from initial conditions such as y = 2 when x = 0. To verify a solution, differentiate it and substitute into the equation. Never omit the constant C when integrating, and substitute the condition only after integrating.; Worked example
Verify that $y = A e^{x} + B e^{- x}$ is a solution of $\frac{d ^{2} y}{d x ^{2}} = y$ .
Try it first, then show the solutionSolution
$y^{'} = A e^{x} - B e^{- x}$
$y^{''} = A e^{x} + B e^{- x} = y$
Answer: Verified (two constants, as expected for a second-order equation)
Variables separable: If the equation can be written as $\frac{d y}{d x} = g (x) h (y)$ , separate: $\frac{d y}{h ( y )} = g (x) d x$ , integrate both sides and add one constant. Simplify logarithms into a neat form (e.g. log|y| = … ⇒ y = Ce^{…}). Then apply any given condition.; Worked example
Solve $\frac{d y}{d x} = x y$ , given y = 2 when x = 0.
Try it first, then show the solutionSolution
$\frac{d y}{y} = x d x \Rightarrow lo g ∣ y ∣ = \frac{x ^{2}}{2} + C$ , so $y = A e^{x^{2} /2}$ .
y(0) = 2 gives A = 2.
Answer: $y = 2 e^{x^{2} /2}$
Equations reducible to variables separable: Some equations become separable after a substitution. For $\frac{d y}{d x} = f (a x + b y + c)$ , put v = ax + by + c, so $\frac{d v}{d x} = a + b \frac{d y}{d x} = a + b f (v)$ , which separates. Write the final answer in terms of x and y.; Worked example
Solve $\frac{d y}{d x} = (x + y)^{2}$ .
Try it first, then show the solutionSolution
v = x + y: $\frac{d v}{d x} = 1 + \frac{d y}{d x} = 1 + v^{2}$ .
$\int \frac{d v}{1 + v ^{2}} = \int d x \Rightarrow tan^{- 1} v = x + C$ .
Answer: $tan^{- 1} (x + y) = x + C$ , i.e. $x + y = tan (x + C)$
Homogeneous differential equations: A function f(x, y) is homogeneous of degree n if $f (λ x, λ y) = λ^{n} f (x, y)$ . The equation $\frac{d y}{d x} = \frac{f ( x , y )}{g ( x , y )}$ is homogeneous if f and g are homogeneous of the same degree, so the right side depends only on y/x. Method: put y = vx, $\frac{d y}{d x} = v + x \frac{d v}{d x}$ ; the equation separates in v and x; integrate and replace v by y/x. If the equation is more natural as $\frac{d x}{d y} = F (\frac{x}{y})$ , put x = vy.; Worked example
Solve $\frac{d y}{d x} = \frac{y}{x} + tan \frac{y}{x}$ .
Try it first, then show the solutionSolution
y = vx: $v + x \frac{d v}{d x} = v + tan v \Rightarrow cot v d v = \frac{d x}{x}$ .
$lo g ∣ sin v ∣ = lo g ∣ x ∣ + lo g C$
Answer: $sin \frac{y}{x} = C x$
Linear equations dy/dx + Py = Q: If P and Q are functions of x (or constants), the equation is linear in y. Multiply by the integrating factor $I F = e^{\int P d x}$ ; the left side becomes $\frac{d}{d x} (y \cdot I F)$ , so $y \cdot I F = \int Q \cdot I F d x + C$ . Simplify the IF: $e^{l o g x} = x$ , $e^{- l o g x} = \frac{1}{x}$ , $e^{\int t a n x d x} = sec x$ . First rewrite the equation so the coefficient of $\frac{d y}{d x}$ is 1.; Worked example
Solve $\frac{d y}{d x} + \frac{y}{x} = x^{2}$ , given y = 1 when x = 1.
Try it first, then show the solutionSolution
P = 1/x: IF = $e^{l o g x} = x$ .
$x y = \int x^{3} d x = \frac{x ^{4}}{4} + C$ .
x = 1, y = 1: $1 = \frac{1}{4} + C$ , C = ¾; so $y = \frac{x ^{3}}{4} + \frac{3}{4 x}$ .
Answer: $y = \frac{x ^{3}}{4} + \frac{3}{4 x}$
Linear equations dx/dy + Px = Q: If an equation is not linear in y but is linear in x (P and Q functions of y), solve for $\frac{d x}{d y}$ : $\frac{d x}{d y} + P x = Q$ with IF = $e^{\int P d y}$ , and $x \cdot I F = \int Q \cdot I F d y + C$ . A sign that you need this form: the equation contains $y^{2}$ or another non-linear y term alongside $\frac{d y}{d x}$ , as in $(x + 2 y^{2}) \frac{d y}{d x} = y$ .; Worked example
Solve $(x + 2 y^{2}) \frac{d y}{d x} = y$ (y > 0).
Try it first, then show the solutionSolution
Invert: $\frac{d x}{d y} = \frac{x}{y} + 2 y \Rightarrow \frac{d x}{d y} - \frac{1}{y} x = 2 y$ .
IF = $e^{- l o g y} = \frac{1}{y}$ : $\frac{x}{y} = \int 2 d y = 2 y + C$ .
Answer: $x = 2 y^{2} + C y$

