CBSE Class 12 Maths Revision Notes Chapter 9

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The Class 12 Mathematics Chapter 9 notes- Differential Equations help students to understand their importance in modelling physical systems. This Chapter holds utmost importance not only in academics but also in real life. With the help of some set of formulas, differential equations describe the relationship between a function and its derivative. The Chapter helps students to gain knowledge of the fundamental topics and learn to find solutions for the differential equations.

To get access to Class 12 Mathematics Chapter 9 Notes, in addition to all the latest information and updates regarding CBSE exams, competitive exams, etc., visit the Extramarks web portal and mobile app.

Class 12 Mathematics Chapter 9: Key Topics

The topics explained in the Class 12 Mathematics Chapter 9 notes are as follows:

Introduction
Basic Concepts of Differential equations
Order and degree of an ordinary differential equation
General and Particular Solutions
Formation of a Differential Equation
Different Procedures for Solving the First Order and First Degree Differential Equations
- Differential equations with variables separable
- Homogeneous differential equations
- Linear differential equations

An overview of the topics covered in Class 12 Mathematics Chapter 9 are given below.

Introduction:

In the Class 12 Mathematics Chapter 9 Notes, students learn how to carry out the differentiation of a given function, i.e., how to find f’(x) in the domain, where f is the function and x is the independent variable. Furthermore, various applications of differential equations in the branch of physics, chemistry, biology, economics, etc., are explained in detail. Some basic concepts introduced in Class 11, as well as Chapter 5 of the NCERT Books related to differential equations, their solutions, formation, and methods of solving, are included in this Chapter.

Basic Concepts

Differential equation

Consider equation x + y = 7 and x ddxy+ y = 0. We notice that the first equation involves two independent and/or dependent variables, x and y but the second equation involves variables and the derivative of the dependent variable y w.r.t the independent variable x. This equation is called the differential equation.

Ordinary Differential Equation:

The equation which involves the differentiation of the dependent variable, say y, for only one independent variable, say x, is known as the ordinary differential equation. In this Chapter, we deal with ordinary differential equations.

NOTE:

Notations:

First order derivative, ddxy = y’
Second order derivative, d2dx2y = y’’
Third order derivative, d3dx3y = y’’’
nth order derivative, dndxny

Order of an ordinary differential equation

The order of the derivative of the highest order of the dependent variable y w.r.t the independent variable x is known as the order of the ordinary differential equation.

NOTE: The order can never be more than the total number of arbitrary constants in the differential equation.

For example:

First order ordinary differential equation is ddxy = ex

Second order ordinary differential equation is d2dx2y + y = 0

Third order ordinary differential equation is d3dx3y + x2 (d2dx2y)3= 0

For order of the first, second and third equations maximum derivative of variable y with respect to variable x is ddxy, d2dx2y, and d3dx3y, respectively.

Degree of a differential equation

The degree of the differential equation is given as the highest exponent of the derivative of the highest order, provided that the exponent of each derivative, as well as the unknown variable, is a non-negative (positive) integer.

NOTE:

(i) The order and degree are always non-negative (positive) integers

(ii) The differential equation is in the form of a polynomial equation

(iii) The degree of a differential equation is not defined if the differential equation is not in the form of a polynomial equation.

Considering the same examples given above, the degree of these equations is 1.

Illustration: Consider d2dx2y = 3ddxy+3, find its order and degree

This will become d2dx2y3 = ddxy+32. Therefore, the order of the equation = 2 and degree = 3

Students may refer to the Class 12 Mathematics Chapter 9 notes to practice various problems to find the order and degree of the equation.

Formation of the ordinary differential equations:

Use the below-given steps to form an ordinary differential equation from a relation.

Step 1: Count the total number of arbitrary constants (n) in the given equation.

Step 2: Carry out differentiation for the equation w.r.t to dependent variable n times, where n is equal to the number of arbitrary constants.

Step 3: After differentiating, all the arbitrary constants are eliminated from the given equation.

Step 4: The equation thus obtained is known as the required differential equation.

Consider, y = mx + c represents a family of straight lines.

We give different values to the parameters m and c.

For eg, when m = 1 and c = 0, then y = x

when m = 1 and c = 1, then y = x + c

when m = -1 and c = 0, then y = -x

when m = -1 and c = -1, then y = -x – c

Differentiating equation with respect to x is

ddxy = m and d2dx2y = 0

Solution of the Differential Equation

The function y = f(x) + c satisfying the given differential equation is known as the solution of the equation. The curve y = f(x) is known as the solution curve or the integral curve.

There are two types of solutions: General and Particular solutions.

General Solution: The solutions containing various arbitrary constants as the order of the equation are the general solution of the ordinary differential equation. A general solution of the ordinary differential equation contains n arbitrary constants of order n.
Particular Solution: This is an expansion of the general solution. When particular values are substituted for arbitrary constants in the general solution, the solution obtained is called the particular solution.

Procedure to form a differential equation that represents the family of curves

If the curve F1(x, y, a) = 0, then it depends on one parameter.

For example: parabola y2= ax is represented as f(x, y, a)= y2= ax.

Differentiating with respect to x, we get g(x, y, y’, a)= 0

Eliminating the arbitrary constants, i.e. a, we get F(x, y, y’)= 0

2. If the curve F2(x, y, a, b) = 0, then it depends on the two parameters a and b.

Differentiating with respect to x, we get g (x, y, y′, a, b) = 0

In this case, we cannot eliminate the parameters a and b, so we need a third equation.

Again differentiating, we get h (x, y, y’, y’’, a, b) = 0

Eliminating the arbitrary constants, i.e. a and b, we get F (x, y, y′, y″) = 0

The Class 12 Mathematics Chapter 9 notes include various problems and their solutions for students to practice sums and learn to form ordinary differential equations.

Methods of Solving the First Order and First Degree Ordinary Differential Equations

There are three methods to solve first order first-degree ordinary differential equations.

METHOD 1:

Variable separable form

Let the first-order differential equation be ddxy = F(x, y)

Let F(x, y) be the product of two functions g(x) and h(y), i.e. F(x, y) = g(x) . h(y), where g(x) and h(y) is a function of x and y respectively, then the differential equation will be of the variable separable type.

Consider h(y) ≠ 0, then the equation will become

1h(y) dy = g(x) dx

On Integrating both the sides, we get

1h(y) dy = g(x) dx

Which can be written as H(y) = G(x) + C, where H(y) and G(x) are said to be the anti-derivatives of 1h(y) and g(x).

The Class 12 Mathematics Chapter 9 notes help students to practice various problems that can be solved using the variable separable form.

2. Solution of the differential equation: ddxy = f(ax+by+c):

The f is some function of ax+by+c

Let z = ax+by+c

On differentiating,

ddxz = a + b ddxy

ddxy = ddxz – ab = f(z)

Therefore, dz b. f(z) + a = dx

On integrating, we get

dz b. f(z) + adx =dx+c or

dz b. f(z) + adx =x+c where z= ax+by+c

This is the solution for any equation in the form of ddxy = f(ax+by+c)

METHOD 2:

Homogeneous differential equation

The function F(x, y) of degree n is a homogeneous function if F(λx, λy) = n .F(x, y) for λ is a constant and λ 0.

F (x, y) is a homogeneous function having degree n if F(x, y) = xn g(yx) or F(x, y) = yn h(xy)

If a function F(x, y) of degree zero is a homogeneous function, then the differential equation ddxy = F (x, y) is also homogenous.

The differential equation can be written as dydx=f(x, y)g(x, y),

dydx= xn f(yx)xn g(yx)=f(yx)g(yx)= F(yx)

Solving a homogeneous differential equation:

To solve an equation of the form dydx= g(yx)……..(#),

we substitute y = v.x ……..(+)

Using the product rule to differentiate the equation with respect to x,

dydx = v + xdvdx ……..(*)

Substituting (*) in (#), we get

v + xdvdx = g(yx), i.e. v + xdvdx = g(v) or

xdvdx = g(v) – v……..(~)

Therefore, reducing eqn (~) into variable separable form

dvg(v) – v = dxx

On integrating both the sides,

dvg(v) – v = dxx + c

If we replace the value of v = yx, we get the general solution of the initial differential equation.

