MathsCBSE Class 12 · Differential Equations

Variables Separable Method for Differential Equations (Class 12)

Q: How do I know if a differential equation is variables separable?

Check whether dy/dx can be written as a product of a function of x alone and a function of y alone. If the right side has x and y added together, as in x + y, it is not separable.

Q: What is the difference between general and particular solutions?

The general solution contains an arbitrary constant and represents a family of curves. A particular solution is one member of that family, found by using a given condition such as y(0) = 3.

Q: Is formation of differential equations in the 2026-27 syllabus?

No. Forming a differential equation from a given family of curves was dropped. Questions ask for order and degree, and for solving separable, homogeneous and linear equations.

Q: Which NCERT section covers this method?

Section 9.4.1 (Differential Equations with Variables Separable) of Chapter 9 in NCERT Mathematics Part II for Class 12.

Q: Why can I write the constant as log C?

Any real number can be written as log C for a suitable positive C, so it is still an arbitrary constant. It lets you combine logarithms into a neat answer like y = Cx.

Updated 15 October 2026 · Fact-checked

A first-order equation is variables separable if it can be written as dy/dx = g(x)h(y). Rearrange to dy/h(y) = g(x) dx, integrate both sides and add one constant C to get the general solution. Use a given condition, such as y(0) = 3, to find C for the particular solution.

Follow the arrows: separate x and y, integrate with one constant C, then use the given condition to fix C.

Illustration drawn for QuestionBuddy.

Understand Variables Separable Method

A differential equation links an unknown function with its derivatives. Its general solution contains arbitrary constants (one for a first-order equation); a particular solution fixes them using given conditions.

When is an equation separable? When $\frac{d y}{d x}$ factors into a function of $x$ times a function of $y$ :
$\frac{d y}{d x} = g (x) h (y)$ .
Examples: $\frac{d y}{d x} = 2 x y$ , $\frac{d y}{d x} = x e^{- y}$ (since $e^{- y}$ is a function of $y$ ), $\frac{d y}{d x} = \frac{x + 1}{2 y}$ , $\frac{d y}{d x} = e^{x + y} = e^{x} e^{y}$ . Not separable: $\frac{d y}{d x} = x + y$ .

The method.

1. Bring all $y$ -terms with $d y$ to one side and all $x$ -terms with $d x$ to the other: $\frac{d y}{h ( y )} = g (x) d x$ (assuming $h (y) \neq = 0$ ).
2. Integrate both sides: $\int \frac{d y}{h ( y )} = \int g (x) d x + C$ . One constant is enough.
3. Simplify, often by writing the constant as $lo g ∣ C ∣$ or $lo g A$ so that logs combine, and solve for $y$ if convenient.
4. For a particular solution, substitute the condition and find $C$ .

Integration techniques from the Integrals chapter (substitution, partial fractions) are often needed in step 2; for instance $\frac{d y}{y ( y - 1 )}$ needs partial fractions.

Exponential growth and decay. If a quantity changes at a rate proportional to itself, $\frac{d Q}{d t} = k Q$ . Separating, $\frac{d Q}{Q} = k d t$ , so $lo g Q = k t + C$ and $Q = Q_{0} e^{k t}$ , where $Q_{0}$ is the starting value. $k > 0$ gives growth (populations, continuous compound interest), $k < 0$ gives decay (medicines, radioactive substances). Doubling time is $\frac{l o g 2}{k}$ .

Newton's law of cooling. A hot object cools at a rate proportional to the excess temperature over the room: $\frac{d T}{d t} = - k (T - T_{s})$ . Separating gives $T - T_{s} = (T_{0} - T_{s}) e^{- k t}$ . One measurement finds $k$ ; then you can predict any time or temperature.

Formation of differential equations is not in the 2026-27 syllabus, but order and degree, general and particular solutions, and the separable, homogeneous and linear methods are. Separable equations are NCERT Section 9.4.1.

Variables Separable Method: formulas to remember

Separable form: $\frac{d y}{d x} = g (x) h (y) \Rightarrow \int \frac{d y}{h ( y )} = \int g (x) d x + C$; One arbitrary constant for a first-order equation.
Growth and decay: $\frac{d Q}{d t} = k Q \Rightarrow Q = Q_{0} e^{k t}$; $k > 0$ 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 log.
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 surrounding temperature.
Useful integrals: $\int \frac{d y}{y} = lo g ∣ y ∣, \int \frac{d y}{1 + y ^{2}} = tan^{- 1} y, \int e^{y} d y = e^{y}$; Most separable problems reduce to these.
Constant as a log: $lo g ∣ y ∣ = lo g ∣ x ∣ + lo g ∣ C ∣ \Rightarrow y = C x$; Choose the constant's form to simplify.

How to solve Variables Separable Method questions

A dependable routine for separable equations and their word problems.