Maths in real life: Differential Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Exponential growth

Suppose a town's population grows at a rate proportional to its size and doubles in 20 years.

How long will it take to triple?

Try it first, then show the solutionSolution

$\frac{d P}{d t} = k P \Rightarrow P = P_{0} e^{k t}$ ; doubling: $e^{20 k} = 2$ , so $k = \frac{l o g 2}{20}$ .
Tripling: $e^{k t} = 3 \Rightarrow t = \frac{l o g 3}{k} = \frac{20 l o g 3}{l o g 2} \approx 31.7$

Answer: About 31.7 years

Memory hook: Rate proportional to amount: the answer is always exponential.

Newton's law of cooling

Imagine a cup of tea at 85 °C left in a room at 25 °C; after 10 minutes it is at 65 °C.

Assuming the rate of cooling is proportional to the temperature difference, find its temperature after 20 minutes.

Try it first, then show the solutionSolution

$T - 25 = 60 e^{- k t}$ ; at t = 10: $40 = 60 e^{- 10 k} \Rightarrow e^{- 10 k} = \frac{2}{3}$ .
At t = 20: $T - 25 = 60 (\frac{2}{3})^{2} = 26.7$ , so T ≈ 51.7 °C.

Answer: About 51.7 °C

Memory hook: Equal time steps multiply the temperature difference by the same factor.

Continuous compounding

Suppose ₹10 000 is invested at 8% per year compounded continuously, so dA/dt = 0.08A.

Find the amount after 5 years.

Try it first, then show the solutionSolution

$A = 10000 e^{0.08 t}$
At t = 5: $10000 e^{0.4} \approx 14918$

Answer: About ₹14 918

Memory hook: dA/dt = rA gives A = A₀e^{rt}.

