MathsISC Class 12 · Differential Equations

Variables Separable Method: Separate, Integrate, Apply

Updated 15 October 2026 · Fact-checked

A differential equation is variables separable if it can be written as g(y) dy = f(x) dx, with all y-terms on one side and all x-terms on the other. Integrate both sides and add one constant. Equations like dy/dx = f(ax + by + c) become separable with the substitution v = ax + by + c.

Move all y terms with dy to one side and all x terms with dx to the other, integrate, and write one constant C.

Illustration drawn for QuestionBuddy.

Understand Variables Separable Method

The idea. If $\frac{d y}{d x} = f (x) g (y)$ , divide by $g (y)$ and multiply by $d x$ :
$\frac{d y}{g ( y )} = f (x) d x .$
Each side now involves only one variable, so integrate each side with respect to its own variable:
$\int \frac{d y}{g ( y )} = \int f (x) d x + C .$
Only one constant is needed; constants from the two sides combine. The ISC 2025 examiners noted candidates who omitted the constant $C$ altogether, which turns a general solution into one particular curve.

Spotting separability. Look for a product or quotient of a function of $x$ and a function of $y$ . Useful rewrites:

$e^{x + y} = e^{x} e^{y}$ , $e^{x - y} = \frac{e ^{x}}{e ^{y}}$ .
$(1 + x^{2}) d y + (1 + y^{2}) d x = 0$ gives $\frac{d y}{1 + y ^{2}} = - \frac{d x}{1 + x ^{2}}$ .
Factorise: $x y + x + y + 1 = (x + 1) (y + 1)$ .

An equation like $\frac{d y}{d x} = x + y$ is not separable; it is linear.

Equations reducible to separable form (in the syllabus). If the right side is a function of a linear combination, $\frac{d y}{d x} = f (a x + b y + c)$ , put $v = a x + b y + c$ . Then $\frac{d v}{d x} = a + b \frac{d y}{d x} = a + b f (v)$ , which is separable: $\frac{d v}{a + b f ( v )} = d x$ . For example $\frac{d y}{d x} = (x + y)^{2}$ becomes $\frac{d v}{1 + v ^{2}} = d x$ .

Integration after separating often needs partial fractions (ISC 2024 set $x (x^{2} - 1) \frac{d y}{d x} = 1$ , where the examiners noted candidates who failed to integrate the right side correctly), or standard results such as $\int \frac{d y}{1 + y ^{2}} = tan^{- 1} y$ .

General and particular solutions. A condition such as $y = 0$ when $x = 2$ fixes $C$ . The 2024 examiners again noted candidates who did not obtain the particular solution.

Growth and decay. The most important separable model is
$\frac{d y}{d t} = k y \Rightarrow \frac{d y}{y} = k d t \Rightarrow y = y_{0} e^{k t} .$
With $k > 0$ it describes growth (bacteria, populations, continuously compounded money); with $k < 0$ , decay (medicines leaving the blood, radioactive substances). The time to double is $\frac{l o g 2}{k}$ ; the half-life is $\frac{l o g 2}{∣ k ∣}$ .

Newton's law of cooling. A body cools at a rate proportional to the difference between its temperature $T$ and the surroundings $T_{s}$ : $\frac{d T}{d t} = - k (T - T_{s})$ . Separating, $T - T_{s} = (T_{0} - T_{s}) e^{- k t}$ . Two readings give $k$ ; then any time or temperature can be found.

Variables Separable Method: formulas to remember

Separable form: $\frac{d y}{d x} = f (x) g (y) \Rightarrow \int \frac{d y}{g ( y )} = \int f (x) d x + C$; One constant only.
Reducible form: $\frac{d y}{d x} = f (a x + b y + c) : v = a x + b y + c, \frac{d v}{d x} = a + b f (v)$; Then separate v and x.
Growth and decay: $\frac{d y}{d t} = k y \Rightarrow y = y_{0} e^{k t}$; k > 0 growth, k < 0 decay.
Doubling time and half-life: $t_{double} = \frac{l o g 2}{k}, t_{1/2} = \frac{l o g 2}{∣ k ∣}$; Natural logarithms.
Newton's law of cooling: $\frac{d T}{d t} = - k (T - T_{s}) \Rightarrow T = T_{s} + (T_{0} - T_{s}) e^{- k t}$; T_s is the temperature of the surroundings.
Useful integrals: $\int \frac{d y}{y} = lo g ∣ y ∣, \int \frac{d y}{1 + y ^{2}} = tan^{- 1} y, \int e^{- y} d y = - e^{- y}$; Add one constant C after integrating both sides.

How to solve Variables Separable Method questions

Use these steps for separable equations and their reducible forms.

