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MATHEMATICS-IMATHEMATICS-I

CONTENTSCONTENTS Ordinary Differential Equations of First Order and First DegreeOrdinary Differential Equations of First Order and First Degree Linear Differential Equations of Second and Higher OrderLinear Differential Equations of Second and Higher Order Mean Value TheoremsMean Value Theorems Functions of Several VariablesFunctions of Several Variables Curvature, Evolutes and EnvelopesCurvature, Evolutes and Envelopes Curve TracingCurve Tracing Applications of IntegrationApplications of Integration Multiple IntegralsMultiple Integrals Series and SequencesSeries and Sequences Vector Differentiation and Vector OperatorsVector Differentiation and Vector Operators Vector IntegrationVector Integration Vector Integral TheoremsVector Integral Theorems Laplace transformsLaplace transforms

TEXT BOOKSTEXT BOOKS A text book of Engineering Mathematics, Vol-I A text book of Engineering Mathematics, Vol-I

T.K.V.Iyengar, B.Krishna Gandhi and Others, T.K.V.Iyengar, B.Krishna Gandhi and Others, S.Chand & CompanyS.Chand & Company

A text book of Engineering Mathematics, A text book of Engineering Mathematics, C.Sankaraiah, V.G.S.Book LinksC.Sankaraiah, V.G.S.Book Links

A text book of Engineering Mathematics, Shahnaz A A text book of Engineering Mathematics, Shahnaz A Bathul, Right PublishersBathul, Right Publishers

A text book of Engineering Mathematics, A text book of Engineering Mathematics, P.Nageshwara Rao, Y.Narasimhulu & N.Prabhakar P.Nageshwara Rao, Y.Narasimhulu & N.Prabhakar Rao, Deepthi PublicationsRao, Deepthi Publications

REFERENCESREFERENCES

A text book of Engineering Mathematics, A text book of Engineering Mathematics, B.V.Raman, Tata Mc Graw HillB.V.Raman, Tata Mc Graw Hill

Advanced Engineering Mathematics, Irvin Advanced Engineering Mathematics, Irvin Kreyszig, Wiley India Pvt. Ltd.Kreyszig, Wiley India Pvt. Ltd.

A text Book of Engineering Mathematics, A text Book of Engineering Mathematics, Thamson Book collectionThamson Book collection

UNIT-IUNIT-I

ORDINARY DIFFERENTIAL ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER EQUATIONS OF FIRST ORDER

AND FIRST DEGREEAND FIRST DEGREE

UNIT HEADERUNIT HEADER

Name of the Course:B.TechName of the Course:B.TechCode No:07A1BS02Code No:07A1BS02Year/Branch:I Year Year/Branch:I Year

CSE,IT,ECE,EEE,ME,CIVIL,AEROCSE,IT,ECE,EEE,ME,CIVIL,AEROUnit No: I Unit No: I

No.of slides:26No.of slides:26

S.No.S.No. ModuleModule LectureLectureNo. No.

PPT Slide No.PPT Slide No.

11 Introduction,Exact Introduction,Exact differential equationsdifferential equations

L1-10L1-10 8-198-19

22 Linear and Bernoulli’s Linear and Bernoulli’s equations,Orthogonal equations,Orthogonal trajectoriestrajectories

L11-13L11-13 20-2320-23

33 Newton’s law of Newton’s law of cooling and decaycooling and decay

L14-15L14-15 24-2624-26

UNIT INDEXUNIT INDEXUNIT-I UNIT-I

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Lecture-1Lecture-1INTRODUCTIONINTRODUCTION

An equation involving a dependent variable An equation involving a dependent variable and its derivatives with respect to one or more and its derivatives with respect to one or more independent variables is called a Differential independent variables is called a Differential Equation. Equation.

