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ELEMENTARY DIFFERENTIAL

EQUATIONMATH 2103

Engr. Ernesto P. Pucyutan


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TOPICSSecond Quarter

1. Linear Equation of Higher Order

2. Homogenous Linear Equation with Constant

Coefficients3. Non-Homogenous Linear Equation with Constant

Coefficients

4. Laplace Transform


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FIRST QUARTER


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INTRODUCTIONThe construction of mathematical models to appropriate real-

world problems has been one of the most important aspects of

the theoretical development of each of the branches of

science. It is often the case that these mathematical models

involve an equation in which a function and its derivatives playimportant roles. Such equations are called differential

equations.

The differential equation is one which contains within at leastone derivative. Sometimes, for analytical convenience, the

differential equation is written in terms of differentials. It may

also be given either in explicit or implicit form.

EXAMPLES NEXT


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EXAMPLESOFDIFFERENTIALEQUATION

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INTRODUCTION

When an equation involves one or more derivatives with

respect to a particular variable, that variable is called

independent variable. A variable is called dependent if a

derivative of that variable occurs.

In the equation iis the dependent variable, tthe

independent variable and l, r, c, e and are calledparameters.

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CLASSIFICATIONS OF D.E

1. The order of a differential equation is the order of the

highest-ordered derivative appearing in the equation.

2. The degreeof a differential equation is the largest

power or exponent the highest-ordered derivative present

in the equation.

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CLASSIFICATIONS OF D.E

3. The type of a differential equation may be ordinary or

partial as to the type of derivatives or differentials

appearing in the equation , that is, if it contains

ordinary derivatives, it is ordinary differential equation

and if the derivatives are partial, the equation is a partialdifferential equation.

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SOLUTIONS OF D.E

1. General Solution- involving 1 or more arbitrary constant

Ex: y(t)=C1ekt

+ C2ekt

2. Particular Solution- no arbitrary constant

Ex: p= 3.9ekt

3. Complete Solution- combination of two solutions

(particular and a complimentary solution)

Y=Yp+Yc

4. Computer Solution- using computer software

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ELIMINATION OF ARBITRARY

CONSTANTS

Methods for the elimination of arbitrary constants vary

with the way in which the constants enter the given

relation. A method that is efficient for one problem may

be poor for another. One fact persists throughout. Because

each differentiation yields a new relation, the number ofderivatives that need be used is the same as the number

of arbitrary constants to be eliminated. We shall in each

case determine the differential equation that is

(a) Of order equal to the number of arbitrary constants

given relation.

b) Consistent with that relation.

(c) Free from arbitrary constants.

EXAMPLE


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EXAMPLE

Example 1.

y = C1e-2x

+ C2e3x (1)

Y = -2C1e-2x

+3 C2e3x (2)

Y = 4C1e-2x+9 C2e3x (3)

Elimination of equations 1 and 2 yields to

y+2y= 15 C2e3x;

The elimination of C1from equation 1 and 2 yieldsto y + 2y = 5 C2e

3x

Hence, y + 2y = 3(y + 2y) or y cy 6y = 0

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EXAMPLE

Example 2: Find the solution of xsiny + x2y = c

Solution:

xcosy dy + siny dx + x2dy + y2xdx = 0

(siny + 2xy)dx + (xcosy + x2)dy = 0

Example 3: Find the solutionof 3x2xy2= c

Solution:

6xdx(x2ydy + y2dx)=0

6xdx2xydyy2dx = 0

(6xy2)dx2xydy = 0

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FAMILIES OF CURVES

Obtain the differential equation of the family plane curves

described

1.

Straight lines through the origin. Answer2. Straight lines through the fixed (h,k); h and k not to be

eliminated. Answer

3. Straight lines with slope and x-intercept equal. Answer

4. Straight lines with slope and y-intercept equal. Answer5. Straight lines with the algebraic sum of the intercept

fixed as k. Answer

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FAMILIES OF CURVES

1. General equation:

y = mx

m = slope

y = m or m = dy/dxSubstitute m,

Y = (dy/dx)x

ydx = xdy

ydx-xdy = 0

2. General equation:

(yk) = m ( xh )

dy = mdx

m = dy/dxSustitute

(yk) = dy/dx(x-h)

(yk)dx = (x - h)dy

(yk) dx( xh) dy =0

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FAMILIES OF CURVES3. General Equation:

y = m(x - a)

m = slope;

a = x-intercept

y = m(xm)

dy = mdx

m=dy/dx =y

Substitute,

y=y (x y)

y= xy - (y)2

4. General Equation:

y = mx + b

m = slope

b =y-intercept;b = m

y = mx + m

dy = mdx

m = dy/dx

Sustitute,

y = (dy/dx)x + dy/dx

ydx = xdy + dy

ydx(x+1)dy = 0

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FAMILIES OF CURVES

5. For xintercept:

y = m(xa)

y = m

y = y (x a)y = xy ay

a = (xy y)/y

For y- intercept:

y = mx + b

y = m

Y = yx+ b

b = yxy

But, k = a + bK = (xy y)/y + (yxy)