1Rewrite the equation as $\frac{d y}{d x} = \dots$ and check whether the right side factorises into (function of $x$ ) × (function of $y$ ).
2If the right side is a function of $a x + b y + c$ , substitute $v = a x + b y + c$ and rewrite in terms of $v$ and $x$ .
3Separate: put every $y$ (or $v$ ) term with $d y$ on one side and every $x$ term with $d x$ on the other.
4Integrate both sides, using partial fractions or standard forms where needed, and add one constant $C$ .
5Simplify, combining logarithms if possible; replace $v$ by its expression in $x$ and $y$ if you substituted.
6If a condition is given, find $C$ and write the particular solution.
7For models (growth, decay, cooling), use the given data to find the rate constant first, then answer the question asked with units.

Variables Separable Method in real life

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 decay

Suppose a patient is given 500 mg of a medicine, and the amount $m$ (mg) left in the body after $t$ hours satisfies $\frac{d m}{d t} = - 0.2 m$ .

Find $m (t)$ , the amount left after 5 hours, and the half-life of the medicine.

Try it first, then show the solutionSolution

Separate: $\frac{d m}{m} = - 0.2 d t \Rightarrow lo g m = - 0.2 t + C$ .
At $t = 0$ , $m = 500$ , so $m = 500 e^{- 0.2 t}$ .
After 5 hours: $m = 500 e^{- 1} \approx 183.9$ mg.
Half-life: $e^{- 0.2 t} = \frac{1}{2} \Rightarrow t = \frac{l o g 2}{0.2} \approx 3.47$ hours.

Answer: $m = 500 e^{- 0.2 t}$ ; about 184 mg after 5 hours; half-life about 3.47 hours.

Memory hook: Rate proportional to amount: separate, integrate, get an exponential.

Exponential growth from two readings

Imagine a culture of bacteria that grows at a rate proportional to its size. There are 1000 bacteria at the start and 4000 after 2 hours.

How many bacteria will there be after 5 hours?

Try it first, then show the solutionSolution

$\frac{d N}{d t} = k N \Rightarrow N = 1000 e^{k t}$ .
$4000 = 1000 e^{2 k} \Rightarrow e^{2 k} = 4 \Rightarrow k = lo g 2$ per hour.
$N (5) = 1000 e^{5 l o g 2} = 1000 \times 2^{5} = 32000$ .

Answer: 32 000 bacteria (the population doubles every hour).

Memory hook: Two readings fix k; after that the exponential does the rest.

Tricks to remember Variables Separable Method

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

y with dy, x with dx
Every y-term goes with dy, every x-term with dx, before you integrate.
Use it when: Setting up the separation.
One C is enough
Integrate both sides, add C once, on the x-side.
Use it when: Avoiding the omitted-constant error.
Inside the bracket? Call it v
dy/dx = f(x + y): let v = x + y, then dv/dx = 1 + dy/dx.
Use it when: Reducible equations.
log 2 over k
Doubling time and half-life are both log 2 divided by the rate constant.
Use it when: Growth and decay questions.

Variables Separable Method: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Splitting an exponential · 1 mark · MCQBoard level

The general solution of $\frac{d y}{d x} = e^{2 x + y}$ is:
(a) $e^{2 x} + 2 e^{- y} = C$
(b) $e^{2 x} - 2 e^{- y} = C$
(c) $e^{2 x} + e^{y} = C$
(d) $2 e^{2 x} + e^{- y} = C$

Try it first, then show the solutionSolution

$e^{- y} d y = e^{2 x} d x \Rightarrow - e^{- y} = \frac{e ^{2 x}}{2} + C_{1}$ . Multiplying by 2 and renaming the constant: $e^{2 x} + 2 e^{- y} = C$ .

Answer: (a).

Example 2: A simple separable equation · 2 marks · Short answerBoard level

Solve $\frac{d y}{d x} = x y^{2}$ , given $y = 1$ when $x = 0$ .

Try it first, then show the solutionSolution

$\frac{d y}{y ^{2}} = x d x \Rightarrow - \frac{1}{y} = \frac{x ^{2}}{2} + C$ . (1 mark)
At $(0, 1)$ : $C = - 1$ . So $\frac{1}{y} = 1 - \frac{x ^{2}}{2}$ , that is $y = \frac{2}{2 - x ^{2}}$ . (1 mark)

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

Example 3: Partial fractions after separating · 3 marks · Short answerHarder board level

Solve $x (x + 1) \frac{d y}{d x} = 1$ ( $x > 0$ ), given $y = 0$ when $x = 1$ .

