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MATHEMATICS-I
CONTENTS
   Ordinary Differential Equations of First Order and First Degree
   Linear Differential Equations of Second and Higher Order
   Mean Value Theorems
   Functions of Several Variables
   Curvature, Evolutes and Envelopes
   Curve Tracing
   Applications of Integration
   Multiple Integrals
   Series and Sequences
   Vector Differentiation and Vector Operators
   Vector Integration
   Vector Integral Theorems
   Laplace transforms
TEXT BOOKS
   A text book of Engineering Mathematics, Vol-I
    T.K.V.Iyengar, B.Krishna Gandhi and Others,
    S.Chand & Company
   A text book of Engineering Mathematics,
    C.Sankaraiah, V.G.S.Book Links
   A text book of Engineering Mathematics, Shahnaz A
    Bathul, Right Publishers
   A text book of Engineering Mathematics,
    P.Nageshwara Rao, Y.Narasimhulu & N.Prabhakar
    Rao, Deepthi Publications
REFERENCES
 A text book of Engineering Mathematics,
  B.V.Raman, Tata Mc Graw Hill
 Advanced Engineering Mathematics, Irvin
  Kreyszig, Wiley India Pvt. Ltd.
 A text Book of Engineering Mathematics,
  Thamson Book collection
UNIT-I

 ORDINARY DIFFERENTIAL
EQUATIONS OF FIRST ORDER
   AND FIRST DEGREE
UNIT HEADER
  Name of the Course:B.Tech
      Code No:07A1BS02
      Year/Branch:I Year
CSE,IT,ECE,EEE,ME,CIVIL,AERO
          Unit No: I
        No.of slides:26
UNIT INDEX
                        UNIT-I
S.No.           Module           Lecture   PPT Slide No.
                                 No.
  1     Introduction,Exact       L1-10     8-19
        differential equations
  2     Linear and Bernoulli’s   L11-13    20-23
        equations,Orthogonal
        trajectories
  3     Newton’s law of          L14-15    24-26
        cooling and decay
Lecture-1
             INTRODUCTION
 An equation involving a dependent variable
  and its derivatives with respect to one or more
  independent variables is called a Differential
  Equation.
 Example 1: y″ + 2y = 0
 Example 2: y2 – 2y1+y=23
 Example 3: d2y/dx2 + dy/dx – y=1


L1-3:
TYPES OF A DIFFERENTIAL
               EQUATION
   ORDINARY DIFFERENTIAL EQUATION: A
    differential equation is said to be ordinary, if the
    derivatives in the equation are ordinary derivatives.
   Example:d2y/dx2-dy/dx+y=1
   PARTIAL DIFFERENTIAL EQUATION: A
    differential equation is said to be partial if the
    derivatives in the equation have reference to two or
    more independent variables.
   Example:∂4y/∂x4+∂y/∂x+y=1

L1-3:
Lecture-2
                    DEFINITIONS
   ORDER OF A DIFFERENTIAL EQUATION: A
    differential equation is said to be of order n, if the nth
    derivative is the highest derivative in that equation.
   Example: Order of d2y/dx2+dy/dx+y=2 is 2
   DEGREE OF A DIFFERENTIAL EQUATION: If
    the given differential equation is a polynomial in y(n),
    then the highest degree of y(n) is defined as the degree
    of the differential equation.
   Example: Degree of (dy/dx)4+y=0 is 4

L1-3:
SOLUTION OF A DIFFERENTIAL
            EQUATION
   SOLUTION: Any relation connecting the variables of an
    equation and not involving their derivatives, which satisfies
    the given differential equation is called a solution.
   GENERAL SOLUTION: A solution of a differential equation
    in which the number of arbitrary constant is equal to the order
    of the equation is called a general or complete solution or
    complete primitive of the equation.
   Example: y = Ax + B
   PARTICULAR SOLUTION: The solution obtained by giving
    particular values to the arbitrary constants of the general
    solution, is called a particular solution of the equation.
   Example: y = 3x + 5

L1-3:
Lecture-3
 EXACT DIFFERENTIAL EQUATION
 Let M(x,y)dx + N(x,y)dy = 0 be a first order
  and first degree differential equation where M
  and N are real valued functions for some x, y.
  Then the equation Mdx + Ndy = 0 is said to be
  an exact differential equation if ∂M/∂y=∂N/∂x
 Example:
  (2y sinx+cosy)dx=(x siny+2cosx+tany)dy