ExampleExample 1: y 1: y″ + 2y = 0″ + 2y = 0 ExampleExample 2: y 2: y2 2 – 2y– 2y11+y=23+y=23 ExampleExample 3: d 3: d22y/dxy/dx22 + dy/dx – y=1 + dy/dx – y=1

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TYPES OF A DIFFERENTIAL TYPES OF A DIFFERENTIAL EQUATIONEQUATION

ORDINARY DIFFERENTIAL EQUATIONORDINARY DIFFERENTIAL EQUATION: A : A differential equation is said to be ordinary, if the differential equation is said to be ordinary, if the derivatives in the equation are ordinary derivatives.derivatives in the equation are ordinary derivatives.

ExampleExample:d:d22y/dxy/dx22-dy/dx+y=1-dy/dx+y=1 PARTIAL DIFFERENTIAL EQUATIONPARTIAL DIFFERENTIAL EQUATION: A : A

differential equation is said to be partial if the differential equation is said to be partial if the derivatives in the equation have reference to two or derivatives in the equation have reference to two or more independent variables.more independent variables.

ExampleExample::∂∂44y/∂xy/∂x44+∂y/∂x+y=1+∂y/∂x+y=1

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Lecture-2Lecture-2DEFINITIONSDEFINITIONS

ORDER OF A DIFFERENTIAL EQUATIONORDER OF A DIFFERENTIAL EQUATION: A : A differential equation is said to be of order differential equation is said to be of order nn, if the , if the nnthth derivative is the highest derivative in that equation.derivative is the highest derivative in that equation.

ExampleExample: Order of d: Order of d22y/dxy/dx22+dy/dx+y=2 is 2+dy/dx+y=2 is 2 DEGREE OF A DIFFERENTIAL EQUATIONDEGREE OF A DIFFERENTIAL EQUATION: If : If

the given differential equation is a polynomial in the given differential equation is a polynomial in yy((nn)), , then the highest degree of then the highest degree of yy((nn)) is defined as the degree is defined as the degree of the differential equation.of the differential equation.

ExampleExample: Degree of (dy/dx): Degree of (dy/dx)44+y=0 is 4+y=0 is 4

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SOLUTION OF A DIFFERENTIAL SOLUTION OF A DIFFERENTIAL EQUATIONEQUATION

SOLUTIONSOLUTION: Any relation connecting the variables of an : Any relation connecting the variables of an equation and not involving their derivatives, which satisfies equation and not involving their derivatives, which satisfies the given differential equation is called a solution.the given differential equation is called a solution.

GENERAL SOLUTIONGENERAL SOLUTION: A solution of a differential equation : A solution of a differential equation in which the number of arbitrary constant is equal to the order in which the number of arbitrary constant is equal to the order of the equation is called a general or complete solution or of the equation is called a general or complete solution or complete primitive of the equation.complete primitive of the equation.

ExampleExample: : yy = = AxAx + + BB PARTICULAR SOLUTIONPARTICULAR SOLUTION: The solution obtained by giving : The solution obtained by giving

particular values to the arbitrary constants of the general particular values to the arbitrary constants of the general solution, is called a particular solution of the equation.solution, is called a particular solution of the equation.

ExampleExample: : yy = 3 = 3xx + 5 + 5

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Lecture-3Lecture-3EXACT DIFFERENTIAL EQUATIONEXACT DIFFERENTIAL EQUATION

Let M(x,y)dx + N(x,y)dy = 0 be a first order Let M(x,y)dx + N(x,y)dy = 0 be a first order and first degree differential equation where M and first degree differential equation where M and N are real valued functions for some x, y. and N are real valued functions for some x, y. Then the equation Mdx + Ndy = 0 is said to be Then the equation Mdx + Ndy = 0 is said to be an exact differential equation if an exact differential equation if ∂M/∂y=∂N/∂x∂M/∂y=∂N/∂x

ExampleExample: : (2y sinx+cosy)dx=(x siny+2cosx+tany)dy(2y sinx+cosy)dx=(x siny+2cosx+tany)dy

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Lecture-4Lecture-4Working rule to solve an exact equationWorking rule to solve an exact equation

STEP 1: Check the condition for exactness, STEP 1: Check the condition for exactness,

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

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Lecture-5Lecture-5INTEGRATING FACTORINTEGRATING FACTOR

Let Mdx + Ndy = 0 be not an exact differential Let Mdx + Ndy = 0 be not an exact differential equation. Then Mdx + Ndy = 0 can be made equation. Then Mdx + Ndy = 0 can be made exact by multiplying it with a suitable function exact by multiplying it with a suitable function uu is called an integrating factor. is called an integrating factor.

