Class 12 Maths · Original EZT practice
Differential Equations (Class 12 Maths, NCERT, Chapter 9)
Differential Equations is Chapter 9 of the NCERT Class 12 Mathematics Part 2 textbook. Algebra equations tie down numbers; this chapter's equations tie down functions - 'a differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation'. It starts with vocabulary - 'order of a differential equation is defined as the order of the highest order derivative', with degree alongside - then general vs particular solutions, and three workhorse methods for first-order, first-degree equations: variables separable, homogeneous equations, and linear equations with integrating factor. (Note: this is original practice, not a textbook answer key.)
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 26 October 2026:
1. What the chapter covers
9.1 Introduction: why equations with derivatives arise (dy/dx = g(x) and beyond). 9.2 Basic Concepts: order and degree of a differential equation, with polynomial-in-derivatives caveats. 9.3 General and Particular Solutions: arbitrary constants; a particular solution fixes them. 9.4 Methods of Solving First Order, First Degree Differential Equations: 9.4.1 variables separable, 9.4.2 homogeneous equations (substitute y = vx), 9.4.3 linear equations dy/dx + Py = Q with integrating factor e^(integral P dx). EXERCISES 9.1-9.5 plus the Miscellaneous Exercise close the chapter.
2. Key ideas at a glance
Order = highest derivative present; degree = its power after clearing fractions and radicals in derivatives - degree is undefined if the equation is not polynomial in derivatives. The general solution of a first-order equation carries one arbitrary constant; initial conditions pick a particular solution. Variables separable: get all x terms with dx and all y terms with dy, then integrate both sides. Homogeneous test: F(ax, ay) = a^n F(x, y); the substitution y = vx makes it separable. Linear: multiply through by the integrating factor and the left side becomes d/dx(y x IF).
3. Traps this chapter sets
Reading degree off an equation that is not polynomial in derivatives: sin of a derivative or a derivative inside a root makes degree undefined, not high. Forgetting the arbitrary constant on integration - and then wondering why no initial condition can fix the curve. Substituting y = vx but writing dy/dx = v dv/dx: the product rule gives v + x dv/dx, and the missing x is the whole mechanism. Stopping at the integrating factor as if it were the answer: the IF only works when you multiply through and collapse the left side into d/dx(y x IF).
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Order counts data
Choose the method: Read order, count constants.
EZT practice question
A falling raindrop with air resistance obeys m dv/dt = mg - kv, where v(t) is its velocity and m, g, k are positive constants. State the order and degree of this differential equation, and explain what the order implies about the information needed to predict the motion completely.
Show the worked solution
- Order 1 - only the first derivative dv/dt appears.
- Degree 1 - the equation is polynomial in dv/dt and the highest derivative is raised to the first power.
- First order means the general solution carries one arbitrary constant, so one further fact - the velocity at a single instant, such as v = 0 at release - fixes the motion.
Answer: Order 1, degree 1. Because the equation is first order, its general solution has one arbitrary constant, so one piece of initial data (the velocity at one instant) is enough to predict the motion completely.
Check: Only first derivatives present, the equation polynomial in dv/dt, and the one-constant-to-one-datum link stated.
Watch for this: Do not call it second order because acceleration is 'like' a second derivative of position - the equation involves v and dv/dt only, so as an equation in v(t) it is first order; the variable you choose decides the order you read.
Now try it yourself
How many initial facts would a second-order equation need?
Check your answer
Two - a second-order general solution carries two arbitrary constants, so two conditions (for example position and velocity at one instant) are needed to pin them down.
Order counts constants, and constants count the data required.
Verify, don't solve
Choose the method: Differentiate and substitute.
EZT practice question
Verify that y = Ae^x + Be^(-x) is a solution of d^2y/dx^2 - y = 0.
Show the worked solution
- Differentiate twice: y' = Ae^x - Be^(-x), then y'' = Ae^x + Be^(-x).
- But that is exactly y.
- So y'' - y = 0 identically, for every A and B.
Answer: y'' = Ae^x + Be^(-x) = y, so y'' - y = 0 identically - the given family satisfies the equation, and its two arbitrary constants match the second order.
Check: Two differentiations, the identification of y'' with y, and the constant count matching the order.
Watch for this: Do not lose the minus sign on the first derivative and then 'fix' the second - differentiating e^(-x) brings down a factor of -1 each time, and two of them cancel, which is exactly why the verification works.
Now try it yourself
Why must a general solution of a second-order equation carry two arbitrary constants?
Check your answer
Solving means integrating twice, and each integration introduces one constant - so a family short of two constants cannot cover every solution of a second-order equation.
The constant count is the fingerprint of the order.
Separate, then name
Choose the method: x with dx, y with dy.
EZT practice question
Solve dy/dx = x/y by separating variables, and describe the family of curves.
Show the worked solution
- y dy = x dx.
- Integrate: y^2/2 = x^2/2 + C.
- So y^2 - x^2 = k - a family of hyperbolas.
Answer: y^2 - x^2 = k, a family of hyperbolas - variables separated and both sides integrated, with the two halves of the constant merged into one k.
Check: All y terms with dy, all x terms with dx, and the constants merged into a single arbitrary k.
Watch for this: Do not carry two separate constants C1 and C2 into the final answer - their difference is one arbitrary constant, and leaving both makes the family look two-parameter when it is one.
Now try it yourself
Why does a first-order equation produce a family rather than one curve?
Check your answer
One integration gives one arbitrary constant, and each value of that constant picks a different curve - the equation constrains the shape, the constant picks the member.
The general solution is the whole family at once.
Diagnose the step
Choose the method: Product rule check.
