Differential Equations

ELEMENTARY DIFFERENTIAL EQUATIONMATH 2103

Engr. Ernesto P. Pucyutan

TOPICSFirst Quarter1. Introduction2. Definitions, Classifications and Solution of

D.E and Elimination of Arbitrary Constant3. Families of Curves 4. Equations of order 15. Homogenous Functions6. Exact Equations7. Linear Equation of Order 18. Elementary applications

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TOPICSSecond Quarter1. Linear Equation of Higher Order2. Homogenous Linear Equation with Constant

Coefficients3. Non-Homogenous Linear Equation with Con

stant Coefficients4. Laplace Transform

FIRST QUARTER

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 play important roles. Such equations are called differential equations.

The differential equation is one which contains within at least one 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

EXAMPLES OF DIFFERENTIAL EQUATION

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INTRODUCTIONWhen 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 i is the dependent variable, t the independent variable and l, r, c, e and ω are called parameters.

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CLASSIFICATIONS OF D.E1. The order of a differential equation is the order of the

highest-ordered derivative appearing in the equation.

2. The degree of a differential equation is the largest power or exponent the highest-ordered derivative present in the equation.

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CLASSIFICATIONS OF D.E3. 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 partial differential equation.

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SOLUTIONS OF D.E1. General Solution- involving 1 or more arbitrary constant

Ex: y(t)=C1ekt + C2 ekt

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 CONSTANTSMethods 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 of derivatives 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

EXAMPLEExample 1.

y = C1e-2x + C2 e3x (1)

Y’ = -2C1e-2x +3 C2 e3x (2)

Y” = 4C1e-2x +9 C2 e3x (3)

Elimination of equations 1 and 2 yields to y”+2y’= 15 C2 e3x;

The elimination of C1 from equation 1 and 2 yields to y’ + 2y = 5 C2 e3x

Hence, y” + 2y’ = 3(y’ + 2y) or y” c y’ – 6y = 0NEXT

EXAMPLEExample 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 3x2 – xy2 = cSolution:6xdx – (x2ydy + y2dx)=06xdx – 2xydy – y2dx = 0(6x – y2)dx – 2xydy = 0

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FAMILIES OF CURVESObtain the differential equation of the family plane curves

described

1. Straight lines through the origin. Answer 2. Straight lines through the fixed (h,k); h and k not to be

eliminated. Answer3. Straight lines with slope and x-intercept equal. Answer4. 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 CURVES1. General equation:

y = mxm = slopey’ = m or m = dy/dx

Substitute m,Y = (dy/dx)xydx = xdyydx-xdy = 0

2. General equation: (y – k) = m ( x – h )dy = mdxm = dy/dx

Sustitute (y – k) = dy/dx(x-

h) (y – k)dx = (x -

h)dy(y – k) dx – ( x – h) dy =0

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FAMILIES OF CURVES3. General Equation:

y = m(x - a) m = slope;a = x-intercepty = m(x – m)dy = mdxm=dy/dx =y’

Substitute,y=y’ (x – y’) y= xy’ - (y’)2

4. General Equation:y = mx + bm = slopeb =y-intercept;b = my = mx + mdy = mdxm = dy/dx

Sustitute,y = (dy/dx)x + dy/dxydx = xdy + dyydx – (x+1)dy = 0

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FAMILIES OF CURVES5. For x – intercept:

y = m(x – a)y’ = my = y’ (x – a)y = xy’ – ay’a = (xy’ – y)/y’

For y- intercept: y = mx + by’ = m

Y = y’x + bb = y –xy’But, k = a + bK = (xy’ – y)/y + (y – xy’)Multiply by y’,ky’ = xy’ –y + y’ (y =xy’)ky’= (1 – y’)(xy’ – y)ky’ – (1 – y’)(xy’ – y) = 0

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EQUATIONS OF ORDER 1General Equation :

M(x,y)dx + N(x,y)dy = 0It can be solved by:1. Separation of Variables

2. Homogenous Equations

3. Linear Coefficients of two variables

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SEPARATION OF VARIABLESSolve the following:

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

2. 2xyy’ = 1 + y2; when x = 2 , y = 3 Answer3. xyy’ = 1 + y2 ; when x = 2, y =3 Answer4. 2ydx = 3xdy; when x = 2, y = 1 Answer5. 2ydx =3xdy; when x = -2, y= 1 Answer