Multiply by y,

ky = xy y + y (y =xy)

ky= (1 y)(xy y)

ky (1y)(xy y) = 0

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EQUATIONS OF ORDER 1

General Equation :

M(x,y)dx + N(x,y)dy = 0

It can be solved by:

1. Separation of Variables

2. Homogenous Equations

3. Linear Coefficients of two variables

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SEPARATIONOFVARIABLES

Solve the following:

1. dr/dt = - 4rt ; when t = 0, r = ro Answer

2. 2xyy = 1 + y2; when x = 2 , y = 3 Answer

3. xyy = 1 + y2

; when x = 2, y =3 Answer4. 2ydx = 3xdy; when x = 2, y = 1 Answer

5. 2ydx =3xdy; when x = -2, y= 1 Answer

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1. dr/dt = - 4rt ; when t = 0, r = ro


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2. 2xyy = 1 + y2; when x = 2 , y = 3

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3. xyy = 1 + y2 ; when x = 2, y =3

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4. 2ydx = 3xdy; when x = 2, y = 1

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5. 2ydx =3xdy; when x = -2, y= 1

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HOMOGENOUSEQUATION

When the equation is Mdx + Ndy = 0Where: M & N are homogenous functions of the same

degree in x and y.

If M is simpler than N use x = uy otherwise use y = vx

HOMOGENOUS FUNCTION

Example:

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HOMOGENOUSEQUATION

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1. 3(3x2+ y2)dx2xydy = 0c


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HOMOGENOUSEQUATION

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3. 2(2x2+ y2)dxxydy = 0 )


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HOMOGENOUSFUNCTIONDetermine in each exercise whether or not the function is

homogenous; if it is homogenous, state the degree offunction.

1. 4x23xy + y2 Answer: Homogenous 2nddegree

Solution

2. x3xy +y3 Answer: Not homogenousSolution

3. 2y + (x2 + y2)1/2 Answer: Homogenous 1stdegree

Solution

4. (xy )1/2 Answer: Homogenous degree

Solution5. ex Answer: Not homogenous

Solution

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HOMOGENOUSFUNCTION

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EXACTEQUATION

The equation M(x,y)dx + N(x,y)dy = 0 is an exact equation if:

Then,

TOPICS EXAMPLES


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EXACTEQUATIONTest for the exactness and find the complete solution of

the following.

1. (2xy3x2)dx + (x2+ 2y)dy = 0

Answer: x2y + y2x3= c Solution

2. (cos2y3x2y2)dx + (cos2y2xsin2y2x3y)dy = 0

Answer: xcos2yx3y3+ sin2y = c Solution

3. (yexy2y3)dx + (xexy6xy22y) dy = 0

Answer: exy2xy3y2 + 3 =0 Solution

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EXACTEQUATION

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EXACTEQUATION

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LINEAREQUATIONOFORDER1

TOPICS

GENERAL EQUATION:

EXAMPLES


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LINEAREQUATIONOFORDER1Find the general solution of the following:

1. (x4+ 2y) dxxdy = 0 Solution

2. y = cscxycotx Solution

3. (3x1 ) y = 6y10(3x1)1/3 Solution

Answers:

1. 2y = x4+ C1x2 (C1= 2C)

2. y sin x = x + C

3. y = 2 (3x1)1/3+ C (3x1)2

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ELEMENTARYAPPLICATION

Isogonal and Orthogonal Trajectories

Newtons Law of Cooling

Exponential Growth or Decay

Mixture Problem

TOPICS


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BACK

The 2ndterm is first reduced to a

proper fraction by the method of

partial fraction. Thus


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BACK


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NEWTONSLAWOFCOOLINGThe rate of change in the temperature of a body is directly

proportional to the difference in the temperature

between the body and the environment.

BACK

Example: The thermometer reading 18 F brought into

a room where the temperature is 70 F ; 1 min later the

thermometer reading is 31 F . Determine thetemperature reading as a function of time and, in

particular, find the temperature reading 5 min after the

thermometer is brought into the room. Answer


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EXPONENTIALGROWTHORDECAY

Problems:

1. Radium decomposes at a rate proportional to the

amount present. In 100 years, 100 mg of radium

decompose to 96 mg. How many mg will be left after

another 100 years? What is the half-lifeof Radium?

Answer

2. The population of a certain community follows the law

of exponential change. If the present population of the

community is 144,000 and 10 years ago was 100000when will the population double? In 10 years what will

be the population of the community?Answer

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Solution to No. 1:


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Solution to No. 2:


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MIXTUREPROBLEMExample:

1. A tank initially contains 200 L of fresh water. Brine

containing 2.5 N/L of dissolved salt runs into the tank

at the rate of 8 L/min and the mixture kept uniform by

stirring runs out at the same rate. How long will it takefor the quantity of salt in the tank to be 180 N? In 10

min, determine the concentration of salt in the

mixture.