ExampleExample 1:ydx-xdy=0 is not an exact equation. 1:ydx-xdy=0 is not an exact equation. Here 1/xHere 1/x22 is an integrating factor is an integrating factor

ExampleExample 2:y(x 2:y(x22yy22+2)dx+x(2-2x+2)dx+x(2-2x22yy22)dy=0 is not )dy=0 is not an exact equation. Here 1/(3xan exact equation. Here 1/(3x33yy33) is an ) is an integrating factorintegrating factor

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Lecture-6Lecture-6METHODS TO FIND INTEGRATING METHODS TO FIND INTEGRATING

FACTORSFACTORS METHOD 1: With some experience METHOD 1: With some experience

integrating factors can be found by inspection. integrating factors can be found by inspection. That is, we have to use some known That is, we have to use some known differential formulae.differential formulae.

ExampleExample 1:d(xy)=xdy+ydx 1:d(xy)=xdy+ydx ExampleExample 2:d(x/y)=(ydx-xdy)/y 2:d(x/y)=(ydx-xdy)/y22

ExampleExample 3:d[log(x 3:d[log(x22+y+y22)]=2(xdx+ydy)/(x)]=2(xdx+ydy)/(x22+y+y22))

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FACTORSFACTORS METHOD 2: If Mdx + Ndy = 0 is a non-exact METHOD 2: If Mdx + Ndy = 0 is a non-exact

but homogeneous differential equation and but homogeneous differential equation and Mx + Ny Mx + Ny ≠ 0 then 1/(Mx + Ny) is an ≠ 0 then 1/(Mx + Ny) is an integrating factor of Mdx + Ndy = 0.integrating factor of Mdx + Ndy = 0.

ExampleExample 1:x 1:x22ydx-(xydx-(x33+y+y33)dy=0 is a non-exact )dy=0 is a non-exact homogeneous equation. Here I.F.=-1/yhomogeneous equation. Here I.F.=-1/y44

ExampleExample 2:y 2:y22dx+(xdx+(x22-xy-y-xy-y22)dy=0 is a non-exact )dy=0 is a non-exact homogeneous equation. Here I.F.=1/(xhomogeneous equation. Here I.F.=1/(x22y-yy-y33))

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FACTORSFACTORS METHOD 3: If the equation Mdx + Ndy = 0 is of the METHOD 3: If the equation Mdx + Ndy = 0 is of the

form y.f(xy) dx + x.g(xy) dy = 0 and Mx – Ny form y.f(xy) dx + x.g(xy) dy = 0 and Mx – Ny ≠ 0 ≠ 0 then 1/(Mx – Ny) is an integrating factor of Mdx + then 1/(Mx – Ny) is an integrating factor of Mdx + Ndy = 0.Ndy = 0.

ExampleExample 1:y(x 1:y(x22yy22+2)dx+x(2-2x+2)dx+x(2-2x22yy22)dy=0 is non-exact )dy=0 is non-exact and in the above form. Here I.F=1/(3xand in the above form. Here I.F=1/(3x33yy33))

ExampleExample 2:(xysinxy+cosxy)ydx+(xysinxy- 2:(xysinxy+cosxy)ydx+(xysinxy-cosxy)xdy=0 is non-exact and in the above form. cosxy)xdy=0 is non-exact and in the above form. Here I.F=1/(2xycosxy)Here I.F=1/(2xycosxy)