NOTE: If the differential equation F(x, y) of degree zero is homogeneous, then we substitute v = xy and follow the above-mentioned steps to find the general solution of the equation.

2. Reducing the differential equations to homogeneous forms

Consider an equation of the form dydx=ax+by+ca’x+b’y+c’ where aa’bb’. ….. (1)

This equation can be reduced to the homogenous form by substituting x=x′+h and y=y′+k, where h and k are constants.

Let dx = dx’ and dy = dy’

The equation will become

dy’dx’=a(x’+h)+b(y′+k)+ca'(x’+h)+b'(y′+k)+c’ = ax’+by′+(ah+bk+c)a’x’+b’y′+(a’h+b’k+c’)

Let us choose h and k such that ah+bk+c=0 and a′h+b′k+c′=0

Therefore the equation (1) reduces to

dydx=ax’+by’a’x’+b’y’ which is of the homogeneous form.

With the help of the Class 12 Mathematics Chapter 9 notes, students can solve homogeneous equations easily and within a limited time- time.

METHOD 3: Linear differential equations

Consider a differential equation of the form dydx +Py = Q, where P and Q are functions of x. This represents the linear differential equation of first order. We can also write it as,

dxdy +P1y = Q1, where P1 and Q1 are functions of x.

Solving a homogeneous differential equation:

To solve an equation of the form dydx +Py = Q

Multiply both the sides by say the function g (x), we get

g (x) dydx +P.(g (x)).y = Q.g (x) which becomes

g (x) dydx +P.(g (x)).y = ddx [y. g (x)] which on further solving reduces to

P = g'(x)g(x)

On integrating with respect to x, we get

P dx = g'(x)g(x) dx

Therefore, g(x) = eP dx. It is called the Integrating factor of the differential equation.

Substituting the value of g(x) in our initial equation we get,

eP dxdydx +P.(eP dx).y = Q. eP dxor

dydxy. eP dx= Q. eP dx

On integrating again with respect to x, we get

y eP dx= Q. eP dxdx + c

This equation gives the general solution of the given differential equation.

Steps to solve:

Step 1: Find the Integrating factor = eP dx

Step 2: Write the solution in the form of y × IF = ∫(Q × IF) dx + C

For dxdy +P1y = Q1 as the differential equation, the solution will be in the form x × IF = ∫ (Q’ × IF) dy + C

NOTE:

If P = k, where k is a constant, then I.F. is ek
The general solution is y = c. e–P dxwhen Q = 0
The linear differential equation does not contain any product or powers of y or dydx.

Students may access the Class 12 Mathematics Chapter 9 notes to practice problems in linear differential equations.

Differential equation reducible to the linear form:

If the equations are not in the linear form, then, in that case, they can be reduced to linear equations with the help of a certain transformation.

Consider the equation of the form f’(y) + f(y) P(x) = Q(x) …. 1

If f(y) = u then f’(y) dx = du

Then (1) is reduced to

dudx + u.P(x) = Q(x) is in the linear differential equation form.

2. An extended form of linear differential equations:

Bernoulli’s equation:

Consider the equation of the form dydx + P(y) = Q(yn), where P and Q are functions x, n 0, 1.

This equation is known as Bernoulli’s equation.

For dydx + P(y) = Qyn, divide both sides by yn

1yn dydx + P. 1yn-1 = Q

Let us substitute v = 1yn-1 and different both sides

– (n-1)yn dydx = dvdx

1yn dydx = -1(n-1) dvdx

Therefore, the equation becomes dvdx= (1-n) y-ndydx or

dvdx + (1-n) P.v = Q (1-n) where v is an independent variable in the form of a linear equation.

Clairaut form:

Consider the equation of the form y = Px + f(p), where P = dydx . This equation is the Clairaut form of differential equation

Substituting value of P we get

y = x.dpdx + f’(P)dpdx= 0

[x + f’(P)] dpdx= 0

If dpdx= 0 then p = C

Orthogonal Trajectory

The orthogonal trajectory is any curve that cuts every member of a family of curves at every right angle.

Method to find the Orthogonal Trajectory:

Step 1: Consider f(x, y, c)=0, c is a parameter, is the equation of the family of curves.

Step 2: Differentiate the linear equation with respect to x. This will eliminate 0 from the equation.

Step 3: Substitute dydx= -dxdy in the equation. It will give the equation of Orthogonal Trajectory

Step 4: By further solving this equation, we will get the Orthogonal Trajectories.

METHOD 4: Exact Differential Equation:

Necessary and Sufficient Conditions for the differential equation Mdx + Ndy = 0, where M and N are functions of x and y, respectively, will be an exact differential equation if dNdy=dNdx

An exact differential equation can be derived from the solution directly by differentiating without any elimination, multiplication, etc.

For example, consider an equation xdy+ydx=0. It is an exact differential equation because it can be derived from the function xy=c through direct differentiation.

Application of Differential Equations:

The differential equation has several applications in the physical science and engineering sector.

Consider a point P(x, y). Let PT and PN be tangent and normal at point P. The tangent at point P(x, y) makes an angle θ with an x-axis.

Then the slope of the tangent at P = tan θ =(dydx)P at P

Then the slope of normal= -1(dydx)P

Equation of the tangent at point P is given as

Y – y = (dydx)P(X – x)

Equation of the normal at point P is given as

Y – y = -1(dydx)P(X – x)

In ΔPGT, sin θ = PGTG=yTG and cos θ = PGPN=yPN

From this we can say that tan θ = GNyinformation

The Class 12 Mathematics Chapter 9 notes explain the applications of differential equations in a detailed and well-explained manner. Students may refer to it to clear their doubts.

Some Differential results are:

d(xy) = x. dy + y. dx
d(x+y) = dx + dy
d(yx) = x. dy – y. dxx2
x. dy – y. dxxy= log (x.y)
x. dy – y. dxxy= log (yx)
dx + dy.x+y= d log (x+y)
d(1x–1y) = d(1x)-d(1y)

NCERT Class 12 Mathematics Chapter 9 Notes- Differential Equations: Chapter Weightage

Chapter 9 Differential Equations of CBSE Class 12 Mathematics belongs to the Unit Calculus. To score a total of 35 marks in the unit, students must understand the importance of the Chapter. The six exercises and miscellaneous problems provide sufficient practice for the Class 12 examinations. The Class 12 Mathematics Chapter 9 notes also include CBSE extra questions and elaborates all the concepts clearly.

Students learn about the following:

A differential equation is defined as the equation which involves the differentiation of a dependent variable, say x, with respect to some independent variable, say y.
Order is said to be the highest order derivative of the differential equation, and degree is defined if and only if the differential equation is in the form of a polynomial equation. A degree is the power of the highest order derivative.
The solution of a differential equation is a function that satisfies the equation. It can be either a general solution or a particular solution.
Formation of a differential equation by differentiating and eliminating the arbitrary constants.
Different methods to solve the differential equations.
Applications of differential equations and problems used in the physical science and engineering sector.

NCERT Class 12 Mathematics Chapter 9 Notes: Exercises & Solutions

The Class 12 Mathematics Chapter 9 notes are essential for the in-depth study of differential equations. Students will be asked to solve various questions based on the order, degree, General and particular Solutions of the given Differential Equation, etc. With the help of the Class 12 Mathematics Chapter 9 notes, students can ensure in-depth knowledge and consistent practice of all important questions. The key points and summary included in the notes help students to revise and retain their knowledge.

Chapter 9 Mathematics Class 12 notes provide a helping hand to students of Class 12 for their exam preparation. The notes contain all the important questions and exercise problems that are included in the CBSE books with their practical solutions. Each solution is well explained in a practical, detailed and step-wise manner in the Class 12 Mathematics Chapter 9 notes in order to help students to gain a clear knowledge of the differential equation. The questions which are included in the notes are most likely to come in the examinations, thus making it essential for students to practice all the questions. It will also boost their confidence to attain good marks.