Try it first, then show the solutionSolution

$d y = \frac{d x}{x ( x + 1 )} = (\frac{1}{x} - \frac{1}{x + 1}) d x$ . (1 mark)
$y = lo g x - lo g (x + 1) + C = lo g \frac{x}{x + 1} + C$ . (1 mark)
At $x = 1$ , $y = 0$ : $0 = lo g \frac{1}{2} + C$ , so $C = lo g 2$ and $y = lo g \frac{2 x}{x + 1}$ . (1 mark)

Answer: $y = lo g \frac{2 x}{x + 1}$ . (ISC 2024 set the same type with $x (x^{2} - 1)$ .)

Example 4: Reducible to separable · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

Put $v = x + y + 1$ : $\frac{d v}{d x} = 1 + \frac{d y}{d x} = 1 + v^{2}$ . (1 mark)
$\frac{d v}{1 + v ^{2}} = d x \Rightarrow tan^{- 1} v = x + C$ . (1 mark)
At $x = 0$ , $y = 0$ : $v = 1$ , so $C = tan^{- 1} 1 = \frac{π}{4}$ . Hence $x + y + 1 = tan (x + \frac{π}{4})$ , that is $y = tan (x + \frac{π}{4}) - x - 1$ . (1 mark)

Answer: $y = tan (x + \frac{π}{4}) - x - 1$ .

Example 5: Rewriting exponentials · 3 marks · Short answerHarder board level

Solve $\frac{d y}{d x} = e^{x + y} + x^{2} e^{y}$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = e^{y} (e^{x} + x^{2})$ , so $e^{- y} d y = (e^{x} + x^{2}) d x$ . (1 mark)
Integrate: $- e^{- y} = e^{x} + \frac{x ^{3}}{3} + C$ . (1 mark)
So $e^{x} + e^{- y} + \frac{x ^{3}}{3} = C$ (renaming the constant). (1 mark)

Answer: $e^{x} + e^{- y} + \frac{x ^{3}}{3} = C$ .

Example 6: Newton's law of cooling · 5 marks · Case-basedHOTS / competency

Suppose a cup of tea at 85 °C is left in a room at 25 °C, and it cools according to $\frac{d T}{d t} = - k (T - 25)$ . After 10 minutes it is at 55 °C.
(i) Solve the equation. (1 mark)
(ii) Find $k$ . (2 marks)
(iii) Find when the tea reaches 35 °C. (2 marks)

Try it first, then show the solutionSolution

(i) $\frac{d T}{T - 25} = - k d t \Rightarrow lo g (T - 25) = - k t + C$ , so $T - 25 = 60 e^{- k t}$ using $T (0) = 85$ .
(ii) $55 - 25 = 60 e^{- 10 k} \Rightarrow e^{- 10 k} = \frac{1}{2} \Rightarrow k = \frac{l o g 2}{10} \approx 0.0693$ per minute.
(iii) $35 - 25 = 60 e^{- k t} \Rightarrow e^{- k t} = \frac{1}{6} \Rightarrow t = \frac{l o g 6}{k} = \frac{10 l o g 6}{l o g 2}$ .
$t \approx \frac{10 \times 1.792}{0.693} \approx 25.85$ minutes.

Answer: $T = 25 + 60 e^{- k t}$ with $k = \frac{l o g 2}{10}$ ; the tea reaches 35 °C after about 25.9 minutes.

Where students lose marks in Variables Separable Method

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

Mistake: Omitting the constant of integration
The ISC 2025 examiners noted candidates who left out C, so their "general solution" was just one curve.
Do this: Write + C as soon as you integrate, and keep it until a condition fixes it.
Usually costs: 1 mark.
Mistake: Integrating the separated side wrongly
In ISC 2024 candidates failed to integrate 1/(x(x² − 1)), which needs partial fractions.
Do this: Recognise the type of integral on each side before integrating.
Usually costs: 2 to 3 marks.
Mistake: Forgetting dv/dx = a + b dy/dx
In reducible equations, differentiating v = x + y gives 1 + dy/dx, not dy/dx.
Do this: Differentiate the substitution explicitly as a separate line.
Usually costs: The whole question.
Mistake: Not finding the particular solution
The 2024 examiners reported candidates who stopped at the general solution even when a condition was given.
Do this: Substitute the given values, find C, and write the final equation.
Usually costs: 1 to 2 marks.

Exam tips

Show the separated form on its own line, for example $\frac{d y}{1 + y ^{2}} = (1 + x) d x$ ; it earns the method mark.
Keep logarithmic constants as $lo g C$ when convenient; the final answer is then a neat product.
In model questions, define the variables and units in one line before writing the equation.
Round only at the final step, and give units (minutes, mg, °C).
If the equation does not separate, test whether it is homogeneous or linear before trying harder algebra.