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Lecture-9Lecture-9METHODSMETHODS TO FIND INTEGRATING TO FIND INTEGRATING

FACTORSFACTORS METHOD 4: If there exists a continuous single METHOD 4: If there exists a continuous single

variable functionvariable function f(x f(x) such that ) such that ∂M/∂y-∂N/∂x=Nf(x) ∂M/∂y-∂N/∂x=Nf(x) then ethen e∫f(x)dx∫f(x)dx is an integrating is an integrating factor of factor of Mdx + NdyMdx + Ndy = 0 = 0

ExampleExample 1:2xydy-(x 1:2xydy-(x22+y+y22+1)dx=0 is non-exact and+1)dx=0 is non-exact and ∂ ∂M/∂y - ∂N/∂x=N(-2/x). Here I.F=1/xM/∂y - ∂N/∂x=N(-2/x). Here I.F=1/x22

ExampleExample 2:(3xy-2ay 2:(3xy-2ay22)dx+(x)dx+(x22-2axy)=0 is non-exact -2axy)=0 is non-exact and ∂M/∂y - ∂N/∂x=N(1/x). Here I.F=xand ∂M/∂y - ∂N/∂x=N(1/x). Here I.F=x

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FACTORSFACTORS METHOD 5: If there exists a continuous single METHOD 5: If there exists a continuous single

variable functionvariable function f(y f(y) such that ) such that ∂∂N/∂x - ∂M/∂y=Mg(y)N/∂x - ∂M/∂y=Mg(y) then e then e∫g(y)dy∫g(y)dy is an integrating is an integrating

factor of factor of Mdx + NdyMdx + Ndy = 0 = 0 ExampleExample 1:(xy 1:(xy33+y)dx+2(x+y)dx+2(x22yy22+x+y+x+y44)dy=0 is a non-)dy=0 is a non-

exact equation and exact equation and ∂N/∂x - ∂M/∂y=M(1/y). Here ∂N/∂x - ∂M/∂y=M(1/y). Here I.F=yI.F=y

ExampleExample 2:(y 2:(y44+2y)dx+(xy+2y)dx+(xy33+2y+2y44-4x)dy=0 is a non--4x)dy=0 is a non-exact equation and ∂N/∂x - ∂M/∂y=M(-3/y).Here exact equation and ∂N/∂x - ∂M/∂y=M(-3/y).Here I.F=1/yI.F=1/y33

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Lecture-11Lecture-11LEIBNITZ LINEAR EQUATIONLEIBNITZ LINEAR EQUATION

An equation of the form yAn equation of the form y′ + Py = Q is called a ′ + Py = Q is called a linear differential equation.linear differential equation.

Integrating Factor(I.F.)=eIntegrating Factor(I.F.)=e∫pdx∫pdx

Solution is y(I.F) = ∫Q(I.F)dx+CSolution is y(I.F) = ∫Q(I.F)dx+C ExampleExample 1:xdy/dx+y=logx. Here I.F=x and solution 1:xdy/dx+y=logx. Here I.F=x and solution

is xy=x(logx-1)+Cis xy=x(logx-1)+C ExampleExample 2:dy/dx+2xy=e 2:dy/dx+2xy=e-x-x.Here I.F=e.Here I.F=exx and solution and solution

is yeis yexx=x+C=x+C

Lecture-12Lecture-12BERNOULLI’S LINEAR EQUATIONBERNOULLI’S LINEAR EQUATION

An equation of the form yAn equation of the form y′ + Py = Qy′ + Py = Qynn is called a is called a Bernoulli’s linear differential equation. This Bernoulli’s linear differential equation. This differential equation can be solved by reducing it to differential equation can be solved by reducing it to the Leibnitz linear differential equation. For this the Leibnitz linear differential equation. For this dividing above equation by ydividing above equation by ynn

