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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
TOPICS
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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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SEPARATIONOFVARIABLES
BACK
1. dr/dt = - 4rt ; when t = 0, r = ro
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SEPARATIONOFVARIABLES
2. 2xyy = 1 + y2; when x = 2 , y = 3
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SEPARATIONOFVARIABLES
3. xyy = 1 + y2 ; when x = 2, y =3
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SEPARATIONOFVARIABLES
4. 2ydx = 3xdy; when x = 2, y = 1
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SEPARATIONOFVARIABLES
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
BACK
1. 3(3x2+ y2)dx2xydy = 0c
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HOMOGENOUSEQUATION
BACK
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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LINEAREQUATIONOFORDER1
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LINEAREQUATIONOFORDER1
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ELEMENTARYAPPLICATION
Isogonal and Orthogonal Trajectories
Newtons Law of Cooling
Exponential Growth or Decay
Mixture Problem
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ELEMENTARYAPPLICATION
BACK
The 2ndterm is first reduced to a
proper fraction by the method of
partial fraction. Thus
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ELEMENTARYAPPLICATION
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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EXPONENTIALGROWTHORDECAY
BACK
Solution to No. 1:
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EXPONENTIALGROWTHORDECAY
BACK
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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HOMOGENOUSL.E W/ CONSTANT
COEFFICIENTS(YC).Case 2: roots are real, repeated and distinct
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HOMOGENOUSL.E W/ CONSTANT
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
UNDETERMINEDCOEFFICIENTS
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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
UNDETERMINEDCOEFFICIENTS
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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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