Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2 – General and Particular Solutions | Boundless Maths
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Ex 9.2 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2 – General and Particular Solutions

This Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2 page covers all 12 questions, solved step-by-step — differentiating a given explicit or implicit function (sometimes more than once) and substituting back into the differential equation to confirm that L.H.S. equals R.H.S.

Questions 1–10 all follow the same verification pattern, but the functions get progressively trickier — from a simple exponential in Q1 to implicit relations that need implicit differentiation in Q7, Q8 and Q9, where you differentiate both sides with respect to x and then use the original equation itself to simplify the result. Questions 11 and 12 are conceptual MCQs on a rule worth memorising: a general solution of an nth order differential equation has exactly n independent arbitrary constants, while any particular solution — obtained by fixing those constants — has none.

12Questions
Easy–MedDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2 — All 12 Questions

1

Verify that the function y=e^x+1 is a solution of y''-y'=0.

Easy +
Solution
  1. Differentiate once: y=e^x+1\ \Rightarrow\ y'=e^x.
  2. Differentiate again: y''=e^x.
  3. Substitute into L.H.S.: y''-y'=e^x-e^x=0, which equals R.H.S.
Answer: Verified — y = eˣ + 1 is a solution of the given differential equation.
2

Verify that the function y=x^2+2x+C is a solution of y'-2x-2=0.

Easy +
Solution
  1. Differentiate: y=x^2+2x+C\ \Rightarrow\ y'=2x+2.
  2. Substitute into L.H.S.: y'-2x-2=(2x+2)-2x-2=0, which equals R.H.S.
Answer: Verified — y = x² + 2x + C is a solution of the given differential equation.
3

Verify that the function y=\cos x+C is a solution of y'+\sin x=0.

Easy +
Solution
  1. Differentiate: y=\cos x+C\ \Rightarrow\ y'=-\sin x.
  2. Substitute into L.H.S.: y'+\sin x=-\sin x+\sin x=0, which equals R.H.S.
Answer: Verified — y = cos x + C is a solution of the given differential equation.
4

Verify that the function y=\sqrt{1+x^2} is a solution of y'=\dfrac{xy}{1+x^2}.

Medium +
Solution
  1. Differentiate (chain rule, since y=(1+x^2)^{1/2}): y'=\dfrac{1}{2}(1+x^2)^{-1/2}\cdot 2x=\dfrac{x}{\sqrt{1+x^2}}.
  2. Compute R.H.S.: \dfrac{xy}{1+x^2}=\dfrac{x\sqrt{1+x^2}}{1+x^2}=\dfrac{x}{\sqrt{1+x^2}}.
  3. Compare: L.H.S. y'=\dfrac{x}{\sqrt{1+x^2}} equals R.H.S., so the equation is satisfied.
Answer: Verified — y = √(1+x²) is a solution of the given differential equation.
5

Verify that the function y=Ax is a solution of xy'=y (x\neq 0).

Easy +
Solution
  1. Differentiate: y=Ax\ \Rightarrow\ y'=A.
  2. Substitute into L.H.S.: xy'=xA=Ax=y, which equals R.H.S.
Answer: Verified — y = Ax is a solution of the given differential equation.
6

Verify that the function y=x\sin x is a solution of xy'=y+x\sqrt{x^2-y^2} (x\neq 0 and x \gt y or x \lt -y).

Hard +
Solution
  1. Differentiate using the product rule: y=x\sin x\ \Rightarrow\ y'=\sin x+x\cos x.
  2. Compute L.H.S.: xy'=x(\sin x+x\cos x)=x\sin x+x^2\cos x.
  3. Simplify \sqrt{x^2-y^2}: since y=x\sin x, x^2-y^2=x^2-x^2\sin^2 x=x^2(1-\sin^2 x)=x^2\cos^2 x, so \sqrt{x^2-y^2}=|x\cos x|. The given domain condition (x\neq 0, and x \gt y or x \lt -y) makes x\cos x non-negative, so this simplifies to x\cos x.
  4. Compute R.H.S.: y+x\sqrt{x^2-y^2}=x\sin x+x\cdot x\cos x=x\sin x+x^2\cos x.
  5. Compare: L.H.S. = R.H.S. =x\sin x+x^2\cos x, so the equation is satisfied.
Answer: Verified — y = x sin x is a solution of the given differential equation.
7

Verify that the function xy=\log y+C is a solution of y'=\dfrac{y^2}{1-xy} (xy\neq 1).