ExampleExample 1: xdy/dx+y=x 1: xdy/dx+y=x22yy66.Here I.F=1/x.Here I.F=1/x55 and solution and solution is 1/(xy)is 1/(xy)55=5x=5x33/2+Cx/2+Cx55

ExampleExample 2: dy/dx+y/x=y 2: dy/dx+y/x=y22xsinx. Here I.F=1/x and xsinx. Here I.F=1/x and solution is 1/xy=cosx+Csolution is 1/xy=cosx+C

Lecture-13Lecture-13ORTHOGONAL TRAJECTORIESORTHOGONAL TRAJECTORIES

If two families of curves are such that each member If two families of curves are such that each member of family cuts each member of the other family at of family cuts each member of the other family at right angles, then the members of one family are right angles, then the members of one family are known as the orthogonal trajectories of the other known as the orthogonal trajectories of the other family.family.

ExampleExample 1: The orthogonal trajectory of the family of 1: The orthogonal trajectory of the family of parabolas through origin and foci on y-axis is parabolas through origin and foci on y-axis is xx22/2c+y/2c+y22/c=1/c=1

ExampleExample 2: The orthogonal trajectory of rectangular 2: The orthogonal trajectory of rectangular hyperbolas is xy=chyperbolas is xy=c22

PROCEDURE TO FIND PROCEDURE TO FIND ORTHOGONAL TRAJECTORIESORTHOGONAL TRAJECTORIES

Suppose Suppose f (x ,y ,cf (x ,y ,c) = 0 is the given family of ) = 0 is the given family of curves, where c is the constant.curves, where c is the constant.STEP 1: Form the differential equation by STEP 1: Form the differential equation by eliminating the arbitrary constant.eliminating the arbitrary constant.STEP 2: Replace ySTEP 2: Replace y′ by -1/y′ in the above ′ by -1/y′ in the above equation.equation.STEP 3: Solve the above differential equation.STEP 3: Solve the above differential equation.

Lecture-14Lecture-14NEWTON’S LAW OF COOLINGNEWTON’S LAW OF COOLING

The rate at which the temperature of a hot body The rate at which the temperature of a hot body decreases is proportional to the difference decreases is proportional to the difference between the temperature of the body and the between the temperature of the body and the temperature of the surrounding air.temperature of the surrounding air.

θ′θ′ ∞∞ ( (θθ – – θθ00)) Example: Example: If a body is originally at 80If a body is originally at 80ooC and C and

cools down to 60cools down to 60ooC in 20 min.If the temperature C in 20 min.If the temperature of the air is at 40of the air is at 40ooC then the temperture of the C then the temperture of the body after 40 min is 50body after 40 min is 50ooCC

Lecture-15Lecture-15LAW OF NATURAL GROWTHLAW OF NATURAL GROWTH

When a natural substance increases in Magnitude as a When a natural substance increases in Magnitude as a result of some action which affects all parts equally, result of some action which affects all parts equally, the rate of increase depends on the amount of the the rate of increase depends on the amount of the substance present.substance present.

NN′ = k N′ = k N Example:Example: If the number N of bacteria in a culture If the number N of bacteria in a culture

grew at a rate proportional to N. The value of N was grew at a rate proportional to N. The value of N was initially 100 and increased to 332 in 1 hour. Then the initially 100 and increased to 332 in 1 hour. Then the value of N after one and half hour is 605value of N after one and half hour is 605

LAW OF NATURAL DECAYLAW OF NATURAL DECAY

The rate of decrease or decay of any substance The rate of decrease or decay of any substance is proportion to N the number present at time.is proportion to N the number present at time.

NN′ = -k N′ = -k N ExampleExample:: A radioactive substance disintegrates A radioactive substance disintegrates

at a rate proportional to its mass. When mass is at a rate proportional to its mass. When mass is 10gms, the rate of disintegration is 0.051gms 10gms, the rate of disintegration is 0.051gms per day. The mass is reduced to 10 to 5gms in per day. The mass is reduced to 10 to 5gms in 136 days.136 days.

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