Practice questions on Variables Separable Method

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 general solution of $\frac{d y}{d x} = y$ is:
(a) $y = x + C$
(b) $y = C e^{x}$
(c) $y = e^{x} + C$
(d) $y = C x$
Hint
Separate: $\frac{d y}{y} = d x$ .
Answer
(b) $y = C e^{x}$ .
Q2 · 2 marks · Short answerBoard level
Solve $\frac{d y}{d x} = \frac{s i n x}{c o s y}$ .
Hint
$cos y d y = sin x d x$ .
Answer
$sin y = - cos x + C$ , that is $sin y + cos x = C$ .
Q3 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} = (1 + y^{2}) (1 + x)$ , given $y = 1$ when $x = 0$ .
Hint
$\frac{d y}{1 + y ^{2}} = (1 + x) d x$ .
Answer
$tan^{- 1} y = x + \frac{x ^{2}}{2} + C$ ; at $(0, 1)$ , $C = \frac{π}{4}$ . So $y = tan (x + \frac{x ^{2}}{2} + \frac{π}{4})$ .
Q4 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} = cos (x + y)$ .
Hint
Put $v = x + y$ and use $1 + cos v = 2 cos^{2} \frac{v}{2}$ .
Answer
$\frac{d v}{d x} = 1 + cos v = 2 cos^{2} \frac{v}{2}$ , so $\frac{1}{2} sec^{2} \frac{v}{2} d v = d x$ and $tan \frac{v}{2} = x + C$ . Solution: $tan \frac{x + y}{2} = x + C$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d y}{d x} = x + y$ is a variables separable equation.
Reason (R): An equation is variables separable if it can be written as $g (y) d y = f (x) d x$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Can $x + y$ be written as a product?
Answer
(d). $x + y$ is a sum, not a product of a function of $x$ and a function of $y$ , so A is false (it is linear); R is the correct definition.
Q6 · 3 marks · Short answerHOTS / competency
A town's population grows at a rate proportional to its size and doubles in 20 years. How long does it take to triple?
Hint
Find $k$ from $e^{20 k} = 2$ .
Answer
$P = P_{0} e^{k t}$ with $k = \frac{l o g 2}{20}$ . Tripling: $e^{k t} = 3 \Rightarrow t = \frac{20 l o g 3}{l o g 2} \approx 31.7$ years.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose water drains from a tank through a small hole so that its depth $h$ (m) after $t$ minutes satisfies $\frac{d h}{d t} = - 0.2 h$ , with $h = 9$ at $t = 0$ .
(i) Separate the variables and integrate. (1 mark)
(ii) Find $h (t)$ . (1 mark)
(iii) Find the depth after 10 minutes. (1 mark)
(iv) When is the tank empty? (1 mark)
Hint
$\int h^{- 1/2} d h = 2 h$ .
Answer
(i) $\frac{d h}{h} = - 0.2 d t \Rightarrow 2 h = - 0.2 t + C$ . (ii) $C = 6$ , so $h = 3 - 0.1 t$ and $h = (3 - 0.1 t)^{2}$ . (iii) $h (10) = 4$ m. (iv) $3 - 0.1 t = 0 \Rightarrow t = 30$ minutes.

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.When is a differential equation variables separable?Show answer
When it can be written as $g (y) d y = f (x) d x$ .
2.Solve $\frac{d y}{d x} = 2 x$ .Show answer
$y = x^{2} + C$ .
3.What substitution reduces $\frac{d y}{d x} = f (x + y)$ to separable form?Show answer
$v = x + y$ .
4.Solve $\frac{d y}{d t} = k y$ with $y (0) = y_{0}$ .Show answer
$y = y_{0} e^{k t}$ .
5.What is the half-life if $\frac{d m}{d t} = - 0.1 m$ ?Show answer
$\frac{l o g 2}{0.1} \approx 6.93$ time units.
6.State Newton's law of cooling as a differential equation.Show answer
$\frac{d T}{d t} = - k (T - T_{s})$ .

Variables Separable Method: frequently asked questions

How do I know whether to use separation of variables?

Rewrite the equation as dy/dx = (something). If that something is a product of a function of x and a function of y, separate. If not, check for a function of ax + by + c (reducible), a function of y/x (homogeneous) or the linear form.

Why is only one constant needed when I integrate both sides?

Each side would produce its own constant, but they can be combined into a single constant on one side. Writing one C is enough and is standard.

What is an equation reducible to variables separable?

It is an equation such as dy/dx = f(ax + by + c) that is not separable as it stands. The substitution v = ax + by + c turns it into a separable equation in v and x.

Where are separable equations used in real life?

In growth and decay (populations, bacteria, medicines, radioactive substances), cooling of objects, draining tanks and many other situations where the rate of change depends on the current amount.

Are equations reducible to variables separable in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus states that equations reducible to variable separable type are included, along with the variables separable method itself.