NCERT Solutions for Differential Equations Exercise 9.2 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.2 is provided in the article. Class 12 Chapter 9 Differential Equations Exercises include questions on Order and Degree of Differential Equations.

Download PDF: NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

Also check other Exercise Solutions of Class 12 Maths Chapter 9 Differential Equations

Also Read:

Chapter 9 Differential Equations Important Topics
Differential Equations Applications	Difference between parametric and nonparametric	Partial Differential Equation
Ordinary differential equations	Differential Equations Formula	Differential Calculus
Partial Derivative	How to Solve Linear Differential Equation	Homogeneous Differential Equation

Also Read:

CBSE Class 12 Mathematics Study Guides
Algebra	Arithmetic	Calculus
Mensuration	Trigonometry	Geometry
Permutation and Combination	Math Study Notes	Math MCQs
NCERT Solutions for Class 12 Maths	Math Formula	Difference between in Maths

CBSE CLASS XII Related Questions

1.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).
View Solution
2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :
- \( -6 \) unit/sec
- \( 0 \) unit/sec
- \( 18 \) unit/sec
- \( 6 \) unit/sec
View Solution
3.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :
- \(\frac{1}{2}\)
- \(\frac{3}{5}\)
- \(\frac{1}{5}\)
- \(\frac{3}{10}\)
View Solution
4.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]
View Solution
5.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :
- \( \log \cos 2x + C \)
- \( \sec^{2} 2x + C \)
- \( e^{x}\tan x + C \)
- \( e^{x}\cot x + C \)
View Solution
6.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]
View Solution

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You can also check

NCERT Solutions for Differential Equations Exercise 9.2 Solutions

CBSE CLASS XII Related Questions

1.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

4.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

5.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

6.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

Similar Mathematics Concepts

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NCERT Solutions for Differential Equations Exercise 9.2 Solutions

CBSE CLASS XII Related Questions

1.Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

2.The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

4.For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :Minimize \(Z = 18x + 10y\)subject to constraints\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

5.\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

6.If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

2.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

3.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

4.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

5.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

6.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]