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SEPARATION OF VARIABLES

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

SEPARATION OF VARIABLES

2. 2xyy’ = 1 + y2; when x = 2 , y = 3

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SEPARATION OF VARIABLES

3. xyy’ = 1 + y2 ; when x = 2, y =3

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SEPARATION OF VARIABLES

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

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SEPARATION OF VARIABLES

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

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HOMOGENOUS EQUATIONWhen 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 = vxHOMOGENOUS FUNCTION

Example:

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HOMOGENOUS EQUATION

Solve the following:1. 3(3x2 + y2)dx – 2xydy = 0 Solution2. (x – 2y)dx + (2x + y)dy = 0 Solution3. 2(2x2 + y2)dx – xydy = 0 Solution

Answers:1. x3 = c (9x2 + y2)2. ln (x2 + y2) + 4 Arctan y/x = c3. x4 = c2 (4x2 + y2)

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HOMOGENOUS EQUATION

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1. 3(3x2 + y2)dx – 2xydy = 0c

HOMOGENOUS EQUATION

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2. (x – 2y)dx + (2x + y)dy = 0

HOMOGENOUS EQUATION

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3. 2(2x2 + y2)dx – xydy = 0 )

HOMOGENOUS FUNCTIONDetermine in each exercise whether or not the function is

homogenous; if it is homogenous, state the degree of function.

1. 4x2 – 3xy + y2 Answer: Homogenous 2nd degreeSolution

2. x3 – xy +y3 Answer: Not homogenous Solution

3. 2y + (x2 + y2)1/2 Answer: Homogenous 1st degree Solution

4. (x – y )1/2 Answer: Homogenous ½ degreeSolution

5. ex Answer: Not homogenous Solution

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HOMOGENOUS FUNCTION

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1. f (x , y) = 4x2 – 3xy + y2

f (λx , λy) = (λx)2 – 3(λx)( λy) + (λy)2

= 4 λ2x2 – 3λ2xy + λ2y2

f (λx , λy) = λ2(4x2 – 3xy + y2)Homogenous, 2nd degree

2. f (x , y) = x3 – xy +y3 f (λx , λy) = (λx)3 – (λx)( λy) + (λy)3

= λ3x3 –λ2xy + λ3y3 f (λx , λy) = λ3x3 –λ2xy + λ3y3

NOT Homogenous

HOMOGENOUS FUNCTION

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

Then,

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EXACT EQUATIONTest for the exactness and find the complete solution of

the following.

1. (2xy – 3x2)dx + (x2 + 2y)dy = 0 Answer: x2y + y2 – x3 = c Solution

2. (cos2y – 3x2y2)dx + (cos2y – 2xsin2y – 2x3y)dy = 0Answer: xcos2y –x3y3 + ½ sin2y = c Solution

3. (yexy – 2y3)dx + (xexy – 6xy2 – 2y) dy = 0Answer: exy – 2xy3 – y2 + 3 =0 Solution

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EXACT EQUATION

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EXACT EQUATION

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EXACT EQUATION

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LINEAR EQUATION OF ORDER 1

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GENERAL EQUATION:

EXAMPLES

LINEAR EQUATION OF ORDER 1Find the general solution of the following:

1. (x4 + 2y) dx – xdy = 0 Solution2. y’ = cscx – ycotx Solution3. (3x – 1 ) y’ = 6y – 10(3x – 1)1/3 Solution

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

2. y sin x = x + C3. y = 2 (3x – 1)1/3 + C (3x – 1)2

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LINEAR EQUATION OF ORDER 1

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LINEAR EQUATION OF ORDER 1

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LINEAR EQUATION OF ORDER 1

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ELEMENTARY APPLICATION

Isogonal and Orthogonal Trajectories

Newton’s Law of Cooling

Exponential Growth or Decay

Mixture Problem

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ELEMENTARY APPLICATIONIsogonal Trajectories

1. Find the isogonal trajectories of the one parameter family of curves (x + c) y2 .= 1 if ϕ = Arctan 4. Answer

Orthogonal Trajectories

2. Find the orthogonal trajectories of the following:

a.) x – 4y = 0Answerb.) x2 + y2 = c2 C.) x2 – y2 = C1

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ELEMENTARY APPLICATION

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The 2nd term is first reduced to a proper fraction by the method of partial fraction. Thus

ELEMENTARY APPLICATION

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NEWTON’S LAW OF COOLINGThe rate of change in the temperature of a body is directly

proportional to the difference in the temperature between the body and the environment.