L1-3:
Lecture-4
 Working rule to solve an exact equation
STEP 1: Check the condition for exactness,

        if exact proceed to step 2.
STEP 2: After checking that the equation is
        exact, solution can be obtained as
       ∫M dx+∫(terms not containing x) dy=c



L1-3:
Lecture-5
        INTEGRATING FACTOR
 Let Mdx + Ndy = 0 be not an exact differential
  equation. Then Mdx + Ndy = 0 can be made
  exact by multiplying it with a suitable function
  u is called an integrating factor.
 Example 1:ydx-xdy=0 is not an exact equation.
  Here 1/x2 is an integrating factor
 Example 2:y(x2y2+2)dx+x(2-2x2y2)dy=0 is not
  an exact equation. Here 1/(3x3y3) is an
  integrating factor
L1-3:
Lecture-6
 METHODS TO FIND INTEGRATING
          FACTORS
 METHOD 1: With some experience
  integrating factors can be found by inspection.
  That is, we have to use some known
  differential formulae.
 Example 1:d(xy)=xdy+ydx
 Example 2:d(x/y)=(ydx-xdy)/y 2
 Example 3:d[log(x2+y2)]=2(xdx+ydy)/(x2+y2)



L1-3:
Lecture-7
 METHODS TO FIND INTEGRATING
          FACTORS
 METHOD 2: If Mdx + Ndy = 0 is a non-exact
  but homogeneous differential equation and
  Mx + Ny ≠ 0 then 1/(Mx + Ny) is an
  integrating factor of Mdx + Ndy = 0.
 Example 1:x2ydx-(x3+y3)dy=0 is a non-exact
  homogeneous equation. Here I.F.=-1/y 4
 Example 2:y2dx+(x2-xy-y2)dy=0 is a non-exact
  homogeneous equation. Here I.F.=1/(x2y-y3)

L1-3:
Lecture-8
 METHODS TO FIND INTEGRATING
          FACTORS
   METHOD 3: If the equation Mdx + Ndy = 0 is of the
    form y.f(xy) dx + x.g(xy) dy = 0 and Mx – Ny ≠ 0
    then 1/(Mx – Ny) is an integrating factor of Mdx +
    Ndy = 0.
   Example 1:y(x2y2+2)dx+x(2-2x2y2)dy=0 is non-exact
    and in the above form. Here I.F=1/(3x3y3)
   Example 2:(xysinxy+cosxy)ydx+(xysinxy-
    cosxy)xdy=0 is non-exact and in the above form.
    Here I.F=1/(2xycosxy)

L1-3:
Lecture-9
 METHODS TO FIND INTEGRATING
          FACTORS
   METHOD 4: If there exists a continuous single
    variable function f(x) such that ∂M/∂y-
    ∂N/∂x=Nf(x) then e∫f(x)dx is an integrating factor of
    Mdx + Ndy = 0
   Example 1:2xydy-(x2+y2+1)dx=0 is non-exact and
    ∂M/∂y - ∂N/∂x=N(-2/x). Here I.F=1/x2
   Example 2:(3xy-2ay2)dx+(x2-2axy)=0 is non-exact
    and ∂M/∂y - ∂N/∂x=N(1/x). Here I.F=x


L1-3:
Lecture-10
    METHODS TO FIND INTEGRATING
             FACTORS
   METHOD 5: If there exists a continuous single
    variable function f(y) such that
    ∂N/∂x - ∂M/∂y=Mg(y) then e∫g(y)dy is an integrating
    factor of Mdx + Ndy = 0
   Example 1:(xy3+y)dx+2(x2y2+x+y4)dy=0 is a non-
    exact equation and ∂N/∂x - ∂M/∂y=M(1/y). Here
    I.F=y
   Example 2:(y4+2y)dx+(xy3+2y4-4x)dy=0 is a non-
    exact equation and ∂N/∂x - ∂M/∂y=M(-3/y).Here
    I.F=1/y3
L1-3:
Lecture-11
        LEIBNITZ LINEAR EQUATION
   An equation of the form y′ + Py = Q is called a
    linear differential equation.
   Integrating Factor(I.F.)=e∫pdx
   Solution is y(I.F) = ∫Q(I.F)dx+C
   Example 1:xdy/dx+y=logx. Here I.F=x and solution
    is xy=x(logx-1)+C
   Example 2:dy/dx+2xy=e-x.Here I.F=ex and solution
    is yex=x+C