Answer: t = 11.2 min c = 0.825 N/ LSolution

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The Wronskian W: A Functional


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The Wronskian W: A Functional

Determinant

Ex. Show that the functions 1, x and x2

arelinearly independent in all intervals.

y1= 1 y1= 0 y1=o

y2= x y2=1 y2=0y3= x2 y3=2x y3=2

= 2 therefore 1, x and x^2 arelenearly dependent.

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HOMOGENOUSL.E W/ CONSTANT

COEFFICIENTS(YC).Case 1: r1, r2 and rn of the auxiliary equation are real and

distinct

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COEFFICIENTS(YC).Case 2: roots are real, repeated and distinct

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COEFFICIENTS(YC).Case 4: has repeated conjugate complex roots

A=0 p=2 b=3

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NON-HOMOGENOUSL.E W/ CONSTANT

COEFFICIENTS(YC+ YP).1. Reduction of Order

a. The roots of the auxiliary equation are all equal

b. The order of the equation is not too largec. The equation is factorable

2. Undetermined Coefficients

3. Variation of Parameters

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REDUCTIONOFORDER

Find the general solution of the following:

1. ( D24 )y = 4x3ex Solution

2. ( D3 2D2+ D )y = x Solution

Answer:

1. y = c1e2x+ c2e

-2x+ exx

2. y = x2/2 + 2x + 3 + ex(+c1+ c2x) + c3

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REDUCTIONOFORDER2. ( D3 2D2+ D )y = x

(a) the auxiliary equationm32m2+ m= 0 or m( m1 )2= 0 hasthe roots r1= r2= 1and r3 = 0

(b) the factored form of the given

equation is D( D1 )(D1)y = x

(c) to get Yc consider

D( D1 )(D1)y = 0;

Yc = ex(+c1+ c2x) + c3

(d) for the particular integral Yp. Use the

method of reduction of order:

Let z = ( D1 )(D1)y and so Dz = xz = x2/2 substitue to the above equation

( D1 )(D1)y = x2/2

Let (D1) y = v (D1)v = x2

/2P = - 1 Q = x2/2 and = e-x

v= Qdx or v e-x = e-x x2/2

which by integration by parts yields to:

v = - x2/2 - x1

Subsitute back ; (D1) y = - x2/2 - x1

Whose solution is y = Yp = x2/2 + 2x + 3

y = Yp + Yc or

y = x

2

/2 + 2x + 3 + e

x

(+c1+ c2x) + c3

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Find the solution of the following:

1. (D24 )y = 4x3ex Solution

2. (D2+ 2D + 5)y = 3e-xsinx10 Solution

3. (D3

D ) y = 4e-x

+ 3e2x

Solution

Answer:

1. Y = -x + ex + C1e2x+ C2e-

2x

2. Y = e-x(C1cos2x + C2sin2x) +exsinx23. Y = 2xe-x+ + e2x + C1+ C2e

x+ C3e-x

UNDETERMINEDCOEFFICIENTS

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1. (D24 )y = 4x3ex

(a) Yc = C1e2x+ C2e-2x(b) For 4x: ( q = 0, an= 4 0, p = 1)

Yp1 = A xp+ Bxp-1 + . + Lx + M or

Yp1 = Ax + B

For3ex : ( q = 1, p = 0, p = 0)Yp2 = eqx(A xp+ Bxp-1 + . + Lx +M)xror

Yp2 = Aex

So Yp = Yp1 + Yp2

Yp = Ax + B + Ce

x

Yp=A + Cex

Yp= Cex

(c) substitute;

(D24 )Yp= 4x3ex

Cex4(Ax + B + Cex) = 4x3ex

Therefore; C = 1. B = 0 and A = -1Yp = Ax + B + Cex ;

Yp = -x + ex

(d) y = Yc + Yp

Y = -x + ex + C1e2x+ C

2e-2x


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2. (D2+ 2D + 5)y = 3e-xsinx10

(a) m2+ 2m + 5 = 0 has the complexroots r1= -1 + 2i r2= -12i

(b) Yc = e-x(C1cos2x + C2sin2x)

(c) For 3e-xsinx: ( q = -1, b = 1, p = 0)

Yp1 = Ae

-x

cosx + Be

-x

sinx

For10: ( q = 0, p = 0, an 0)

Yp2 = C

So Yp = Yp1 + Yp2Yp = Ae-xcosx + Be-xsinx + C

Yp=-(A+ B) e-xsinx(AB ) e-xcosx

Yp= -2Be-xcosx + 2Ae-xsinx

(c) substitute;

(D2+ 2D + 5)Yp = 3e-xsinx10

Therefore; C = -2. B = 1 and A = 0

Yp =e-xsinx2

(d) y = Yc + Yp

Y = e-x(C1cos2x + C2sin2x) +exsinx2


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LAPLACETRANSFORM

From Complex to algebraic

Developed by Pierre Simon de Laplace

Time domain to s domain

It is used in control system and signal analysis

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PROPERTIES OF LAPLACE

1. Constant multiple

2. Linearity

3. Change scale

4. Shifting

5. Unnamed

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LAPLACETRANSFORMTABLE

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INVERSE LAPLACE TRANSFORM


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INVERSELAPLACETRANSFORM

Example:

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