1Write the equation as $\frac{d y}{d x} = g (x) h (y)$ ; factor if needed (for example $e^{x - y} = e^{x} \cdot e^{- y}$ ).
2Separate: $\frac{d y}{h ( y )} = g (x) d x$ .
3Integrate both sides; add a single constant on one side.
4Simplify to an explicit or neat implicit form; combining logs is usually best.
5Substitute the initial condition to find the constant and write the particular solution.
6In word problems, define the variable and units, solve, then answer the question asked (a time, a temperature, an amount).

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 growth: continuous compounding

Suppose a bank compounds interest continuously at 6% a year, so a balance $A$ (in rupees) satisfies $\frac{d A}{d t} = 0.06 A$ .

Solve the equation and find how long it takes for any deposit to double.

Try it first, then show the solutionSolution

Separate: $\frac{d A}{A} = 0.06 d t$ ; integrate: $lo g A = 0.06 t + C$ , so $A = A_{0} e^{0.06 t}$ .
Doubling: $2 A_{0} = A_{0} e^{0.06 t} \Rightarrow e^{0.06 t} = 2$ .
$t = \frac{l o g 2}{0.06} \approx \frac{0.693}{0.06} \approx 11.55$ years.

Answer: $A = A_{0} e^{0.06 t}$ ; any deposit doubles in about 11.55 years.

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

Exponential decay

Imagine a medicine leaves the bloodstream at a rate of 20% of the amount present per hour, continuously: $\frac{d M}{d t} = - 0.2 M$ . A patient starts with 100 mg.

Find the amount after 3 hours.

Try it first, then show the solutionSolution

Separate and integrate: $M = M_{0} e^{- 0.2 t}$ with $M_{0} = 100$ .
$M (3) = 100 e^{- 0.6}$ .
$e^{- 0.6} \approx 0.5488$ , so $M (3) \approx 54.9$ mg.

Answer: About 54.9 mg.

Memory hook: A constant percentage loss per unit time is exponential decay.

Tricks to remember Variables Separable Method

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

x with dx, y with dy
Every x-term goes to the dx side and every y-term to the dy side before integrating.
Use it when: Setting up every separable equation.
One equation, one constant
First order means one arbitrary constant: write $+ C$ on only one side.
Use it when: Avoiding a confusing $C_{1} - C_{2}$ .
Log constant for log answers
If both sides give logs, write the constant as $lo g C$ and combine.
Use it when: Neat general solutions like $y = C x$ .
Proportional rate = exponential
'Rate proportional to amount' means $Q = Q_{0} e^{k t}$ ; 'proportional to the difference from room temperature' means cooling.
Use it when: Word problems.

Variables Separable Method: solved examples

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

Example 1: Separating an exponential · 2 marks · Very short answerBoard level

Find the general solution of $\frac{d y}{d x} = x e^{- y}$ .

Try it first, then show the solutionSolution

Separate: $e^{y} d y = x d x$ . (1 mark)
Integrate: $e^{y} = \frac{x ^{2}}{2} + C$ , i.e. $y = lo g (\frac{x ^{2}}{2} + C)$ . (1 mark)

Answer: $e^{y} = \frac{x ^{2}}{2} + C$

Example 2: A particular solution · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Separate: $\frac{d y}{y} = 2 x d x$ . (1 mark)
Integrate: $lo g ∣ y ∣ = x^{2} + C$ , so $y = A e^{x^{2}}$ . (1 mark)
$y (0) = 3 \Rightarrow A = 3$ . Particular solution: $y = 3 e^{x^{2}}$ . (1 mark)

Answer: $y = 3 e^{x^{2}}$

Example 3: Logs on both sides · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

Separate: $\frac{d y}{1 + y} = \frac{x d x}{1 + x ^{2}}$ . (1 mark)
Integrate: $lo g ∣1 + y ∣ = \frac{1}{2} lo g (1 + x^{2}) + lo g A$ , so $1 + y = A 1 + x^{2}$ . (1 mark)
$x = 0$ , $y = 1$ : $2 = A$ . Hence $y = 2 1 + x^{2} - 1$ . (1 mark)

Answer: $y = 2 1 + x^{2} - 1$

Example 4: Partial fractions after separating · 5 marks · Long answerHarder board level

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

Try it first, then show the solutionSolution

$x \frac{d y}{d x} = y^{2} - y = y (y - 1)$ , so $\frac{d y}{y ( y - 1 )} = \frac{d x}{x}$ . (1 mark)
$\frac{1}{y ( y - 1 )} = \frac{1}{y - 1} - \frac{1}{y}$ . (1 mark)
Integrate: $lo g ∣ y - 1∣ - lo g ∣ y ∣ = lo g ∣ x ∣ + lo g ∣ C ∣$ , so $\frac{y - 1}{y} = C x$ . (1 mark)
$y (1) = 2$ : $\frac{1}{2} = C$ . So $\frac{y - 1}{y} = \frac{x}{2}$ . (1 mark)
Solve for y: $2 y - 2 = x y \Rightarrow y = \frac{2}{2 - x}$ . (1 mark)

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

Example 5: Growth with a doubling time · 3 marks · Short answerHOTS / competency

A bacteria culture grows at a rate proportional to the number present. It doubles in 3 hours. How long does it take to become 10 times its initial size? (Use $lo g 2 = 0.693$ , $lo g 10 = 2.303$ .)