Topics in this chapter

Differential Equations formula sheet

Order and degree: $order = highest derivative; degree = its power (polynomial form)$; Degree undefined if derivatives sit inside sin, log, e^.
Variables separable: $\frac{d y}{d x} = g (x) h (y) \Rightarrow \int \frac{d y}{h ( y )} = \int g (x) d x + C$
Reducible form: $\frac{d y}{d x} = f (a x + b y + c) : v = a x + b y + c, \frac{d v}{d x} = a + b f (v)$
Homogeneous: $\frac{d y}{d x} = F (\frac{y}{x}) : y = v x, \frac{d y}{d x} = v + x \frac{d v}{d x}$; Or x = vy for $\frac{d x}{d y} = F (\frac{x}{y})$ .
Linear in y: $\frac{d y}{d x} + P y = Q : I F = e^{\int P d x}, y \cdot I F = \int Q \cdot I F d x + C$; Coefficient of dy/dx must be 1.
Linear in x: $\frac{d x}{d y} + P x = Q : I F = e^{\int P d y}, x \cdot I F = \int Q \cdot I F d y + C$
Useful integrating factors: $e^{l o g x} = x, e^{- l o g x} = \frac{1}{x}, e^{\int t a n x d x} = sec x, e^{\int \frac{2 x}{1 + x ^{2}} d x} = 1 + x^{2}$
Growth and decay: $\frac{d N}{d t} = k N \Rightarrow N = N_{0} e^{k t}; \frac{d T}{d t} = - k (T - T_{s}) \Rightarrow T - T_{s} = (T_{0} - T_{s}) e^{- k t}$; Separable models.

How the board asks Differential Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Order, degree and type (Section A)	Order and degree of a given equation; which of four equations is homogeneous; value of a constant for which the solution is a circle.	Remove radicals before reading degree; test f(λx, λy) for homogeneity.
Formation and verification	Form the equation of a family; show that a function is the general solution.	Differentiate as often as there are constants; eliminate; verify by substitution.
Separable (2-3 marks)	Solve with or without a condition; reducible forms.	Separate, integrate both sides, add C once, apply the condition.
Homogeneous (5 marks)	Particular solution of an equation like $(x^{2} - 2 y^{2}) d x + 2 x y d y = 0$ or with $e^{x / y}$ .	Write dy/dx (or dx/dy) as F(y/x); substitute y = vx; separate; return to x, y; apply the condition.
Linear (5 marks)	dy/dx + Py = Q or dx/dy + Px = Q with a condition.	Standard form, IF simplified, integrate Q·IF, then the particular solution.
Real-life models	Growth, cooling, interest, draining tanks.	Set up the rate equation, separate, use two data points to find constants.

Tricks to remember Differential Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Type first, then method
Separable? Homogeneous (y/x)? Linear in y? Linear in x? Decide before writing anything else.
Use it when: Avoiding the wrong method examiners report.
IF: e to the integral of P
'Eat Pie': e^{∫P}; then y × IF = ∫Q × IF.
Use it when: Linear equations.
e and log cancel
$e^{l o g f} = f$ : simplify the IF before integrating.
Use it when: Clean integrating factors.
y = vx, then v + xv′
Homogeneous substitution always gives dy/dx = v + x dv/dx.
Use it when: Homogeneous equations.
Constants = order
General solution has as many constants as the order; a family with n constants gives order n.
Use it when: Formation and checking.

Differential Equations: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Order and degree · 1 mark · MCQBoard level

The order and degree of $(\frac{d ^{2} y}{d x ^{2}})^{3} + (\frac{d y}{d x})^{4} + y = 0$ are:
(a) 2 and 4
(b) 2 and 3
(c) 4 and 3
(d) 3 and 2

Try it first, then show the solutionSolution

Highest derivative y″ (order 2) appears to the power 3 (degree 3).

Answer: (b)

Example 2: Recognising a homogeneous equation · 1 mark · Assertion-reasonBoard level

Assertion: $\frac{d y}{d x} = \frac{x + y}{x}$ is a homogeneous differential equation.
Reason: The function $f (x, y) = \frac{x + y}{x}$ satisfies $f (λ x, λ y) = λ^{0} f (x, y)$ .
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

$\frac{λ x + λ y}{λ x} = \frac{x + y}{x}$ : homogeneous of degree 0, which is the test for a homogeneous equation.

Answer: (a)

Example 3: Forming the equation of a family · 2 marks · Short answerHarder board level

Form the differential equation of the family of curves $y = A e^{2 x} + B e^{- x}$ . [2]

Try it first, then show the solutionSolution

$y^{'} = 2 A e^{2 x} - B e^{- x}$ ; $y^{''} = 4 A e^{2 x} + B e^{- x}$ .
$y^{''} - y^{'} = 2 (A e^{2 x} + B e^{- x}) = 2 y$ , so $y^{''} - y^{'} - 2 y = 0$ .