The Class 12 Mathematics Chapter 9 notes aim to strengthen the fundamental concepts and knowledge. In case any complications and queries arise in the Chapter, with the Class 12 Mathematics notes Chapter 9, students can face every complication. These Class 12 Mathematics Chapter 9 notes enable quick revision on the exam day of formulas, definitions, derivations and key points in a descriptive and systematic manner.

Students must refer to different academic material provided on the Extramarks web portal to prepare for the Class 12 boards as well as other national level examinations such as the JEE Main, JEE Advanced, etc.

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NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics is a perfect guide for students to study for the Class 12 board exams. Each topic and sub-topic is covered along with the solutions to exercise problems given at the back of the NCERT textbook. The step-by-step solutions help students to understand each topic and apply it to solve similar problems.

The NCERT Exemplar and Class 12 Mathematics Chapter 9 notes cater to the understanding level of students irrespective of their intelligence quotient. Students can score optimum marks by practising a wide variety of questions. They will be able to tackle complex and tough questions like the one that require higher-order thinking skills (HOTS).

Apart from the NCERT Exemplar and Class 12 Mathematics Chapter 9 notes, students must consider using reference books for solving the questions. They are advised to solve various CBSE sample papers and CBSE previous year question papers available on the Extramarks platform. Using these Class 12 Mathematics Chapter 9 notes, students can analyse and score themselves based on their preparation and, thus, focus on the weaker sections. They will also learn important techniques, tips and tricks as well as test strategies to learn the concepts better in no time.

NCERT Class 12 Mathematics Chapter 9 Notes: Key Features

The key features of Extramarks Class 12 Mathematics Chapter 9 notes are as follows:

The Class 12 Chapter 9 Mathematics notes are the key to achieving success for students. It helps not only to strengthen the core knowledge of every concept but also makes them experts in solving Mathematical problems. .
The NCERT Solutions are prepared based on the rules and latest guidelines issued by the CBSE board. This gives the students a preview of the paper pattern and type of questions that they are going to face in the board exams.
The Class 12 Chapter 9 Mathematics notes are curated by some experienced faculty who have years of experience in Mathematics. The notes ensure that there are no errors and these solutions are adequate. Students can depend on the quality of the solutions confidently.
All important concepts and questions are included, which are most likely to come in the final exams.
The Class 12 Mathematics Chapter 9 notes guide the students and bring flexibility and ease of learning from the comfort of their homes. It saves time and students get enough time to focus on other subjects as well.
These Solutions can be accessed over electronic gadgets such as iPads, tablets, mobile, and desktops to make sure that every student can study, review concepts and enjoy the benefits of the academic notes.
With the help of the Class 12 Mathematics Chapter 9 notes, students can also prepare for other competitive and college entrance tests.
Each and every solution in the Chapter 9 Mathematics Class 12 notes is explained in detail, and each l exercise helps students to gain in-depth knowledge of every aspect of the Chapter to create a sound foundational base for competitive exams.

CBSE Exam Preparation Tips

Here are some useful preparation tips that students may follow to do well in their upcoming Class 12 board examination-

Know the exam details:

It is very important for students to know about the CBSE Syllabus, exam pattern, marking system and weightage of the Chapters in every subject.

2. Follow a proper schedule/ timetable:

Students are advised to plan a study schedule and follow it strictly. This will allow them to study with full concentration as well as have sufficient time to relax, sleep or play. Set a proper schedule for sufficient practice, to learn and also revise accordingly.

3. Do not keep the backlog:

It is crucial to exercise the brain in a constructive way. However, it is also important to make sure to not burden it with a lot of thoughts and stress. Be clear and do not put unnecessary pressure on yourself. Proper planning will help them to cover the portion and clarify their doubts at the earliest. .

4. Use the best academic notes:

It is said that there can never be a better practice than match practice. Solving several practice questions included in the Class 12 Mathematics Chapter 9 notes will help students to analyse their weak areas so that they can rectify them in the upcoming weeks. It’s advisable that the students use the best study materials provided by Extramarks to ace their exams.

Q.1

$Solve \frac{dy}{dx} = (e^{x} + 1) y$

Ans

$\begin{array}{l} \frac{dy}{dx} = (e^{x} + 1) y \\ \Rightarrow \frac{1}{y} dy = (e^{x} + 1) dx \\ \Rightarrow \int \frac{dy}{y} = \int (e^{x} + 1) dx \\ \Rightarrow logy = e^{x} + x + c \end{array}$

Q.2

Find the equation of the curve that passes through the point (1, 2) and satisfies the differential equation

$\frac{dy}{dx} = - \frac{2 xy}{(x^{2} + 1)}$

Ans

$\begin{array}{l} We have, \frac{dy}{dx} = - \frac{2 xy}{(x^{2} + 1)} \\ \Rightarrow \frac{dy}{y} = - \frac{2 x}{(x^{2} + 1)} dx \\ \Rightarrow \int \frac{dy}{y} = - \int \frac{2 x}{(x^{2} + 1)} dx \\ \Rightarrow logy = - \log (x^{2} + 1) = logC \\ \Rightarrow \log [y (x^{2} + 1)] = logC \\ \Rightarrow y (x^{2} + 1) = C \\ Since, the curve passes through (1, 2) so putting x = 1 and \\ y = 2 in above equation, we get C = 4. \\ Therefore, the required equation is y (x^{2} + 1) = 4. \end{array}$

Q.3

$Solve (1 + y^{2}) dy = (\tan^{- 1} y - x) dx .$

Ans

$\begin{array}{l} The given equation can be written as \frac{dy}{dx} + \frac{1}{(1 + y^{2})} . x = \frac{\tan^{- 1} y}{(1 + y^{2})} \\ This is a linear equation of the form \frac{dy}{dx} + Px = Q \\ Where P = \frac{1}{(1 + y^{2})} and Q = \frac{\tan^{- 1}}{(1 + y^{2})} \\ Now, I . F . = e \int \frac{1}{1 + y^{2}} dy = e^{\tan^{- 1} y} \\ So its solution is \\ x \times (I . F .) = [\int Q \times (I . F .)] dy + c \\ {xe}^{\tan^{- 1} y} = \int \frac{\tan^{- 1} y}{(1 + y^{2})} e^{\tan^{- 1} dy} + c \\ = \int {te}^{t} dt + c, where, \tan^{- 1} y = t \\ = {te}^{t} + e^{t} + c \\ = (\tan^{- 1}) e^{\tan^{- 1} y} - e^{\tan^{- 1} y} + c \\ Hence, x = (\tan^{- 1} y) - 1 + {Ce}^{- \tan^{- 1} y} is the required solution . \end{array}$

Q.4

Determine the degree of differential equation

${(\frac{d^{2} y}{{dx}^{2}})}^{2} + \frac{dy}{dx} = y .$

Ans

The power of the highest order is 2. So, it is a differential equation of degree 2.