Try it first, then show the solutionSolution

$\frac{d N}{d t} = k N \Rightarrow N = N_{0} e^{k t}$ . (1 mark)
Doubling in 3 h: $e^{3 k} = 2 \Rightarrow k = \frac{l o g 2}{3}$ . (1 mark)
$10 = e^{k t} \Rightarrow t = \frac{l o g 10}{k} = \frac{3 l o g 10}{l o g 2} = \frac{3 \times 2.303}{0.693} \approx 9.97$ hours. (1 mark)

Answer: About 9.97 hours (just under 10 hours).

Example 6: Case study: a cooling cup of tea · 4 marks · Case-basedHOTS / competency

Suppose a cup of tea at $85°$ C is left in a room at $25°$ C. By Newton's law of cooling, $\frac{d T}{d t} = - k (T - 25)$ , with $t$ in minutes. After 5 minutes the tea is at $65°$ C.
(i) Solve the differential equation. (1 mark)
(ii) Find $e^{- 5 k}$ . (1 mark)
(iii) When will the tea reach $45°$ C? (2 marks)

Try it first, then show the solutionSolution

(i) $\frac{d T}{T - 25} = - k d t \Rightarrow lo g (T - 25) = - k t + C$ ; with $T (0) = 85$ : $T = 25 + 60 e^{- k t}$ . (1 mark)
(ii) $65 = 25 + 60 e^{- 5 k} \Rightarrow e^{- 5 k} = \frac{2}{3}$ . (1 mark)
(iii) $45 = 25 + 60 e^{- k t} \Rightarrow e^{- k t} = \frac{1}{3}$ . Taking logs: $k t = lo g 3$ and $5 k = lo g \frac{3}{2}$ . (1 mark)
$t = \frac{5 l o g 3}{l o g 1.5} \approx \frac{5 \times 1.0986}{0.4055} \approx 13.5$ minutes. (1 mark)

Answer: (i) $T = 25 + 60 e^{- k t}$ (ii) $\frac{2}{3}$ (iii) about 13.5 minutes.

Where students lose marks in Variables Separable Method

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

Mistake: Separating incorrectly
Terms joined by + or − cannot be separated by division.
Wrong working
$\frac{d y}{d x} = x + y \Rightarrow \frac{d y}{y} = x d x$ .
Right working
$\frac{d y}{d x} = x + y$ is not separable; it is linear: $\frac{d y}{d x} - y = x$ .
Do this: Check that the right side factors as $g (x) h (y)$ .
Usually costs: The whole question.
Mistake: Forgetting the constant
Without C the solution is only one particular curve.
Wrong working
$lo g ∣ y ∣ = x^{2}$ .
Right working
$lo g ∣ y ∣ = x^{2} + C$ , so $y = A e^{x^{2}}$ .
Do this: Write $+ C$ immediately after integrating.
Usually costs: 1 mark, and the particular solution.
Mistake: Finding C before simplifying incorrectly
Exponentiating $lo g y = x^{2} + C$ gives $y = e^{C} e^{x^{2}}$ , not $e^{x^{2}} + C$ .
Wrong working
$y = e^{x^{2}} + C$ , $y (0) = 3 \Rightarrow C = 2$ .
Right working
$y = A e^{x^{2}}$ , $A = 3$ .
Do this: When exponentiating, the constant multiplies.
Usually costs: The particular solution.
Mistake: Rounding too early in word problems
Early rounding of $k$ changes the final time noticeably.
Wrong working
$k \approx 0.08$ , so $t \approx 13.7$ minutes.
Right working
Keep $t = \frac{5 l o g 3}{l o g 1.5} \approx 13.5$ minutes.
Do this: Keep exact logs until the last step.
Usually costs: Half a mark.

Exam tips

Separable equations appear every year as 1-mark MCQs and 3- to 4-mark questions (including a 4-mark case study in 2025); show the separated line clearly.
Write the general solution before using the initial condition; both earn marks.
In word problems, define variables with units and end with a sentence answering the question.
Use $lo g$ to mean natural logarithm; give approximate values only at the end.
If a first-order equation is not separable, test whether it is homogeneous or linear.