Answer: $\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0$

Example 4: Separable with a condition · 3 marks · Short answerHarder board level

Find the particular solution of $\frac{d y}{d x} = x y$ given that y = 2 when x = 0. [3]

Try it first, then show the solutionSolution

Separate: $\frac{d y}{y} = x d x$ .
Integrate: $lo g ∣ y ∣ = \frac{x ^{2}}{2} + C \Rightarrow y = A e^{x^{2} /2}$ .
x = 0, y = 2 ⇒ A = 2.

Answer: $y = 2 e^{x^{2} /2}$

Example 5: Linear equation with a condition · 3 marks · Short answerHarder board level

Solve $x \frac{d y}{d x} + y = x^{3}$ , given y = 1 when x = 1. [3]

Try it first, then show the solutionSolution

Divide by x: $\frac{d y}{d x} + \frac{1}{x} y = x^{2}$ ; IF = $e^{l o g x} = x$ .
$x y = \int x^{3} d x = \frac{x ^{4}}{4} + C$ .
x = 1, y = 1: C = ¾, so $y = \frac{x ^{3}}{4} + \frac{3}{4 x}$ .

Answer: $y = \frac{x ^{3}}{4} + \frac{3}{4 x}$

Example 6: Homogeneous equation with a particular solution · 5 marks · Long answerHarder board level

(a) Show that $(x^{2} + y^{2}) d x - 2 x y d y = 0$ is homogeneous. [1]
(b) Find its particular solution given y = 0 when x = 1. [4]

Try it first, then show the solutionSolution

(a) $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y}$ ; numerator and denominator are both homogeneous of degree 2, so the right side is a function of y/x.
(b) y = vx: $v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{2 v} \Rightarrow x \frac{d v}{d x} = \frac{1 - v ^{2}}{2 v}$ .
Separate: $\frac{2 v d v}{1 - v ^{2}} = \frac{d x}{x} \Rightarrow - lo g ∣1 - v^{2} ∣ = lo g ∣ x ∣ + lo g C$ , so $x (1 - v^{2}) = k$ .
With v = y/x: $x (1 - \frac{y ^{2}}{x ^{2}}) = k \Rightarrow x^{2} - y^{2} = k x$ .
x = 1, y = 0 ⇒ k = 1: $x^{2} - y^{2} = x$ .

Answer: $x^{2} - y^{2} = x$

Example 7: Reducible to separable · 3 marks · Short answerHOTS / competency

Solve $\frac{d y}{d x} = (x + y)^{2}$ . [3]

Try it first, then show the solutionSolution

Let v = x + y: $\frac{d v}{d x} = 1 + \frac{d y}{d x} = 1 + v^{2}$ .
$\int \frac{d v}{1 + v ^{2}} = \int d x \Rightarrow tan^{- 1} v = x + C$ .
So $tan^{- 1} (x + y) = x + C$ .

Answer: $tan^{- 1} (x + y) = x + C$

Example 8: Verifying a general solution · 2 marks · Short answerHOTS / competency

Show that $y = A e^{x} + B e^{- x}$ is the general solution of $\frac{d ^{2} y}{d x ^{2}} - y = 0$ . [2]

Try it first, then show the solutionSolution

$y^{''} = A e^{x} + B e^{- x} = y$ , so the equation is satisfied for all A, B.
It has two independent arbitrary constants, matching the order 2, so it is the general solution.

Answer: Verified

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

A cooling cup of tea

Suppose a cup of tea at 85 °C cools in a room at 25 °C. The rate of fall of its temperature T is proportional to (T − 25). After 10 minutes the tea is at 65 °C.