L1-3:
Lecture-12
BERNOULLI’S LINEAR EQUATION
   An equation of the form y′ + Py = Qyn is called a
    Bernoulli’s linear differential equation. This
    differential equation can be solved by reducing it to
    the Leibnitz linear differential equation. For this
    dividing above equation by yn
   Example 1: xdy/dx+y=x2y6.Here I.F=1/x5 and solution
    is 1/(xy)5=5x3/2+Cx5
   Example 2: dy/dx+y/x=y2xsinx. Here I.F=1/x and
    solution is 1/xy=cosx+C
Lecture-13
     ORTHOGONAL TRAJECTORIES
   If two families of curves are such that each member
    of family cuts each member of the other family at
    right angles, then the members of one family are
    known as the orthogonal trajectories of the other
    family.
   Example 1: The orthogonal trajectory of the family of
    parabolas through origin and foci on y-axis is
    x2/2c+y2/c=1
   Example 2: The orthogonal trajectory of rectangular
    hyperbolas is xy=c2
PROCEDURE TO FIND
   ORTHOGONAL TRAJECTORIES
Suppose f (x ,y ,c) = 0 is the given family of
        curves, where c is the constant.
STEP 1: Form the differential equation by
         eliminating the arbitrary constant.
STEP 2: Replace y′ by -1/y′ in the above
         equation.
STEP 3: Solve the above differential equation.
Lecture-14
   NEWTON’S LAW OF COOLING
 The rate at which the temperature of a hot
  body decreases is proportional to the
  difference between the temperature of the
  body and the temperature of the surrounding
  air.
              θ′ ∞ (θ – θ0)
 Example: If a body is originally at 80 oC and
  cools down to 60oC in 20 min.If the
  temperature of the air is at 40oC then the
  temperture of the body after 40 min is 50 oC
Lecture-15
       LAW OF NATURAL GROWTH
   When a natural substance increases in Magnitude as a
    result of some action which affects all parts equally,
    the rate of increase depends on the amount of the
    substance present.
                 N′ = k N
   Example: If the number N of bacteria in a culture
    grew at a rate proportional to N. The value of N was
    initially 100 and increased to 332 in 1 hour. Then the
    value of N after one and half hour is 605
LAW OF NATURAL DECAY
 The rate of decrease or decay of any substance
  is proportion to N the number present at time.
                  N′ = -k N
 Example: A radioactive substance disintegrates
  at a rate proportional to its mass. When mass is
  10gms, the rate of disintegration is 0.051gms
  per day. The mass is reduced to 10 to 5gms in
  136 days.