Q.5

$Solve \frac{dy}{dx} - {xsin}^{2} x = \frac{1}{xlogx} .$

Ans

$\begin{array}{l} \frac{dy}{dx} - {xsin}^{2} x = \frac{1}{xlogx} . \\ \frac{dy}{dx} = {xsin}^{2} x + \frac{1}{xlogx} \\ dy = ({xsin}^{2} x + \frac{1}{xlogx}) dx \\ Integrating, we get \\ \int dy = \int {xsin}^{2} xdx + \int \frac{dx}{xlogx} \\ y = \int x (\frac{1 - \cos 2 x}{2}) dx + \int \frac{dx}{xlogx} \\ [\begin{array}{l} Letlogx = t \\ \frac{1}{x} dx = dt \end{array}] \\ = \frac{1}{2} \int xdx - \frac{1}{2} \int xcos 2 xdx + \int \frac{1}{t} \\ = \frac{x^{2}}{4} - \frac{1}{2} [x (\frac{\sin 2 x}{2}) - \int \frac{\sin 2 x}{2} dx] + \log |t| + c \\ = \frac{x^{2}}{4} - \frac{xsin 2 x}{4} - \frac{\cos 2 x}{8} + \log |logx| + c \end{array}$

Q.6

$\begin{array}{l} Form the differential equation of the curve {(x - a)}^{2} + {(y - b)}^{2} = r^{2} \\ where a, b are arbitrary constants . \end{array}$

Ans

$\begin{array}{l} {(x - a)}^{2} + {(y - b)}^{2} = r^{2} \\ Differentiating w . r . t . x ., we get \\ 2 (x - a) + 2 (y - b) y ‘ = 0 \\ (x - a) + (y - b) y ‘ = 0 \Rightarrow (x - a) = - (y - b) y ‘ \\ Differentiating w . r . t . x \\ 1 + (y - b) y ‘ ‘ + y ‘ y ‘ = 0 \\ (y - b) = - \frac{1 + {(y ‘)}^{2}}{y ‘ ‘} and (x - a) = [\frac{1 + {(y ‘)}^{2}}{y ‘ ‘}] y ‘ \\ ∴ {[1 + {(y ‘)}^{2}]}^{3} = r^{2} {(y ‘ ‘)}^{2} \end{array}$

Q.7

In a bank, the principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?

Ans

$\begin{array}{l} Let P be the principal at any time . According to the given problem, \\ \frac{dp}{dt} = (\frac{5}{100}) \times P = \frac{P}{20} . \dots\dots\dots (1) \\ separating the variables in equation (1) we get \\ \frac{dp}{p} = \frac{dt}{20} \\ Integrating both sides \\ \int \frac{dp}{P} = \int \frac{dt}{20} \\ \log |P| = \frac{1}{20} t + k \\ P = e^{\frac{1}{20} t + k} = e^{\frac{1}{20} t} e^{k} = {Ce}^{\frac{1}{20} t} \\ Now, P = 1000, when t = 0 \\ Substituting the values of P and t in we get C = 1000. \\ P = 1000 e^{\frac{1}{20} t} \\ Let t years be the time required to double the principal . Then \\ 2000 = 1000 e^{\frac{1}{20} t} \\ 2 = e^{\frac{1}{20} t} so, t = 20 \log e^{2} \end{array}$

Q.8

$Solve: \frac{dy}{dx} = \frac{y}{x} - e^{\frac{y}{x}}$

Ans

$\begin{array}{l} \frac{dy}{dx} = \frac{y}{x} - e^{\frac{y}{x}} \\ Let \frac{y}{x} = v \\ \frac{dy}{dx} = v + x \frac{dv}{dx} \\ ∴ v + x \frac{dv}{dx} = v - e^{v} \\ x \frac{dv}{dx} = - e^{v} \\ - \frac{dv}{e^{v}} = \frac{dx}{x} \\ - e^{- v} dv = \frac{dx}{x} \\ Integrating both sides \\ - \int e^{- v} dv = \int \frac{dx}{x} \\ e^{- v} = \log |x| + c \\ e^{\frac{- y}{x}} = \log |x| + c \end{array}$

Q.9

$Solve \frac{dy}{dx} = \sin^{3} {xcos}^{2} x + {xe}^{x}$

Ans

$\begin{array}{l} dy = (\sin^{3} {xcos}^{2} x + {xe}^{x}) dx \\ \int dy = \int (\sin^{3} {xcos}^{2} x + {xe}^{x}) dx \\ y = \int \sin^{3} {xcos}^{2} xdx + \int {xe}^{x} dx \\ = \int (1 - \cos^{2} x) \cos^{2} x . sinxdx + \int {xe}^{x} dx \\ [\begin{array}{l} Let cosx = t \\ - sinxdx = dt \end{array}] \\ = - \int {(1 - t)}^{2} dt + [{xe}^{x} - \int e^{x} dx] \\ = \frac{t^{5}}{5} - \frac{t^{3}}{3} + {xe}^{x} - e^{x} + c \\ = \frac{\cos^{5} x}{5} - \frac{\cos^{3} x}{3} + {xe}^{x} - e^{x} + c \end{array}$

Q.10

$Form the differential equation representing the family of ellipses \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Ans

$\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, differentiating w . r . t . x, we get \\ \frac{2 x}{a^{2}} + \frac{2 yy ‘}{b^{2}} = 0 \\ \frac{x}{a^{2}} - \frac{yy ‘}{b^{2}} \\ \frac{yy ‘}{x} = \frac{b^{2}}{a^{2}} \\ Diff . w . r . t . x \\ \frac{x (yy ‘ ‘ + y ‘ y ‘) - yy ‘}{x^{2}} = 0 \\ x [yy ‘ ‘ + {(y ‘)}^{2}] - yy ‘ = 0 \\ xyy ‘ ‘ + x {(y ‘)}^{2} = yy ‘ \end{array}$

Q.11

$\begin{array}{l} Solve the differential equation \\ xcos (\frac{y}{x}) \frac{dy}{dx} = ycos (\frac{y}{x}) + x \end{array}$

Ans

$\begin{array}{l} xcos (\frac{y}{x}) \frac{dy}{dx} = ycos (\frac{y}{x}) + x \\ Let \frac{y}{x} = v \\ \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \\ xcosv (v + x \frac{dv}{dx}) = vxcosv + x \\ v + x \frac{dv}{dx} = \frac{vxcosv + x}{xcosv} \\ x \frac{dv}{dx} = \frac{vxcosv + x}{xcosv} - v \\ x \frac{dv}{dx} = \frac{vxcosv + x - vxcosv}{xcosv} = \frac{1}{cosv} \\ cosvdv = \frac{1}{x} dx \\ Integrating both sides \\ \int cosvdv = \int \frac{1}{x} dx \\ sinv = \log |x| + c \\ \sin (\frac{y}{x}) = \log |x| + c \end{array}$

Q.12

$\begin{array}{l} Determine the order and degree of differential equation \\ \frac{d^{4} y}{{dx}^{4}} + 2 x {(\frac{dy}{dx})}^{3} = 0 . \end{array}$

Ans

The order of highest order derivative is 4. So, it’s order is 1. The power of the highest order is 1. So, it is a differential equation of degree 1.

Q.13

Verify that the given function y” – y’ = 0 is a solution of the corresponding differential equation y = e^x + 1.

Ans

$\begin{array}{l} y = e^{x} + 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = e^{x} (i) \\ Differentiating w . r . t . x ., we get \\ \frac{d^{2} y}{{dx}^{2}} = e^{x} = \frac{dy}{dx} by (i) \\ ∴ y ‘ ‘ - y ‘ = 0 \end{array}$

Q.14

$Solve \frac{dy}{dx} = e^{x - y} + x^{2} e^{- y}$

Ans

$\begin{array}{l} \frac{dy}{dx} = e^{x - y} + x^{2} e^{- y} \\ \Rightarrow \frac{dy}{dx} = e^{- y} (e^{x} + x^{2}) \\ \Rightarrow e^{y} dy = (e^{x} + x^{2}) dx \\ \Rightarrow \int e^{y} dy = \int (e^{x} + x^{2}) dx \\ \Rightarrow e^{y} = e^{x} + \frac{x^{3}}{3} + c \end{array}$

Q.15

Verify that the given function y’ – 2x -2 = 0 is a solution of the corresponding differential equation y = x² + 2x +c.

Ans

Given that, y = x² + 2x +c

Differentiating w.r.t. x, we get

y’ = 2x + 2 ⇒ y’ – 2x – 2 = 0

Q.16 Show that y = -cosx +x is a solution of the differential equation y” = cosx.

Ans

y = – cosx – x, differentiating w.r.t. x, we get

y’ = sinx …(1)

Again differentiating (1) w. r. t., we get

y” = cosx.