(i) · 1 mark
Write the differential equation and its type.
Answer
Answer: $\frac{d T}{d t} = - k (T - 25)$ ; variables separable
k > 0 because T falls.
(ii) · 1 mark
Solve it with T = 85 at t = 0.
Answer
Answer: $T = 25 + 60 e^{- k t}$
$\int \frac{d T}{T - 25} = - k \int d t \Rightarrow lo g (T - 25) = - k t + C$ ; C = log 60.
(iii) · 1 mark
Use the 10-minute reading to find $e^{- 10 k}$ .
Answer
Answer: 2/3
40 = 60e^{−10k}.
(iv) · 2 marks
Find the temperature after 20 minutes and say whether the tea will ever reach 25 °C.
Answer
Answer: About 51.7 °C; it approaches 25 °C but never exactly reaches it.
$T = 25 + 60 (2/3)^{2} = 51.7$ ; the exponential term is never zero.

Bacteria in a culture

Suppose a culture starts with 1000 bacteria and grows at a rate proportional to the number present, doubling every 3 hours.

(i) · 1 mark
Write and solve the differential equation.
Answer
Answer: $\frac{d N}{d t} = k N$ , $N = 1000 e^{k t}$
Separable; N(0) = 1000.
(ii) · 1 mark
Express N in terms of powers of 2.
Answer
Answer: $N = 1000 \cdot 2^{t /3}$
$e^{3 k} = 2$ .
(iii) · 1 mark
Find the number after 9 hours.
Answer
Answer: 8000
$2^{3} = 8$ .
(iv) · 2 marks
When will there be 50 000 bacteria?
Answer
Answer: About 16.9 hours
$2^{t /3} = 50 \Rightarrow t = 3 lo g_{2} 50 = 3 \frac{l o g 50}{l o g 2} \approx 16.9$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the order and degree of $\frac{d ^{2} y}{d x ^{2}} = 1 + (\frac{d y}{d x})^{2}$ . (2 marks)

Step	Marks
Order 2 (highest derivative y″)	1
Squares to remove the radical; degree 2	1

Solve $\frac{d y}{d x} = x y$ , given y(0) = 2. (3 marks)

Step	Marks
Separates variables correctly	1
Integrates with constant: log\|y\| = x²/2 + C	1
Particular solution y = 2e^{x²/2}	1

Solve $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$ , given y = 0 when x = 0. (5 marks)

Step	Marks
Standard form $\frac{d y}{d x} + \frac{2 x}{1 + x ^{2}} y = \frac{4 x ^{2}}{1 + x ^{2}}$	1
IF = $e^{l o g (1 + x^{2})} = 1 + x^{2}$	1
$y (1 + x^{2}) = \int 4 x^{2} d x$	1
$y (1 + x^{2}) = \frac{4 x ^{3}}{3} + C$	1
C = 0: $y = \frac{4 x ^{3}}{3 ( 1 + x ^{2} )}$	1

Where students lose marks in Differential Equations

The slips examiners see most often, and what to write instead.