EZT practice question
A student solving a homogeneous equation substitutes y = vx and then writes dy/dx = v (dv/dx). Find the error and write the correct substitution identity, then state in one line why it turns the equation separable.
Show the worked solution
- Correct identity: dy/dx = v + x dv/dx, from the product rule on y = vx.
- The student dropped the x factor and the additive v term.
- The v terms then cancel or collect so the equation separates into f(v) dv = g(x) dx.
Answer: The correct identity is dy/dx = v + x dv/dx (product rule on y = vx); with it, the v terms cancel or collect and the homogeneous equation separates into f(v) dv = g(x) dx - which is why the substitution works on every homogeneous equation.
Check: Product rule applied to y = vx, both terms present, and the separation claim tied to homogeneity.
Watch for this: Do not treat v as a constant during the substitution - v is a function of x, and skipping its derivative is precisely the error being diagnosed.
Now try it yourself
Where exactly does homogeneity get used?
Check your answer
After substituting y = vx, a homogeneous right side F(x, y) collapses to a function of v alone (the x powers cancel), so the equation becomes x dv/dx = g(v) - f(v), which separates.
The substitution is only as good as the homogeneity that feeds it.
Build the IF
Choose the method: e to the integral of P.
EZT practice question
For the linear equation dy/dx + 3x^2 y = x^2, write the integrating factor and rewrite the left side as a single derivative - do NOT solve the equation.
Show the worked solution
- P(x) = 3x^2, so IF = e^(integral 3x^2 dx) = e^(x^3).
- Multiplying through: e^(x^3) dy/dx + 3x^2 e^(x^3) y = x^2 e^(x^3).
- The left side is exactly d/dx(y e^(x^3)).
Answer: IF = e^(x^3); multiplying through makes the left side exactly d/dx(y e^(x^3)) - the product-rule collapse that makes every linear equation integrable.
Check: P identified as the coefficient of y, the exponent integrated not copied, and the collapse verified by differentiating y e^(x^3).
Watch for this: Do not write the IF as e^(3x^2) - the exponent is the INTEGRAL of P, not P; and do not 'solve' by stopping at the IF, which is a tool, not an answer.
Now try it yourself
Why does multiplying by e^(integral P dx) always collapse the left side?
Check your answer
Because d/dx(y x IF) = IF(dy/dx) + y(IF)' and (IF)' = P x IF by the chain rule - so the result is IF times (dy/dx + Py), the original left side.
The IF is reverse-engineered from the product rule.
Fix the constant
Choose the method: Substitute the point.
EZT practice question
The general solution of dy/dx = 3x^2 is y = x^3 + C. Given the curve passes through (1, 2), find the particular solution.
Show the worked solution
- Substitute x = 1, y = 2: 2 = 1 + C.
- So C = 1.
- Particular solution: y = x^3 + 1.
Answer: y = x^3 + 1 - the point (1, 2) fixes C = 1, selecting one member of the family y = x^3 + C.
Check: The point substituted into the general solution, C solved, and the particular solution written without any constant.
Watch for this: Do not leave C in the final answer or substitute the point into the derivative instead of the solution - initial conditions apply to the curve, not to its slope equation.
Now try it yourself
Geometrically, what did the point (1, 2) do?
Check your answer
The family y = x^3 + C is the same cubic slid up and down; requiring passage through (1, 2) picks the single slide whose curve contains that point.
A particular solution is one curve out of the family.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Order vs degree?
Show answer and reason
Order is the highest derivative present; degree is its power once the equation is polynomial in derivatives - else degree is undefined.
Vocabulary questions are free marks if precise.
2. Constants in the general solution?
Show answer and reason
As many arbitrary constants as the order of the equation.
Order counts constants; constants count initial data.
3. Variables separable?
Show answer and reason
Collect all x terms with dx and all y terms with dy, integrate both sides.
Separation first, integration second.
4. Integrating factor for dy/dx + Py = Q?
Show answer and reason
e^(integral P dx) - multiplying through collapses the left side to d/dx(y x IF).
The IF is the product rule in disguise.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Explain in your own words why the degree of (y'')^2 + sin(y') = 0 is defined but the degree of y'' + sin(y') = 0 is not.
- Write any first-order differential equation whose general solution is the family of circles x^2 + y^2 = r^2, by differentiating and eliminating r.
- State in one sentence what an initial condition does to a general solution.
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- Which book is Class 12 Maths Chapter 9 Differential Equations from?
- It is Chapter 9 of the NCERT Class 12 Mathematics Part 2 textbook, quoted here from the official NCERT chapter PDF (lemh203.pdf). The chapter pages carry a 'Reprint 2025-26' footer; the book's front matter lists Edition January 2023 with reprints through 'Reprint 2026-27' - check your own copy, as printings can differ.
- What is a differential equation?
- Per the chapter, an equation involving derivatives of the dependent variable with respect to independent variable(s); when there is only one independent variable, it is an ordinary differential equation. Example: dy/dx + y = 0.
- What are order and degree of a differential equation?
- Order is the order of the highest derivative appearing; degree is the power of that highest derivative once the equation is polynomial in the derivatives. If the equation cannot be made polynomial in derivatives (for example it contains sin of a derivative), the degree is not defined.
- What is the difference between a general and a particular solution?
- The general solution contains arbitrary constants - as many as the order of the equation. Giving specific conditions (like a point the curve passes through) fixes those constants and yields a particular solution.
- Which solution methods does this chapter teach?
- For first-order, first-degree equations: separation of variables, the substitution y = vx for homogeneous equations, and the integrating-factor method for linear equations dy/dx + Py = Q.