Q.17

$Solve : \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + xy \frac{dy}{dx} = 0$

Ans

$\begin{array}{l} \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + xy \frac{dy}{dx} = 0 \\ \Rightarrow \sqrt{x^{2} + x^{2} y^{2} + 1 + y^{2}} + xy \frac{dy}{dx} = 0 \\ \Rightarrow \sqrt{x^{2} (1 + y^{2}) + 1 + y^{2}} + xy \frac{dy}{dx} = 0 \\ \Rightarrow \sqrt{(x^{2} + 1) (1 + y^{2})} + xy \frac{dy}{dx} = 0 \\ \Rightarrow \sqrt{(x^{2} + 1) (1 + y^{2})} + = - xy \frac{dy}{dx} \\ \Rightarrow \frac{\sqrt{x^{2} + 1 dx}}{x} = - \frac{ydy}{\sqrt{1 + y^{2}}} \\ Integrating both sides \\ \int \frac{\sqrt{x^{2} + 1}}{x} dx = - \int \frac{ydy}{\sqrt{1 + y^{2}}} \\ [Let 1 + y^{2} = t \Rightarrow 2 ydy = dt] \\ \int \frac{\sqrt{x^{2} + 1}}{x} \times \frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}} dx = - \int \frac{dt}{2 \sqrt{t}} \\ \int \frac{x^{2} + 1}{x \sqrt{x^{2} + 1}} dx = - \sqrt{t} \\ \int \frac{x^{2}}{x \sqrt{x^{2} + 1}} dx + \int \frac{1}{x \sqrt{x^{2} + 1}} dx = - \sqrt{1 + y^{2}} \\ \int \frac{x}{\sqrt{x^{2} + 1}} dx + \int \frac{1}{x^{2} \sqrt{1 + \frac{1}{x^{2}}}} dx = - \sqrt{1 + y^{2}} \\ [\begin{array}{l} Let x^{2} + 1 = u \Rightarrow 2 xdx = du \\ and \frac{1}{x} = v \Rightarrow - \frac{1}{2} dx = dv \end{array}] \\ \frac{1}{2} \int \frac{du}{\sqrt{u}} - \int \frac{1}{\sqrt{1 + v^{2}}} dv = - \sqrt{1 + y^{2}} \\ \sqrt{u} - \log |v + \sqrt{1 + v^{2}}| + c = - \sqrt{1 + y^{2}} \\ \sqrt{x^{2} + 1} - \log |\frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}}| + c = - \sqrt{1 + y^{2}} \\ \sqrt{x^{2} + 1} + \sqrt{1 + y^{2}} - \log |1 + \sqrt{\frac{1 + x^{2}}{x}}| + c = 0 \end{array}$

Q.18 Determine the order of differential equation y’ + 5y = 4.

AnsThe order of highest order derivative is 1. So, it’s order is 1.

Q.19

$If y = xlog (\frac{x}{a + bx}), prove that x^{3} . \frac{d^{2 y}}{{dx}^{2}} = {(x \frac{dy}{dx} - y)}^{2}$

Ans

$\begin{array}{l} The given relation may be written as \frac{y}{x} = logx - \log (a + bx) \\ (\frac{x . \frac{dy}{dx} - y . 1}{x^{2}}) = (\frac{1}{x} - \frac{b}{a + bx}) = \frac{a}{x (a + bx)} \\ (x . \frac{dy}{dx} - y) = \frac{ax}{(a + bx)} \\ Now differentiating above equation, we get \\ x \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} . 1 - \frac{dy}{dx} = \frac{\{(a + bx) . a - ax . b\}}{{(a + bx)}^{2}} \\ x \frac{d^{2} y}{{dx}^{2}} = \frac{a^{2}}{{(a + bx)}^{2}} \\ x^{3} \frac{d^{2} y}{{dx}^{2}} = {(\frac{ax}{a + bx})}^{2} = {(x \frac{dy}{dx} - y)}^{2} \end{array}$

Q.20

$\begin{array}{l} Find the particular solution of the differential equation \\ \log (\frac{dy}{dx}) = 3 x + 4 y at (0) = 0 . \end{array}$

Ans

$\begin{array}{l} \log (\frac{dy}{dx}) = 3 x + 4 y \Rightarrow \frac{dy}{dx} = e^{3 x + 4 y} \\ \Rightarrow \frac{dy}{dx} = e^{3 x} e^{4 y} \\ \Rightarrow \frac{dy}{e^{4 y}} = e^{3 x} dx \\ Integrating both sides \\ \Rightarrow \int \frac{dy}{e^{4 y}} = \int e^{3 x} dx \\ \Rightarrow \int e^{- 4 y} dy = \int e^{3 x} dx \\ \Rightarrow \frac{e^{- 4 y}}{- 4} = \frac{e^{3 x}}{3} + c \\ Now at x = 0 and y = 0 \\ \Rightarrow \frac{1}{- 4} = \frac{1}{3} + c \\ \Rightarrow - \frac{1}{3} - \frac{1}{4} = c \end{array}$

Q.21

Form the differential equation representing the family of curves given by y² = 4ax.

Ans

Given that y² = 4ax

Differentiating w.r.t. x, we get

2yy’ = 4a or 2yy’ = y²/x

⇒ 2xyy’ – y² = 0

⇒ y(2xy’ – y) = 0

⇒ 2xy’ – y = 0.

Q.22

$Solve \cos^{- 1} (\frac{dy}{dx}) = x .$

Ans

$\begin{array}{l} We have \cos^{- 1} (\frac{dy}{dx}) = x \\ \cos^{- 1} (\frac{dy}{dx}) = x \Rightarrow \frac{dy}{dx} = cosx \\ dy = cosxdx \\ On integrating, we get \\ \int dy = \int cosx dx \\ y = sinx + c \end{array}$

Q.23

$Solve xyy ‘ = 1+ x + y + xy .$

Ans

$\begin{array}{l} xyy ‘ = 1+ x + y + xy \\ xy \frac{dy}{dx} = (1 + x) (1 + y) \\ \frac{y}{(1 + y)} dy = \frac{(1 + x)}{x} dx \\ Integrating both sides \\ \int \frac{y}{(1 + y)} dy = \int \frac{(1 + x)}{x} dx \\ \int \frac{y + 1 - 1}{(1 + y)} dy = \int (\frac{1}{x} + 1) dx \\ \int 1 - \frac{1}{(1 + y)} dy = \int (\frac{1}{x} + 1) dx \\ y - \log |1 + y| = \log |x| + x + c \end{array}$

Q.24

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.

Ans

Let P denote the family of above said parabolas and let (a, 0) be the focus of a member of the given family, where a is an arbitrary constant.

Therefore, equation of family P is y² = 4ax

y² = 4ax, differentiating w. r. t. x., we get

2yy’ = 4a or 2yy’ = y²/x

⇒ 2xyy’ – y² = 0

⇒ y(2xy’ – y) = 0

⇒ 2xy’ – y = 0.

Q.25

$\begin{array}{l} Solve the differential equation \\ (e^{x} + x^{e}) dy - (e^{x - 1} + x^{e - 1}) dx = 0, x = 0, y = 0 \end{array}$

Ans

$\begin{array}{l} We have, (e^{x} + x^{e}) dy - (e^{x - 1} + x^{e - 1}) dx = 0 \\ \Rightarrow (e^{x} + x^{e}) dy = (e^{x - 1} + x^{e - 1}) dx \\ \Rightarrow dy = \frac{(e^{x - 1} + x^{e - 1}) dx}{(e^{x} + x^{e})} \\ On integrating both sides, we get \\ \int dy = \int \frac{(e^{x - 1} + x^{e - 1}) dx}{(e^{x} + x^{e})} \\ \Rightarrow y = \int \frac{e^{x - 1} + x^{e - 1} dx}{e^{x} + x^{e}} \\ Put e^{x} + x^{e} = t \\ On differentiating \\ (e^{x - 1} + x^{e - 1}) dx = dt \\ \Rightarrow e (e^{x - 1} + x^{e - 1}) dx = dt \\ \Rightarrow e (e^{x - 1} + x^{e - 1}) dx = \frac{dt}{e} \\ Now, \\ y = \frac{1}{e} \int \frac{dt}{t} \\ \Rightarrow y = \frac{1}{e} \log |t| + c \\ \Rightarrow y = \frac{1}{e} \log |e^{x} + x^{e}| + c . \dots.. (1) \\ Put x = 0 and y = 0 in (1), we get \\ 0 = \frac{1}{e} \log 1 + c \\ \Rightarrow c = 0 - 0 [∵ \log 1 = 0] \\ \Rightarrow c = 0 \\ From (1), the required solution is given by \\ y = \frac{1}{e} \log |e^{x} + x^{e}| \end{array}$

Q.26

Find the equation of the curve which passes through the origin and at any point slope of the curve is reciprocal of (1+cotx).