Mistake: Not identifying the type of equation
CISCE examiners' comments, 2024 and 2025: candidates could not identify homogeneous equations, tried the homogeneous method on an equation linear in x, and did not recognise the form dx/dy + Px = Q.
Wrong working
$(x + 5 y^{2}) \frac{d y}{d x} = y$ treated as homogeneous.
Right working
$\frac{d x}{d y} - \frac{x}{y} = 5 y$ : linear in x with IF = 1/y.
Do this: Check homogeneity with f(λx, λy) and look for linearity in x as well as y.
Usually costs: Most of a 5-mark question.
Mistake: Leaving the integrating factor unsimplified
CISCE examiners' comments, 2025: candidates did not simplify integrating factors of the type $e^{l o g f (x)}$ .
Wrong working
IF = $e^{- l o g y}$ kept in the integral.
Right working
$e^{- l o g y} = \frac{1}{y}$ .
Do this: Simplify the IF on its own line before multiplying.
Usually costs: 1-2 marks.
Mistake: Stopping at the general solution
CISCE examiners' comments, 2024 and 2025: candidates did not obtain the particular solution even when the condition was given, and omitted the constant C.
Wrong working
Answer: $x^{2} - y^{2} = k x$ (condition unused), or no C at all.
Right working
Use y = 0 when x = 1 to get k = 1: $x^{2} - y^{2} = x$ .
Do this: Underline the condition in the question and finish with it.
Usually costs: 1-2 marks.
Mistake: Integration errors after reduction
CISCE examiners' comments, 2024: after reducing an equation, candidates used wrong integration methods and could not integrate the right side of $x (x^{2} - 1) \frac{d y}{d x} = 1$ correctly.
Wrong working
$\int \frac{d x}{x ( x ^{2} - 1 )} = lo g ∣ x (x^{2} - 1) ∣$
Right working
Partial fractions: $- \frac{1}{x} + \frac{1/2}{x - 1} + \frac{1/2}{x + 1}$ , giving $- lo g ∣ x ∣ + \frac{1}{2} lo g ∣ x^{2} - 1∣$ .
Do this: Revise integration methods; they decide this chapter.
Usually costs: 2-3 marks.
Mistake: Reading the degree before removing radicals
A frequent Section A slip: the degree is read from an equation that still contains a square root of a derivative.
Wrong working
$y^{''} = 1 + y^{'2}$ has degree 1.
Right working
Square first: $(y^{''})^{2} = 1 + y^{'2}$ , degree 2.
Do this: Clear radicals and fractions in the derivatives first.
Usually costs: 1 mark.

Practice questions on Differential Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of arbitrary constants in the general solution of a third-order differential equation is:
(a) 0
(b) 1
(c) 2
(d) 3
Hint
Equal to the order.
Answer
(d)
Q2 · 1 mark · MCQBoard level
The integrating factor of $\frac{d y}{d x} + 2 y = e^{x}$ is:
(a) $e^{x}$
(b) $e^{2 x}$
(c) $e^{- 2 x}$
(d) 2x
Hint
$e^{\int 2 d x}$ .
Answer
(b)
Q3 · 1 mark · MCQBoard level
Which of these is a homogeneous differential equation?
(a) $\frac{d y}{d x} = x + y^{2}$
(b) $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$
(c) $\frac{d y}{d x} = \frac{x + 1}{y}$
(d) $\frac{d y}{d x} = e^{x + y}$
Hint
Same degree on top and bottom.
Answer
(b)
Q4 · 1 mark · Very short answerBoard level
Write the order and degree of $(\frac{d ^{2} y}{d x ^{2}})^{2} + \frac{d y}{d x} = 0$ .
Hint
Highest derivative and its power.
Answer
Order 2, degree 2
Q5 · 2 marks · Short answerBoard level
Solve $\frac{d y}{d x} = e^{x - y}$ .
Hint
$e^{y} d y = e^{x} d x$ .
Answer
$e^{y} = e^{x} + C$
Q6 · 2 marks · Short answerHarder board level
Form the differential equation of all straight lines through the origin.
Hint
y = mx.
Answer
y′ = m = y/x, so $x \frac{d y}{d x} = y$ .
Q7 · 2 marks · Short answerHarder board level
Solve $\frac{d x}{d y} - \frac{x}{y} = 2 y$ (y > 0).
Hint
IF = 1/y.
Answer
$\frac{x}{y} = 2 y + C$ , i.e. $x = 2 y^{2} + C y$
Q8 · 2 marks · Short answerHOTS / competency
Show that $y = A e^{x} + B e^{- x}$ satisfies $y^{''} - y = 0$ and explain why it is the general solution.
Hint
Count constants.
Answer
y″ = y; it contains two independent constants, equal to the order.
Q9 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} + y tan x = sec x$ .
Hint
IF = sec x.
Answer
y sec x = ∫ sec²x dx = tan x + C, i.e. y = sin x + C cos x.
Q10 · 3 marks · Short answerHarder board level
A cup of tea at 85 °C cools to 65 °C in 10 minutes in a 25 °C room. Assuming Newton's law of cooling, find its temperature after 20 minutes.
Hint
Temperature difference falls by the same factor each 10 minutes.
Answer
$T = 25 + 60 e^{- k t}$ with $e^{- 10 k} = \frac{2}{3}$ ; after 20 min T ≈ 51.7 °C.
Q11 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} = \frac{y}{x} + tan \frac{y}{x}$ .
Hint
y = vx.
Answer
$cot v d v = \frac{d x}{x}$ ⇒ $sin \frac{y}{x} = C x$
Q12 · 5 marks · Long answerHarder board level
Solve $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$ , given y = 0 when x = 0.
Hint
IF = 1 + x².
Answer
Standard form with P = 2x/(1 + x²); IF = 1 + x²; $y (1 + x^{2}) = \frac{4 x ^{3}}{3} + C$ ; C = 0; $y = \frac{4 x ^{3}}{3 ( 1 + x ^{2} )}$ .
Q13 · 2 marks · Short answerHarder board level
Solve $\frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}}$ .
Hint
Separate.
Answer
$tan^{- 1} y = tan^{- 1} x + C$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define the order of a differential equation.Show answer
The order of the highest derivative in it.
2.When is the degree defined?Show answer
When the equation is a polynomial in its derivatives.
3.How many constants does the general solution of an nth-order equation have?Show answer
n.
4.Substitution for a homogeneous equation?Show answer
y = vx, so dy/dx = v + x dv/dx.
5.Integrating factor of dy/dx + Py = Q?Show answer
$e^{\int P d x}$ .
6.Solution formula for a linear equation?Show answer
y·IF = ∫Q·IF dx + C.
7.What is a particular solution?Show answer
A solution with the arbitrary constants fixed by given conditions.
8.Are second-order equations solved in ISC?Show answer
No; they are excluded (only formation and verification may involve them).

