# NCERT Class 12 Maths Differential Equations

## Exercise 9.1

## Exercise 9.2

## Exercise 9.3

## Exercise 9.4

## Exercise 9.5

## Exercise 9.6

## Miscellaneous Exercise

Download FREE PDF of Chapter-9 Differential Equations

## Chapter 9 Ex.9.1 Question 1

Determine order and degree (if defined) of differential equation \(\frac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right) = 0\).

**Solution**

\[\begin{align} &\Rightarrow\; \frac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right) = 0\\ &\Rightarrow\; y'''' + \sin \left( {y'''} \right) = 0\end{align}\]

Highest order derivative in the differential equation is \(y''''\). Its order is four.

Differential equation is not a polynomial equation in its derivatives. Its degree is not defined.

## Chapter 9 Ex.9.1 Question 2

Determine order and degree (if defined) of differential equation \(y' + 5y = 0\).

**Solution**

\(y' + 5y = 0\)

Highest order derivative in the differential equation is \(y'\). Its order in one.

It is a polynomial equation in \(y'\). Highest power \(y'\) is \(1\). Its degree is one.

## Chapter 9 Ex.9.1 Question 3

Determine order and degree (if defined) of differential equation \({\left( {\frac{{ds}}{{dt}}} \right)^4} + 3s\frac{{{d^2}s}}{{d{t^2}}} = 0\).

**Solution**

\({\left( {\frac{{ds}}{{dt}}} \right)^4} + 3s\frac{{{d^2}s}}{{d{t^2}}} = 0\)

Highest order derivative in the given differential equation is \(\frac{{{d^2}s}}{{d{t^2}}}\). Its order is two.

It is a polynomial equation in \(\frac{{{d^2}s}}{{d{t^2}}}\) and \(\frac{{ds}}{{dt}}\).

The power \(\frac{{{d^2}s}}{{d{t^2}}}\) is \(1\). Degree is one.

## Chapter 9 Ex.9.1 Question 4

Determine order and degree (if defined) of differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} + \cos \left( {\frac{{dy}}{{dx}}} \right) = 0\).

**Solution**

\({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} + \cos \left( {\frac{{dy}}{{dx}}} \right) = 0\)

Highest order derivative in the given differential equation is \(\frac{{{d^2}y}}{{d{x^2}}}\). Order is \(2.\)

Given differential equation is not a polynomial equation in its derivatives. Degree is not defined.

## Chapter 9 Ex.9.1 Question 5

Determine order and degree (if defined) of differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} = \cos 3x + \sin 3x\).

**Solution**

\(\begin{align}&{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} = \cos 3x + \sin 3x\\ &\Rightarrow\; {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} - \cos 3x - \sin 3x = 0\end{align}\)

Highest order derivative in the given differential equation is \(\frac{{{d^2}y}}{{d{x^2}}}\). Its order is two.

It is a polynomial equation in \(\frac{{{d^2}y}}{{d{x^2}}}\) and the power is \(1\). Its degree is \(1\).

## Chapter 9 Ex.9.1 Question 6

Determine order and degree (if defined) of differential equation \({\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0\).

**Solution**

\({\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0\)

Highest order derivative present in the differential equation is \(y'''\). Its order is three.

Given differential equation is a polynomial equation in \(y''',y''\) and \(y'\).

Highest power raised to \(y'''\) is \(2\). Its degree is \({\text{2}}\).

## Chapter 9 Ex.9.1 Question 7

Determine order and degree (if defined) of differential equation \(y''' + 2y'' + y' = 0\).

**Solution**

\(y''' + 2y'' + y' = 0\)

Highest order derivative present in the differential equation is \(y'''\). Its order is \(3.\)

It is a polynomial equation in \(y''',y''\) and \(y'\). The highest power \(y'''\) is \(1\). Its degree is \(1\).

## Chapter 9 Ex.9.1 Question 8

Determine order and degree (if defined) of differential equation \(y' + y = e'\).

**Solution**

\(\begin{align}&y' + y = e'\\& \Rightarrow\; y' + y - e' = 0\end{align}\)

Highest order derivative present in the differential equation is \(y'\). Its order is one.

Given differential equation is a polynomial equation in \(y'\) and the highest power is one.

Its degree is one.

## Chapter 9 Ex.9.1 Question 9

Determine order and degree (if defined) of differential equation \(y'' + {\left( {y'} \right)^2} + 2y = 0\).

**Solution**

\(y'' + {\left( {y'} \right)^2} + 2y = 0\)

Highest order derivative present in the differential equation is \(y''\). Its order is two.

Given differential equation is a polynomial equation in \(y''\) and \(y'\), the highest power\(y''\) is one. Its degree is one.

## Chapter 9 Ex.9.1 Question 10

Determine order and degree (if defined) of differential equation \(y'' + 2y' + \sin y = 0\).

**Solution**

\(y'' + 2y' + \sin y = 0\)

Highest order derivative present in the differential equation is \(y''\). Its order is two.

This is a polynomial equation in \(y''\)and \(y'\) the highest power \(y''\)is one. Its degree is one.

## Chapter 9 Ex.9.1 Question 11

The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin \left( {\frac{{dy}}{{dx}}} \right) + 1 = 0\) is

(A) \(3\)

(B) \(2\)

(C) \(1\)

(D) not defined

**Solution**

\({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin \left( {\frac{{dy}}{{dx}}} \right) + 1 = 0\)

Differential equation is not a polynomial equation in its derivatives. Its degree is not defined.

Thus, the Correct option is D.

## Chapter 9 Ex.9.1 Question 12

The order of the differential equation \(2{x^2}\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + y = 0\).

(A) \(2\)

(B) \(1\)

(C) \(0\)

(D) not defined

**Solution**

\(2{x^2}\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + y = 0\)

Highest order derivative present in the given differential equation is \(\frac{{{d^2}y}}{{d{x^2}}}\). Its order is two.

Thus, the correct option is A.