Ans

$\begin{array}{l} The slope of the curve at any point p (x, y) is m = \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx} = \frac{1}{1 + cotx} \\ \Rightarrow dy = \frac{1}{1 + cotx} dx \\ On integrating both sides \\ \int dy = \int \frac{1}{1 + cotx} dx \\ y = \int \frac{1}{1 + \frac{cosx}{sinx}} \\ = \int \frac{sinx}{(sinx + cosx)} dx \\ = \frac{1}{2} \int \frac{2 sinx}{(sinx + cosx)} dx \\ = \frac{1}{2} \int \frac{(sinx + cosx) + (sinx - cosx)}{(sinx + cosx)} dx \\ = \int \frac{(sinx + cosx)}{(sinx + cosx)} dx + \int \frac{(sinx - cosx)}{(sinx + cosx)} dx \\ y = \int dx + \int \frac{(sinx - cosx)}{(sinx + cosx)} dx \\ Put (sinx + cosx) = t \\ On differentiating, \\ - (sinx - cosx) dx = dt \\ \Rightarrow y = x - \int \frac{dt}{t} \\ \Rightarrow y = x - \log |t| + c \\ \Rightarrow y = x - \log |sinx + cosx| + c . \dots\dots (1) \\ It is passing through the origin (0, 0) . \\ 0 = 0 \log |\sin 0 + \cos 0| + c \\ \Rightarrow 0 = \log 1 + c \\ \Rightarrow c = 0 - 0 = 0 [∵ \log 1 = 0] \\ Put in (1), we get \\ y = x - \log |sinx + cosx| + 0 \\ Thus, required equation of the curve is, \\ y = x - \log |(sinx + cosx)| \end{array}$

Q.27

$\begin{array}{l} Solve the differential equation . \\ xdy - ydx = \sqrt{x^{2} + y^{2}} dx \end{array}$

Ans

$\begin{array}{l} We have, \\ xdy - ydx = \sqrt{x^{2} + y^{2}} dx \\ \Rightarrow x dy = [y + \sqrt{x^{2} + y^{2}}] dx \\ \Rightarrow \frac{dy}{dx} = \frac{y + \sqrt{x^{2} + y^{2}}}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{y}{x} \sqrt{1 + {(\frac{y}{x})}^{2}} . \dots\dots\dots (1) \\ Put y = vx . \dots\dots. (2) \\ On differentiating both sides w . r . t . x, \\ we get - \\ \frac{dy}{dx} = v + x \frac{dv}{dx} . \dots\dots. (3) \\ On putting the balue of y and \frac{dy}{dx} from (2) and (3) \\ to (1), (1) becomes - \\ v + x \frac{dv}{dx} = v + \sqrt{1 + {(v)}^{2}} \\ \Rightarrow x \frac{dv}{dx} = \sqrt{1 + {(v)}^{2}} \\ On separating variables, we get \\ \frac{dv}{dx} = \frac{dv}{\sqrt{1 + v^{2}}} = \frac{dx}{x} \\ On integrating both sides, we get \\ \int \frac{dv}{\sqrt{1 + v^{2}}} = \int \begin{array}{l} \frac{dx}{x} \end{array} \\ \Rightarrow \log [v + \sqrt{1 + v^{2}}] = logx + \log c \\ \Rightarrow \log [v + \sqrt{1 + v^{2}}] = logcx \\ \Rightarrow v + \sqrt{1 + v^{2}} = cx \\ \Rightarrow \frac{y}{x} + \sqrt{1 + {(\frac{y}{x})}^{2}} = cx \\ \Rightarrow \frac{y + \sqrt{x^{2} + y^{2}}}{x} = cx \\ \Rightarrow y + \sqrt{x^{2} + y^{2}} = {cx}^{2} \\ Hence, solution of the differential equation is - \\ y + \sqrt{x^{2} + y^{2}} - {cx}^{2} = 0 \end{array}$

Q.28

$\begin{array}{l} Determine the order and degree of each of the following \\ differential rquations . Sate also if they are linear or non - linear . \\ (i) \frac{{\{1 + {(\frac{dy}{dx})}^{2}\}}^{\frac{3}{2}}}{\frac{d^{2} y}{{dx}^{2}}} = k \\ (ii) \frac{d^{2} y}{{dx}^{2}} = 1 + \sqrt{\frac{dy}{dx}} \\ (iii) y = \frac{dy}{dx} + \frac{c}{\frac{dy}{dx}} \\ (iv) y + \frac{dy}{dx} = \frac{1}{4} \int y dx \end{array}$

Ans

$\begin{array}{l} (i) The given differential equation when written as polynomial in \\ derivatives becomes \\ k^{2} {(\frac{dy}{dx})}^{2} = {\{1 + {(\frac{dy}{dx})}^{2}\}}^{3} \\ The highest order differential coefficient in this equation is \frac{d^{2} y}{{dx}^{2}} and \\ its power is 2. \\ Therefore, the given differential equation is a non - linear differential \\ equation of second order and second degree . \\ (ii) The given differential equation when written as polynomial in \\ derivatives becomes \\ {(\frac{d^{2} y}{{dx}^{2}} - 1)}^{2} = \frac{dy}{dx} \Rightarrow {(\frac{d^{2} y}{{dx}^{2}})}^{2} - 2 \frac{d^{2} y}{{dx}^{2}} - \frac{dy}{dx} + 1 = 0 \\ Clearly, it is a non - linear differential equation of second \\ order and second degree . \\ (iii) The given differential equation when written as polynomial in \frac{dy}{dx} is \\ {(\frac{dy}{dx})}^{2} - y \frac{dy}{dx} + c = 0 \\ Clearly, it is a non - linear differential equation of order 1 and degree 2. \\ (iv) we have, \\ y + \frac{dy}{dx} = \frac{1}{4} \int y dx \\ \Rightarrow \frac{dy}{dx} + \frac{d^{2} y}{{dx}^{2}} = \frac{1}{4} y \\ Clearly, this is a differential equation of order 2 and degree 1. \\ Also, it is a linear differential equation . \end{array}$

Q.29

$Solve the initial value problem e^{\frac{dy}{dx}} = x + 1; y (0) = 5.$

Ans

$\begin{array}{l} We are given that \\ e^{\frac{dy}{dx}} = x + 1 \\ \Rightarrow \frac{dy}{dx} = \log (x + 1) \\ \Rightarrow dy = \log (x + 1) dx \\ Integrating both sides, we get \\ \int 1 . dy = \int \log (x + 1) \times 1 dx \\ \Rightarrow y = x \log (x + 1) - \int \frac{x}{x + 1} dx \\ \Rightarrow y = x \log (x + 1) - \int \begin{array}{l} \frac{x + 1 - 1}{x + 1} dx \end{array} \\ \Rightarrow y = x \log (x + 1) - \int (1 - \frac{1}{x + 1}) dx \\ \Rightarrow y = x \log (x + 1) - x + \log (x + 1) + C . \dots\dots\dots (i) \\ It is given that y (0) = 5 i . e ., when x = 0, we have y = 5. \\ ∴ 5 = \log 1 - 0+ \log 1 + C \\ \Rightarrow C = 5 [Substituting x = 0, y = 5 in (i)] \\ Substituting the value of C in (i), we get \\ y = x \log (x + 1) - x + \log (x + 1) + 5 \\ We observe that x \log (x + 1) - xlog (x + 1) + 5 is defined for all x \in (- 1, \infty) . \\ Hence, y = xlog (x + 1) - x + \log (x + 1) + 5, where x \in (- 1, \infty) is the solution of \\ the given initial value proble . \end{array}$

Q.30

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x² + 1) dx (x≠0).