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M1 unit i-jntuworld

  • 2. CONTENTS  Ordinary Differential Equations of First Order and First Degree  Linear Differential Equations of Second and Higher Order  Mean Value Theorems  Functions of Several Variables  Curvature, Evolutes and Envelopes  Curve Tracing  Applications of Integration  Multiple Integrals  Series and Sequences  Vector Differentiation and Vector Operators  Vector Integration  Vector Integral Theorems  Laplace transforms
  • 3. TEXT BOOKS  A text book of Engineering Mathematics, Vol-I T.K.V.Iyengar, B.Krishna Gandhi and Others, S.Chand & Company  A text book of Engineering Mathematics, C.Sankaraiah, V.G.S.Book Links  A text book of Engineering Mathematics, Shahnaz A Bathul, Right Publishers  A text book of Engineering Mathematics, P.Nageshwara Rao, Y.Narasimhulu & N.Prabhakar Rao, Deepthi Publications
  • 4. REFERENCES  A text book of Engineering Mathematics, B.V.Raman, Tata Mc Graw Hill  Advanced Engineering Mathematics, Irvin Kreyszig, Wiley India Pvt. Ltd.  A text Book of Engineering Mathematics, Thamson Book collection
  • 5. UNIT-I ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
  • 6. UNIT HEADER Name of the Course:B.Tech Code No:07A1BS02 Year/Branch:I Year CSE,IT,ECE,EEE,ME,CIVIL,AERO Unit No: I No.of slides:26
  • 7. UNIT INDEX UNIT-I S.No. Module Lecture PPT Slide No. No. 1 Introduction,Exact L1-10 8-19 differential equations 2 Linear and Bernoulli’s L11-13 20-23 equations,Orthogonal trajectories 3 Newton’s law of L14-15 24-26 cooling and decay
  • 8. Lecture-1 INTRODUCTION  An equation involving a dependent variable and its derivatives with respect to one or more independent variables is called a Differential Equation.  Example 1: y″ + 2y = 0  Example 2: y2 – 2y1+y=23  Example 3: d2y/dx2 + dy/dx – y=1 L1-3:
  • 9. TYPES OF A DIFFERENTIAL EQUATION  ORDINARY DIFFERENTIAL EQUATION: A differential equation is said to be ordinary, if the derivatives in the equation are ordinary derivatives.  Example:d2y/dx2-dy/dx+y=1  PARTIAL DIFFERENTIAL EQUATION: A differential equation is said to be partial if the derivatives in the equation have reference to two or more independent variables.  Example:∂4y/∂x4+∂y/∂x+y=1 L1-3:
  • 10. Lecture-2 DEFINITIONS  ORDER OF A DIFFERENTIAL EQUATION: A differential equation is said to be of order n, if the nth derivative is the highest derivative in that equation.  Example: Order of d2y/dx2+dy/dx+y=2 is 2  DEGREE OF A DIFFERENTIAL EQUATION: If the given differential equation is a polynomial in y(n), then the highest degree of y(n) is defined as the degree of the differential equation.  Example: Degree of (dy/dx)4+y=0 is 4 L1-3:
  • 11. SOLUTION OF A DIFFERENTIAL EQUATION  SOLUTION: Any relation connecting the variables of an equation and not involving their derivatives, which satisfies the given differential equation is called a solution.  GENERAL SOLUTION: A solution of a differential equation in which the number of arbitrary constant is equal to the order of the equation is called a general or complete solution or complete primitive of the equation.  Example: y = Ax + B  PARTICULAR SOLUTION: The solution obtained by giving particular values to the arbitrary constants of the general solution, is called a particular solution of the equation.  Example: y = 3x + 5 L1-3:
  • 12. Lecture-3 EXACT DIFFERENTIAL EQUATION  Let M(x,y)dx + N(x,y)dy = 0 be a first order and first degree differential equation where M and N are real valued functions for some x, y. Then the equation Mdx + Ndy = 0 is said to be an exact differential equation if ∂M/∂y=∂N/∂x  Example: (2y sinx+cosy)dx=(x siny+2cosx+tany)dy L1-3:
  • 13. Lecture-4 Working rule to solve an exact equation STEP 1: Check the condition for exactness, if exact proceed to step 2. STEP 2: After checking that the equation is exact, solution can be obtained as ∫M dx+∫(terms not containing x) dy=c L1-3:
  • 14. Lecture-5 INTEGRATING FACTOR  Let Mdx + Ndy = 0 be not an exact differential equation. Then Mdx + Ndy = 0 can be made exact by multiplying it with a suitable function u is called an integrating factor.  Example 1:ydx-xdy=0 is not an exact equation. Here 1/x2 is an integrating factor  Example 2:y(x2y2+2)dx+x(2-2x2y2)dy=0 is not an exact equation. Here 1/(3x3y3) is an integrating factor L1-3:
  • 15. Lecture-6 METHODS TO FIND INTEGRATING FACTORS  METHOD 1: With some experience integrating factors can be found by inspection. That is, we have to use some known differential formulae.  Example 1:d(xy)=xdy+ydx  Example 2:d(x/y)=(ydx-xdy)/y 2  Example 3:d[log(x2+y2)]=2(xdx+ydy)/(x2+y2) L1-3:
  • 16. Lecture-7 METHODS TO FIND INTEGRATING FACTORS  METHOD 2: If Mdx + Ndy = 0 is a non-exact but homogeneous differential equation and Mx + Ny ≠ 0 then 1/(Mx + Ny) is an integrating factor of Mdx + Ndy = 0.  Example 1:x2ydx-(x3+y3)dy=0 is a non-exact homogeneous equation. Here I.F.=-1/y 4  Example 2:y2dx+(x2-xy-y2)dy=0 is a non-exact homogeneous equation. Here I.F.=1/(x2y-y3) L1-3:
  • 17. Lecture-8 METHODS TO FIND INTEGRATING FACTORS  METHOD 3: If the equation Mdx + Ndy = 0 is of the form y.f(xy) dx + x.g(xy) dy = 0 and Mx – Ny ≠ 0 then 1/(Mx – Ny) is an integrating factor of Mdx + Ndy = 0.  Example 1:y(x2y2+2)dx+x(2-2x2y2)dy=0 is non-exact and in the above form. Here I.F=1/(3x3y3)  Example 2:(xysinxy+cosxy)ydx+(xysinxy- cosxy)xdy=0 is non-exact and in the above form. Here I.F=1/(2xycosxy) L1-3:
  • 18. Lecture-9 METHODS TO FIND INTEGRATING FACTORS  METHOD 4: If there exists a continuous single variable function f(x) such that ∂M/∂y- ∂N/∂x=Nf(x) then e∫f(x)dx is an integrating factor of Mdx + Ndy = 0  Example 1:2xydy-(x2+y2+1)dx=0 is non-exact and ∂M/∂y - ∂N/∂x=N(-2/x). Here I.F=1/x2  Example 2:(3xy-2ay2)dx+(x2-2axy)=0 is non-exact and ∂M/∂y - ∂N/∂x=N(1/x). Here I.F=x L1-3:
  • 19. Lecture-10 METHODS TO FIND INTEGRATING FACTORS  METHOD 5: If there exists a continuous single variable function f(y) such that ∂N/∂x - ∂M/∂y=Mg(y) then e∫g(y)dy is an integrating factor of Mdx + Ndy = 0  Example 1:(xy3+y)dx+2(x2y2+x+y4)dy=0 is a non- exact equation and ∂N/∂x - ∂M/∂y=M(1/y). Here I.F=y  Example 2:(y4+2y)dx+(xy3+2y4-4x)dy=0 is a non- exact equation and ∂N/∂x - ∂M/∂y=M(-3/y).Here I.F=1/y3 L1-3:
  • 20. Lecture-11 LEIBNITZ LINEAR EQUATION  An equation of the form y′ + Py = Q is called a linear differential equation.  Integrating Factor(I.F.)=e∫pdx  Solution is y(I.F) = ∫Q(I.F)dx+C  Example 1:xdy/dx+y=logx. Here I.F=x and solution is xy=x(logx-1)+C  Example 2:dy/dx+2xy=e-x.Here I.F=ex and solution is yex=x+C L1-3:
  • 21. Lecture-12 BERNOULLI’S LINEAR EQUATION  An equation of the form y′ + Py = Qyn is called a Bernoulli’s linear differential equation. This differential equation can be solved by reducing it to the Leibnitz linear differential equation. For this dividing above equation by yn  Example 1: xdy/dx+y=x2y6.Here I.F=1/x5 and solution is 1/(xy)5=5x3/2+Cx5  Example 2: dy/dx+y/x=y2xsinx. Here I.F=1/x and solution is 1/xy=cosx+C
  • 22. Lecture-13 ORTHOGONAL TRAJECTORIES  If two families of curves are such that each member of family cuts each member of the other family at right angles, then the members of one family are known as the orthogonal trajectories of the other family.  Example 1: The orthogonal trajectory of the family of parabolas through origin and foci on y-axis is x2/2c+y2/c=1  Example 2: The orthogonal trajectory of rectangular hyperbolas is xy=c2
  • 23. PROCEDURE TO FIND ORTHOGONAL TRAJECTORIES Suppose f (x ,y ,c) = 0 is the given family of curves, where c is the constant. STEP 1: Form the differential equation by eliminating the arbitrary constant. STEP 2: Replace y′ by -1/y′ in the above equation. STEP 3: Solve the above differential equation.
  • 24. Lecture-14 NEWTON’S LAW OF COOLING  The rate at which the temperature of a hot body decreases is proportional to the difference between the temperature of the body and the temperature of the surrounding air. θ′ ∞ (θ – θ0)  Example: If a body is originally at 80 oC and cools down to 60oC in 20 min.If the temperature of the air is at 40oC then the temperture of the body after 40 min is 50 oC
  • 25. Lecture-15 LAW OF NATURAL GROWTH  When a natural substance increases in Magnitude as a result of some action which affects all parts equally, the rate of increase depends on the amount of the substance present. N′ = k N  Example: If the number N of bacteria in a culture grew at a rate proportional to N. The value of N was initially 100 and increased to 332 in 1 hour. Then the value of N after one and half hour is 605
  • 26. LAW OF NATURAL DECAY  The rate of decrease or decay of any substance is proportion to N the number present at time. N′ = -k N  Example: A radioactive substance disintegrates at a rate proportional to its mass. When mass is 10gms, the rate of disintegration is 0.051gms per day. The mass is reduced to 10 to 5gms in 136 days.