NCERT Solutions For Class 12 Mathematics Chapter 9: Differential Equations

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The NCERT Solutions for class 12 mathematics chapter 9 Differential Equations are given in the article. Differential equation means the derivatives of a mathematical equation. The chapter Differential Equations belongs to the unit Calculus, that adds up to 35 marks of the total marks.

Chapter 9 of NCERT Solutions for Class 12 Maths covers the concepts of order and degree of differential equations, the method of solving a differential equation, their properties and much more.

Download: NCERT Solutions for Class 12 Mathematics Chapter 9 pdf

Class 12 Maths NCERT Solutions Chapter 9 Differential Equations

NCERT Solutions

Important Topics in Class 12 Mathematics Chapter 8 Applications of Integrals

Important concepts of Class 12 Maths covered in Chapter 9 Differential Equations of NCERT Solutions are:

Order of a differential equation

The order of a differential equation is defined to be of the highest order derivative it contains. Degree of a differential equation is defined as the power to which the highest order derivative is raised.

The equation (f‴)² + (f″)⁴ + f = x is an example of a second-degree, third-order differential equation.

How to Find Order of the Differential Equation?

The order of differential equation can be found by identifying the derivatives in the given expression of the differential equation. The different derivatives in a differential equation are as follows:

First Derivative:dy/dx or y'
Second Derivative: d²y/dx², or y''
Third Derivative: d³y/dx³, or y'''
nth derivative: dny/dxn, or y''''.....n times

Further, the highest derivative present in the differential equation defines the order of the differential equation, and the exponent of the highest derivative represents the degree of the differential equation.

Formation of a Differential Equation whose General Solution is given

For any given differential equation, the solution is of the form f(x,y,a¹,a², …….,an) = 0 where x and y are the variables and a¹ , a² ……. an are the arbitrary constants.

Methods of Solving First Order, First Degree Differential Equations

Different methods of solving first order, first degree differential equations are as follows:

Differential equations with variables separable
Homogeneous differential equations
Linear differential equations

Exercise Solutions of Class 12 Maths Chapter 9 Differential Equations

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

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CBSE CLASS XII Related Questions

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$
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If \[ \begin{bmatrix} 4 + x & x - 1 \\ -2 & 3 \end{bmatrix} \] is a singular matrix, then the value of $ x $ is:
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3.
Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the lines \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad \frac{x - 4}{5} = \frac{y - 1}{2} = z. \]
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4.
If $ \mathbf{a} $ and $ \mathbf{b} $ are position vectors of two points $ P $ and $ Q $ respectively, then find the position vector of a point $ R $ in $ QP $ produced such that \[ QR = \frac{3}{2} QP. \]
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5.
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find $ A^{-1} $. Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
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6.
Verify that lines given by $ \vec{r} = (1 - \lambda) \hat{i + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} $ and $ \vec{r} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} $ are skew lines. Hence, find shortest distance between the lines.}
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CBSE CLASS XII Related Questions

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.
If \[ \begin{bmatrix} 4 + x & x - 1 \\ -2 & 3 \end{bmatrix} \] is a singular matrix, then the value of \( x \) is:

3.
Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the lines \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad \frac{x - 4}{5} = \frac{y - 1}{2} = z. \]

4.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

5.
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]

6.
Verify that lines given by \( \vec{r} = (1 - \lambda) \hat{i + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \) and \( \vec{r} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \) are skew lines. Hence, find shortest distance between the lines.}

Similar Mathematics Concepts

Comments

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NCERT Solutions For Class 12 Mathematics Chapter 9: Differential Equations

Class 12 Maths NCERT Solutions Chapter 9 Differential Equations

Important Topics in Class 12 Mathematics Chapter 8 Applications of Integrals

Exercise Solutions of Class 12 Maths Chapter 9 Differential Equations

CBSE CLASS XII Related Questions

1.Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.If \[ \begin{bmatrix} 4 + x & x - 1 \\ -2 & 3 \end{bmatrix} \] is a singular matrix, then the value of \( x \) is:

3.Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the lines \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad \frac{x - 4}{5} = \frac{y - 1}{2} = z. \]

4.If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

5.If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]

6. Verify that lines given by \( \vec{r} = (1 - \lambda) \hat{i + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \) and \( \vec{r} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \) are skew lines. Hence, find shortest distance between the lines.}

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.
If \[ \begin{bmatrix} 4 + x & x - 1 \\ -2 & 3 \end{bmatrix} \] is a singular matrix, then the value of \( x \) is:

3.
Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the lines \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad \frac{x - 4}{5} = \frac{y - 1}{2} = z. \]

4.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

5.
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]

6.
Verify that lines given by \( \vec{r} = (1 - \lambda) \hat{i + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \) and \( \vec{r} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \) are skew lines. Hence, find shortest distance between the lines.}