Ans

$\begin{array}{l} We have, \\ x dy = (2 x^{2} + 1) dx \\ \Rightarrow dy = (\frac{2 x^{2} + 1}{x}) dx \\ \Rightarrow dy = (2 x + \frac{1}{x}) dx \\ On integrating both sides, we get \\ \int dy = \int (2 x + \frac{1}{x}) dx \\ \Rightarrow y = x^{2} + \log |x| + c \\ This equation represents the family of solution curves \\ of the given differential equation . \\ We have to find a particular member of this family which \\ passes through the point (1, 1) . \\ Substituting x = 1, y = 1 in (i), we get \\ 1 = 1 + 0 + c \\ \Rightarrow c = 0 \\ Putting c = 0 in (i), we get \\ y = x^{2} + \log |x| as the equation of the required curve . \end{array}$

Q.31

$\begin{array}{l} It is known that, if the interest is compounded continuously, \\ the principal changes at the rate equal to the product of the \\ rate of bank interest per annum and the principal . \\ (i) If the interest is compounded continuously at 5% per \\ annum, in how many years will Rs . 100 double itself ? \\ (ii) At what interest rate will Rs . 100 double itself in 10 years ? \\ (\log_{e} 2 = 0.6931) \\ (iii) How much will Rs . 1000 be worth at 5% interest after 10 years ? \\ (e^{0 .5} = 1.648) \end{array}$

Ans

$\begin{array}{l} If P denotes the principal at any time t and the rate of interest be r % \\ per annum compounded continuously, then according to the law \\ given in the problem, we have \\ \frac{dP}{dt} = \frac{\Pr}{100} \\ \Rightarrow \frac{dP}{P} = \frac{r}{100} dt \\ \Rightarrow \int \frac{1}{P} dP = \frac{r}{100} \int dt \\ \Rightarrow Log P = \frac{rt}{100} + C . \dots\dots. (i) \\ Let P_{0} be the intial principal i . e . at t = 0, P = P_{0} \\ Putting P = P_{0} in (i), we get \\ \log P_{0} = C \\ Putting C = \log P_{0} in (i), we get \\ Log P = \frac{rt}{100} + {logP}_{0} \\ \Rightarrow \log (\frac{P}{P_{0}}) = \frac{rt}{100} . \dots\dots (ii) \\ (i) In this case, we have \\ r = r, P_{0} = Rs . 100 and P = 200 = 2 P_{0} \\ Substituting these values in (ii), we have \\ \log 2 \frac{5}{100} t \Rightarrow t = 20 \log_{e} 2 \\ = 20 \times 0 .6931 years \\ = 13 .862 years . \\ (ii) In this case, we have \\ P_{0} = Rs . 100, P = Rs . 200 = 2 P_{0} and t = 10 years . \\ Substituting these values in (ii), we get \\ \log 2 = \frac{10 r}{100} \Rightarrow r = 10 \log 2 \\ = 10 \times 0 .6931 \\ = 6.931 \\ Hence, r = 6.931% per annum . \\ (iii) In this case, we have \\ P_{0} = Rs . 1000, r = 5 and t = 10 \\ Substituting these values in (ii), we get \\ \log (\frac{P}{1000}) = \frac{5 \times 10}{100} = \frac{1}{2} = 0 .5 \\ \Rightarrow \frac{P}{1000} = e^{0 .5} \\ \Rightarrow P = 1000 \times 1 .648 = 1648 \\ Hence, P = Rs . 1648. \end{array}$

Q.32

$\begin{array}{l} Solve the differential equation (x + y) dy (x - y) dx = 0, given \\ that y =1 when x = 1. \end{array}$

Ans

$\begin{array}{l} The given differential equation is \\ (x + y) dy + (x - y) dx = 0 \\ \Rightarrow \frac{dy}{dx} = \frac{x - y}{x + y} \\ \Rightarrow \frac{dy}{dx} = \frac{x - y}{x + y} . \dots.. (i) \\ Since each of the function y - x and x + y is a homogeneous function \\ of degree 1, Therefore, equation (i) is a homogeneous differential \\ equation . \\ Putting y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx} in (i), we get \\ v + x \frac{dv}{dx} = \frac{vx - x}{x + vx} \\ \Rightarrow v + x \frac{dv}{dx} = \frac{v - 1}{v + 1} \\ \Rightarrow x \frac{dv}{dx} = \frac{v - 1}{v + 1} - v \\ \Rightarrow x \frac{dv}{dx} = \frac{v - 1 - v^{2} - v}{v + 1} \\ \Rightarrow x \frac{dv}{dx} = (\frac{v^{2} + 1}{v + 1}) \\ \Rightarrow \frac{v^{2} + 1}{v + 1} dv = \frac{dx}{x}, x \neq 0 [By separating the variables] \\ \Rightarrow \int \frac{v + 1}{v^{2} + 1} dv = - \int \frac{dx}{x} \\ \Rightarrow \int \frac{v + 1}{v^{2} + 1} dv + \int \frac{1}{v^{2} + 1} dv = - \int \frac{dx}{x} \\ \Rightarrow \frac{1}{2} \int \frac{2 v}{v^{2} + 1} dv + \int \frac{1}{v^{2} + 1} dv = - \int \frac{dx}{x} \\ \Rightarrow \frac{1}{2} \log (v^{2} + 1) + \tan^{- 1} v = - \log |x| + C \\ \Rightarrow \log (v^{2} + 1) + 2 \log |x| + 2 \tan^{- 1} v = 2 C \\ \Rightarrow \log (v^{2} + 1) + {logx}^{2} + 2 \tan^{- 1} v = k, where k = 2 C \\ \Rightarrow \log [(v^{2} + 1) x^{2}] + 2 \tan^{- 1} v = k \\ \Rightarrow \log [(\frac{y^{2}}{x^{2}} + 1) x^{2}] + 2 \tan^{- 1} (\frac{y}{x}) = k [∵ v = \frac{y}{x}] \\ \Rightarrow \log [x^{2} + y^{2}] + 2 \tan^{- 1} (\frac{y}{x}) = k . \dots. (ii) \\ It is given that y = 1, when x = 1. \\ Putting x = 1, y = 1 in (ii), we get \\ \log 2 + 2 \tan^{- 1} (1) = k \\ \Rightarrow \log 2 + 2 (\frac{π}{4}) = (\frac{π}{2}) + \log 2 \\ Substituting the value of k in (ii), we get \\ \Rightarrow \log [x^{2} + y^{2}] + 2 \tan^{- 1} (\frac{y}{x}) = (\frac{π}{2}) + \log 2 \\ Hence, \log (x^{2} + y^{2}) + 2 \tan^{- 1} (\frac{y}{x}) = (\frac{π}{2}) + \log 2, x \neq 0 \\ is the required solution of the given differential equation . \end{array}$

Q.33

$\begin{array}{l} A thermometer reading 40^{ο} F . is taken outside . Five minutes \\ later the thermometer reads 30^{ο} F . After anothe 5 minutes the \\ thermometer reads 25^{ο} F . What is the temperature outside ? \end{array}$