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 $\frac{d y}{d x} = \frac{y}{x}$ is:
(a) $y = x + C$
(b) $y = C x$
(c) $x y = C$
(d) $y = C e^{x}$
Hint
$\frac{d y}{y} = \frac{d x}{x}$ .
Answer
(b) $y = C x$ .
Q2 · 2 marks · Very short answerBoard level
Find the general solution of $\frac{d y}{d x} = \frac{s e c ^{2} x}{t a n y}$ .
Hint
$tan y d y = sec^{2} x d x$ .
Answer
$\int tan y d y = \int sec^{2} x d x \Rightarrow - lo g ∣ cos y ∣ = tan x + C$ (or $lo g ∣ sec y ∣ = tan x + C$ ).
Q3 · 2 marks · Very short answerBoard level
Find the general solution of $\frac{d y}{d x} = (1 + x) (1 + y^{2})$ .
Hint
$\frac{d y}{1 + y ^{2}} = (1 + x) d x$ .
Answer
$tan^{- 1} y = x + \frac{x ^{2}}{2} + C$ .
Q4 · 3 marks · Short answerHarder board level
Find the particular solution of $\frac{d y}{d x} = y cot x$ , given $y = 2$ when $x = \frac{π}{6}$ .
Hint
$\int cot x d x = lo g ∣ sin x ∣$ .
Answer
$\frac{d y}{y} = cot x d x \Rightarrow lo g ∣ y ∣ = lo g ∣ sin x ∣ + lo g C \Rightarrow y = C sin x$ . At $x = \frac{π}{6}$ : $2 = \frac{C}{2} \Rightarrow C = 4$ , so $y = 4 sin x$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d y}{d x} + y = 1$ can be solved by separating the variables.
Reason (R): It can be written as $\frac{d y}{1 - y} = d x$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Rearrange $\frac{d y}{d x} = 1 - y$ .
Answer
(a). $\frac{d y}{d x} = 1 - y$ separates as in R, giving $- lo g ∣1 - y ∣ = x + C$ .
Q6 · 3 marks · Short answerHarder board level
Find the equation of the curve through $(1, 2)$ whose slope at any point $(x, y)$ is $\frac{x + 1}{2 y}$ .
Hint
$2 y d y = (x + 1) d x$ .
Answer
$y^{2} = \frac{x ^{2}}{2} + x + C$ . At $(1, 2)$ : $4 = \frac{1}{2} + 1 + C \Rightarrow C = \frac{5}{2}$ . Curve: $y^{2} = \frac{x ^{2}}{2} + x + \frac{5}{2}$ .
Q7 · 3 marks · Short answerHOTS / competency
A radioactive substance decays at a rate proportional to the amount present and has a half-life of 1600 years. What percentage of a sample remains after 400 years?
Hint
$Q = Q_{0} e^{- k t}$ with $e^{- 1600 k} = \frac{1}{2}$ , so $e^{- 400 k} = (\frac{1}{2})^{1/4}$ .
Answer
$\frac{Q ( 400 )}{Q _{0}} = e^{- 400 k} = (e^{- 1600 k})^{1/4} = 2^{- 1/4} \approx 0.8409$ , so about 84.1% remains.

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 $\frac{d y}{d x} = f (x, y)$ variables separable?Show answer
When $f (x, y) = g (x) h (y)$ .
2.Solve $\frac{d y}{d x} = k y$ .Show answer
$y = A e^{k x}$ .
3.How many arbitrary constants does the general solution of a first-order equation have?Show answer
One.
4.Doubling time for $\frac{d N}{d t} = k N$ ?Show answer
$\frac{l o g 2}{k}$ .
5.Is $\frac{d y}{d x} = e^{x + y}$ separable?Show answer
Yes: $e^{- y} d y = e^{x} d x$ .
6.Solution of Newton's law of cooling?Show answer
$T = T_{s} + (T_{0} - T_{s}) e^{- k t}$ .

Variables Separable Method: frequently asked questions

How do I know if a differential equation is variables separable?

Check whether dy/dx can be written as a product of a function of x alone and a function of y alone. If the right side has x and y added together, as in x + y, it is not separable.

What is the difference between general and particular solutions?

The general solution contains an arbitrary constant and represents a family of curves. A particular solution is one member of that family, found by using a given condition such as y(0) = 3.

Is formation of differential equations in the 2026-27 syllabus?

No. Forming a differential equation from a given family of curves was dropped. Questions ask for order and degree, and for solving separable, homogeneous and linear equations.

Which NCERT section covers this method?

Section 9.4.1 (Differential Equations with Variables Separable) of Chapter 9 in NCERT Mathematics Part II for Class 12.

Why can I write the constant as log C?

Any real number can be written as log C for a suitable positive C, so it is still an arbitrary constant. It lets you combine logarithms into a neat answer like y = Cx.