Best order to study it

Learn order and degree, and practise formation by eliminating constants.
Practise verifying solutions and the meaning of general and particular solutions.
Master variables separable and the reducible form v = ax + by + c.
Practise homogeneous equations in both dy/dx and dx/dy forms.
Practise linear equations, simplifying integrating factors, including the dx/dy form.
Finish with growth, decay and cooling models.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find order and degree after removing radicals?
Can you form the equation of a family with two constants?
Can you solve a separable equation with a condition?
Can you solve dy/dx = (x + y)² by substitution?
Can you test for and solve a homogeneous equation?
Can you write the IF and solve a linear equation in y or in x?
Can you always finish with the particular solution?
Can you model growth or cooling with a differential equation?

Differential Equations: frequently asked questions

Are second-order differential equations in the ISC 2027 syllabus?

No. The syllabus explicitly excludes second-order differential equations from solution methods; you solve first-order, first-degree equations only.

Which types of equations must I be able to solve?

Variables separable (including equations reducible to it), homogeneous equations of first order and first degree, and linear equations dy/dx + Py = Q and dx/dy + Px = Q.

How do I know if an equation is linear in x?

If, after writing dx/dy, the equation has x only to the first power (not multiplied by a function of x) with coefficients in y, it is linear in x: dx/dy + Px = Q with P, Q functions of y.

Do I need to verify that the equation is homogeneous?

It is good practice and often worth a mark: show that the right side is a function of y/x, or that f(λx, λy) = λⁿf(x, y) for numerator and denominator with the same n.

How many marks does this chapter carry?

It is part of Calculus (35 of 80 marks); a 5-mark particular-solution question is common.

Why does my answer differ from the book's form?

Constants can be absorbed in different ways (log C, e^C, k). Your answer is correct if it differs only in the form of the arbitrary constant and satisfies the condition.

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024