Ans

$\begin{array}{l} Let at any time t the therometer reading be T^{ο} F and the outside \\ temperature be S^{ο} F . \\ Then, by newton ‘ s law of cooling \\ \frac{dT}{dt} \propto (T - S) \\ \Rightarrow \frac{dT}{dt} = - λ (T - S) \\ \Rightarrow \frac{dT}{T - S} = λdt \\ \Rightarrow \int \frac{1}{T - S} dT = - λ \int dt \\ \Rightarrow \log (T - S) = λ t + c . \dots.. (i) \\ It is given that T {= 40}^{ο} F at t = 0 \\ ∴ \log (40 - S) = 0 + C \\ \Rightarrow C = \log (40 - S) \\ Putting the value of C in (i) we get \\ \log (T - S) = - λ t + \log (40 - S) \\ \Rightarrow \log (\frac{T - S}{40 - S}) = - λ t . \dots.. (ii) \\ It is given that \\ T {= 30}^{ο} F at t = 5 \\ T {= 25}^{ο} F at t = 10 \\ and \\ Subsitituting these values in (ii), we get \\ \log (\frac{30 - S}{40 - S}) = - 5 λ and \log (\frac{25 - S}{40 - S}) = - 10 λ \\ \Rightarrow 2 \log (\frac{30 - S}{40 - S}) = \log (\frac{25 - S}{40 - S}) \\ \Rightarrow {(\frac{30 - S}{40 - S})}^{2} = \log (\frac{25 - S}{40 - S}) \\ \Rightarrow {(30 - S)}^{2} = (25 - S) (40 - S) \\ \Rightarrow 900 + S^{2} - 60 S = 1000 - 255 - 40 S - S^{2} \\ \Rightarrow 5 S = 100 \\ \Rightarrow S = 20^{ο} F \\ Hence, the out side temperature is 20^{ο} F . \end{array}$

Q.34

$\begin{array}{l} Form a differential equation representing the family \\ of curves given by \frac{x}{a} + \frac{y}{b} = 1 . \end{array}$

Ans

$\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 \\ Differentiating both sides w . r . t . x, we get \\ \frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0 \\ Againg, differentiating both sides w . r . t . x, we get \\ \frac{1}{b} \frac{d^{2} y}{{dx}^{2}} = 0 \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 0 \Rightarrow y ‘ ‘ = 0 \\ Hence, the required differential equation is y ” = 0. \end{array}$

Q.35

$\begin{array}{l} Form a differential equation representing the family \\ of curves given by y^{2} = a (b^{2} - x^{2}) \end{array}$

Ans

$\begin{array}{l} y^{2} = a (b^{2} - x^{2}) \\ Differentiating both sides w . r . t . x, we get \\ 2 y \frac{dy}{dx} = a (0 - 2 x) \\ 2 y \frac{dy}{dx} = - 2 ax . \dots.. (i) \\ Againg, differentiating both sides w . r . t . x, we get \\ 2 \{{(\frac{dy}{dx})}^{2} + y \frac{d^{2 y}}{{dx}^{2}}\} = - 2 a \\ \Rightarrow {(\frac{dy}{dx})}^{2} + yy \frac{d^{2 y}}{{dx}^{2}} = - a . \dots\dots (ii) \\ Putting value of - a from equation (ii) toequation (i), we get \\ 2 y \frac{dy}{dx} = 2 \{{(\frac{dy}{dx})}^{2} + y \frac{d^{2 y}}{{dx}^{2}}\} x \\ \Rightarrow y \frac{dy}{dx} = x {(\frac{dy}{dx})}^{2} + y \frac{d^{2 y}}{{dx}^{2}} \\ xyy ‘ ‘ + x {(y ‘)}^{2} - yy ‘ = 0 \\ Hence, the required differential equation is xyy ” + x {(y)}^{2} - yy ‘ = 0 \end{array}$

Q.36

Form the differential equation of the family of parabolas having vertex at originand axis along positive x-axis.

Ans

The equation of parabolas having vertex at origin
and axisalong positive y-axis is as follows
y² = 4ax …(i)
Differentiating w.r.t. x, we get
2yy’ = 4a

Putting value of 4a in equation (i), we get
y² = 2yy’ x

y = 2y’ x

or2y’x – y =0

Thus, the required differential equation is
2y’x – y =0.

Q.37

$\begin{array}{l} This is the general solution of given differential equation . \\ Fox the general solution differential equation \\ \frac{dy}{dx} = (1 + x^{2}) (1 + y^{2}) \end{array}$

Ans

$\begin{array}{l} The given differential equation is : \\ \frac{dy}{dx} = (1 + x^{2}) (1 + y^{2}) \\ \Rightarrow \frac{dy}{(1 + y^{2})} = (1 + x^{2}) dx \\ Integrating both sides, we get \\ \int \frac{dy}{(1 + y^{2})} = \int (1 + x^{2}) dx \\ \tan^{- 1} y = x + \frac{x^{3}}{3} + C \\ This is the general solution of given differential equation . \end{array}$

Q.38

$\begin{array}{l} For the general solution differential equation \\ y logydx - x dy = 0. \end{array}$

Ans

$\begin{array}{l} The given differential equation is : \\ y \log y dx - x dy = 0 \\ \Rightarrow y \log y dx = x dy \\ \Rightarrow \frac{dy}{y \log y} = \frac{dx}{x} \\ Integrating both sides, we get \\ f \frac{dy}{y \log y} = f \frac{dx}{x} \\ \Rightarrow loglog y = logx + logC \\ \Rightarrow \log logy = logCx \\ \Rightarrow logy = Cx \\ \Rightarrow y = e^{cx} \\ This is the general solution of given differential equation . \end{array}$

Q.39

$\begin{array}{l} Check whether xy = logy +C is the solution of the differential equation \\ y’ = \frac{y^{2}}{1 - x y}, (x y \neq 0) . \end{array}$

Ans

$\begin{array}{l} xy = logy + C \\ Differentiating w . r . t . x, we get \\ xy ‘ + y = \frac{1}{y} y ‘ + 0 \\ xyy ‘ + y^{2} = y ‘ \\ y^{2} = y ‘ - xyy ‘ \\ = y ‘ (1 - xy) \\ y ‘ = \frac{y^{2}}{(1 - xy)} \\ Thus, the given function is the solution of the \\ corresponding differential equation . \end{array}$

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CBSE Class 12 Maths Revision Notes Chapter 9

NCERT Class 12 Mathematics Chapter 9 Notes

Class 12 Mathematics Chapter 9: Key Topics

Introduction:

Basic Concepts

Differential equation

Ordinary Differential Equation:

Order of an ordinary differential equation

Degree of a differential equation

Formation of the ordinary differential equations:

Solution of the Differential Equation

Procedure to form a differential equation that represents the family of curves

Methods of Solving the First Order and First Degree Ordinary Differential Equations

METHOD 1:

METHOD 2:

METHOD 3: Linear differential equations

Application of Differential Equations:

NCERT Class 12 Mathematics Chapter 9 Notes- Differential Equations: Chapter Weightage

NCERT Class 12 Mathematics Chapter 9 Notes: Exercises & Solutions

NCERT Exemplar Class 12 Mathematics

NCERT Class 12 Mathematics Chapter 9 Notes: Key Features

CBSE Exam Preparation Tips

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CBSE Class 12 Maths Revision Notes Chapter 9

NCERT Class 12 Mathematics Chapter 9 Notes

Class 12 Mathematics Chapter 9: Key Topics

Introduction:

Basic Concepts

Differential equation

Ordinary Differential Equation:

Order of an ordinary differential equation

Degree of a differential equation

Formation of the ordinary differential equations:

Solution of the Differential Equation

Procedure to form a differential equation that represents the family of curves

Methods of Solving the First Order and First Degree Ordinary Differential Equations

METHOD 1:

METHOD 2:

METHOD 3: Linear differential equations

Application of Differential Equations:

NCERT Class 12 Mathematics Chapter 9 Notes- Differential Equations: Chapter Weightage

NCERT Class 12 Mathematics Chapter 9 Notes: Exercises & Solutions

NCERT Exemplar Class 12 Mathematics

NCERT Class 12 Mathematics Chapter 9 Notes: Key Features

CBSE Exam Preparation Tips

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