Hard +
Solution
  1. Differentiate implicitly w.r.t. x (product rule on the left, since C is constant): xy=\log y+C\ \Rightarrow\ y+xy'=\dfrac{1}{y}y'.
  2. Collect the y' terms: y=y'\left(\dfrac{1}{y}-x\right)=y'\cdot\dfrac{1-xy}{y}.
  3. Solve for y': y'=\dfrac{y^2}{1-xy}, which matches the given equation.
Answer: Verified — xy = log y + C is a solution of the given differential equation.
8

Verify that the function y-\cos y=x is a solution of (y\sin y+\cos y+x)y'=y.

Hard +
Solution
  1. Differentiate implicitly: y-\cos y=x\ \Rightarrow\ y'+\sin y\cdot y'=1, so y'(1+\sin y)=1.
  2. Simplify the bracket using the given equation x=y-\cos y: y\sin y+\cos y+x=y\sin y+\cos y+(y-\cos y)=y(\sin y+1).
  3. Compute L.H.S.: (y\sin y+\cos y+x)y'=y(\sin y+1)\cdot y'=y(\sin y+1)\cdot\dfrac{1}{1+\sin y}=y, which equals R.H.S.
Answer: Verified — y − cos y = x is a solution of the given differential equation.
9

Verify that the function x+y=\tan^{-1}y is a solution of y^2y'+y^2+1=0.

Hard +
Solution
  1. Differentiate implicitly: x+y=\tan^{-1}y\ \Rightarrow\ 1+y'=\dfrac{y'}{1+y^2}.
  2. Collect the y' terms: 1=y'\left(\dfrac{1}{1+y^2}-1\right)=y'\cdot\dfrac{1-(1+y^2)}{1+y^2}=\dfrac{-y^2y'}{1+y^2}.
  3. Cross-multiply and rearrange: 1+y^2=-y^2y', i.e. y^2y'+y^2+1=0, matching the given equation.
Answer: Verified — x + y = tan⁻¹y is a solution of the given differential equation.
10

Verify that the function y=\sqrt{a^2-x^2}, x\in(-a,a), is a solution of x+y\dfrac{dy}{dx}=0 (y\neq 0).

Medium +
Solution
  1. Differentiate (chain rule, since y=(a^2-x^2)^{1/2}): y'=\dfrac{-x}{\sqrt{a^2-x^2}}=\dfrac{-x}{y}.
  2. Substitute into the equation: x+y\cdot y'=x+y\left(\dfrac{-x}{y}\right)=x-x=0, which equals R.H.S.
Answer: Verified — y = √(a² − x²) is a solution of the given differential equation.

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11

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

Easy +
Solution
(A) 0 (B) 2 (C) 3 (D) 4
  1. Recall the rule: the general solution of a differential equation contains as many independent arbitrary constants as the order of the equation.
  2. Apply it here: the order given is 4, so the general solution has 4 arbitrary constants.
Answer: (D) 4
12

The number of arbitrary constants in the particular solution of a differential equation of third order are:

Easy +
Solution
(A) 3 (B) 2 (C) 1 (D) 0
  1. Recall the rule: a particular solution is obtained by assigning specific numerical values to the arbitrary constants of the general solution.
  2. Apply it here: once those values are fixed, no arbitrary constants remain — regardless of the order of the equation (third order, in this case) — so the count is 0.
Answer: (D) 0
Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2

How many questions are there in Exercise 9.2?

Exercise 9.2 has 12 questions — the first 10 ask you to verify that a given explicit or implicit function is a solution of the corresponding differential equation, and the last 2 are multiple-choice questions on how many arbitrary constants appear in a general versus a particular solution.

What is the difference between a general and a particular solution?

A general solution contains as many independent arbitrary constants as the order of the differential equation — for example, a second order equation's general solution has two arbitrary constants. A particular solution is obtained by assigning specific values to those constants, so it is free of arbitrary constants.

Where can I find the official NCERT textbook for this exercise?

Exercise 9.2 is from Chapter 9, Differential Equations, in the NCERT Class 12 Mathematics textbook (Part II), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

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