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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 the temperature reading as a function of time and, in particular, find the temperature reading 5 min after the thermometer is brought into the room. Answer

NEWTON’S LAW OF COOLING

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Example’s Solution:

When t=0 and u=18

C = -52

When t=1 min and u=31

When t= 5 min

EXPONENTIAL GROWTH OR DECAYProblems: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-life” of 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 100000 when will the population double? In 10 years what will be the population of the community? Answer

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EXPONENTIAL GROWTH OR DECAY

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Solution to No. 1:

EXPONENTIAL GROWTH OR DECAY

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Solution to No. 2:

MIXTURE PROBLEMExample:

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 take for 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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MIXTURE PROBLEM

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Solution to Example:

The Wronskian W: A Functional Determinant

Ex. Show that the functions 1, x and x2 are linearly independent in all intervals.

y1= 1 y’1= 0 y’’1=oy2= x y’2=1 y’’2=0y3= x2 y’3=2x y’’3=2

= 2 therefore 1, x and x^2 are lenearly dependent.

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The Differential operatorD(2x3) = 6x2

D2(2x3)= 12 x

LDE of Higher Order Operator form

Derivatives of Exponential Shift

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HOMOGENOUS L.E W/ CONSTANT COEFFICIENTS (YC).

Case 1: r1, r2 and rn of the auxiliary equation are real and distinct

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HOMOGENOUS L.E W/ CONSTANT COEFFICIENTS (YC).

Case 2: roots are real, repeated and distinct

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HOMOGENOUS L.E W/ CONSTANT COEFFICIENTS (YC).

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Case 3: has two conjugate complex roots

a=2 b=1

HOMOGENOUS L.E W/ CONSTANT COEFFICIENTS (YC).

Case 4: has repeated conjugate complex roots

A=0 p=2 b=3

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NON-HOMOGENOUS L.E W/ CONSTANT COEFFICIENTS (YC + YP).

1. Reduction of Ordera. The roots of the auxiliary equation are all

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

2. Undetermined Coefficients

3. Variation of Parameters

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REDUCTION OF ORDERFind the general solution of the following:

1. ( D2 – 4 )y = 4x – 3ex Solution2. ( D3 – 2D2 + D )y = x Solution

Answer:1. y = c1e2x + c2e-2x + ex – x

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

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REDUCTION OF ORDER1. ( D2 – 4 )y = 4x – 3ex

(a) the auxiliary equation m2 – 4= 0 or

( m – 2 )(m + 2) = 0 has the roots r1 = 2 and r2 = -2

(b) the factored form of the given equation is ( D – 2 )(D + 2)y = 4x – 3ex

(c) to get Yc consider ( D – 2 )(D + 2)y = 0; Yc = c1e2x + c2e-2x

(d) for the particular integral Yp. Use the method of reduction of order:

Let z = (D + 2)y and so (D + 2)z = 4x – 3ex

P = - 2 Q = 4x – 3ex and ϕ = e-2x

z ϕ = ∫ ϕ dx or z e-2x = ∫ e-2x (4x – 3ex)

which by integration by parts yields to:

z = -2x – 1 + 3 ex

Subsitute back to z = (D + 2)y give

dy/dx + 2y = -2x – 1 + 3 ex

P = 2 Q = -2x – 1 + 3 ex and ϕ = e2x

y ϕ = ∫ ϕ Qdx y e2x (-2x – 1 + 3 ex)

y = Yp = ex – x

y = Yp + Yc or

y = c1e2x + c2e-2x + ex – x

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REDUCTION OF ORDER2. ( D3 – 2D2 + D )y = x

(a) the auxiliary equation

m3 – 2m2 + m= 0 or m( m – 1 )2 = 0 has the roots r1 = r2= 1and r3 = 0

(b) the factored form of the given equation is D( D – 1 )(D – 1)y = x

(c) to get Yc consider

D( D – 1 )(D – 1)y = 0;

Yc = ex (+c1 + c2x) + c3

(d) for the particular integral Yp. Use the method of reduction of order:

Let z = ( D – 1 )(D – 1)y and so Dz = x

z = x2/2 substitue to the above equation

( D – 1 )(D – 1)y = x2/2

Let (D – 1) y = v (D – 1)v = x2/2

P = - 1 Q = x2/2 and ϕ = e-x

vϕ = ∫ ϕ Qdx or v e-x = ∫ e-x x2/2

which by integration by parts yields to:

v = - x2/2 - x – 1

Subsitute back ; (D – 1) y = - x2/2 - x – 1

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

y = Yp + Yc or

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

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Find the solution of the following:1. (D2 – 4 )y = 4x – 3ex Solution2. (D2 + 2D + 5)y = 3e-xsinx – 10 Solution3. (D3 – D ) y = 4e-x + 3e2x Solution

Answer:1. Y = -x + ex + C1 e2x + C2 e-2x

2. Y = e-x(C1cos2x + C2sin2x) +exsinx – 2

3. Y = 2xe-x + ½ + e2x + C1 + C2ex+ C3e-x

UNDETERMINED COEFFICIENTS

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1. (D2 – 4 )y = 4x – 3ex

(a) Yc = C1 e2x + C2 e-2x

(b) For 4x: ( q = 0, an = 4 ≠ 0, p = 1)

Yp1 = A xp + Bxp-1 + …. + Lx + M or

Yp1 = Ax + B

For – 3ex : ( q = 1, p = 0, p = 0)

Yp2 = eqx (A xp + Bxp-1 + …. + Lx + M)xr or

Yp2 = Aex

So Yp = Yp1 + Yp2

Yp = Ax + B + Cex

Y’p=A + Cex

Y”p= Cex

(c) substitute;

(D2 – 4 )Yp= 4x – 3ex

Cex – 4(Ax + B + Cex) = 4x – 3ex

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

Yp = Ax + B + Cex ;

Yp = -x + ex

(d) y = Yc + Yp

Y = -x + ex + C1 e2x + C2 e-2x

UNDETERMINED COEFFICIENTS

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2. (D2 + 2D + 5)y = 3e-xsinx – 10

(a) m2 + 2m + 5 = 0 has the complex roots r1 = -1 + 2i r2 = -1 – 2i

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

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

Yp1 = Ae-xcosx + Be-xsinx

For – 10: ( q = 0, p = 0, an ≠ 0)

Yp2 = C

So Yp = Yp1 + Yp2

Yp = Ae-xcosx + Be-xsinx + C

Y’p=-(A+ B) e-xsinx – (A – B ) e-xcosx

Y”p= -2Be-xcosx + 2Ae-xsinx

(c) substitute;

(D2 + 2D + 5)Yp = 3e-xsinx – 10

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

Yp =e-xsinx – 2

(d) y = Yc + Yp

Y = e-x(C1cos2x + C2sin2x) +exsinx – 2

UNDETERMINED COEFFICIENTS

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3. (D3 – D ) y = 4e-x + 3e2x

(a) m (m+1)(m – 1) = 0 has the complex roots r1 = 0r2 = 1 and r3 = -1

(b) Yc = C1 + C2ex+ C3e-x

(c)For 4e-x: ( q = -1, r = 1, p = 0)

Yp1 = Axe-x

For 3e2x: ( q = 2, p = 0, r = 0)

Yp2 = B e2x

So Yp = Yp1 + Yp2

Yp = Axe-x + B e2x

Y’p=A(-xe-x+ e-x)+2 B e2x

Y”p= A(xe-x -2e-x)+ 4B e2x

Y”’p= A(-xe-x+ 3e-x)+8 B e2x

(c) substitute;

(D3 – D ) Yp = 4e-x + 3e2x

Therefore; B = 1/2 and A = 2

Yp =2xe-x + ½ + e2x

(d) y = Yc + Yp

Y = 2xe-x + ½ + e2x + C1 + C2ex+ C3e-x

UNDETERMINED COEFFICIENTS

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LAPLACE TRANSFORM 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 LAPLACE1. Constant multiple

2. Linearity

3. Change scale

4. Shifting

5. Unnamed

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LAPLACE TRANSFORM TABLE

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INVERSE LAPLACE TRANSFORM

INVERSE LAPLACE TRANSFORM

Example:

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LAPLACE TRANSFORM OF DERRIVATIVE

Example:

Use partial fraction…