Engineering Analysis I Second and Higher Order Differential Equations Dr. Omar R. Daoud

2nd Order Differential Equations Introduction 2nd Order Differential Equations For any three numbers a, b and c, the two numbers: are solutions to the quadratic equation: with the properties: 6/21/2012 Part II

2nd Order Differential Equations Introduction 2nd Order Differential Equations The differential equation: can be re-written to read: that is: 6/21/2012 Part II

2nd Order Differential Equations Introduction 2nd Order Differential Equations The differential equation can again be re-written as: where: 6/21/2012 Part II

2nd Order Differential Equations Introduction 2nd Order Differential Equations The differential equation: has solution: This means that: That is: 6/21/2012 Part II

2nd Order Differential Equations Introduction 2nd Order Differential Equations The differential equation: has solution: where: 6/21/2012 Part II

2nd Order Differential Equations Introduction 2nd Order Differential Equations A 2nd Order D.E. will appear in the form of It will be denoted by homogeneous if otherwise its not. Moreover, if one of the solutions is known  The reduction of order technique will be used to find the second linearly independent solution 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique 2nd Order Differential Equations Assume that the 2nd order D.E. has two different basis (fundamental solutions); namely y1 and y2. They called linearly independent on a certain set I if and only if these basis are not proportional; while they called linearly dependent if 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique 2nd Order Differential Equations To obtain the basis if y1(x) is known, y2(x) should be set to As its known that the homogenous 2nd order D.E. is in the form of However, 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique 2nd Order Differential Equations Substitute 1 and 3 in 2 Rearranging and simplifying Let then 5 will be 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique Integrate Take the exponential and manipulate 2nd Order Differential Equations 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique Example: Find the basis for the following D.E. Divide both sides by x2 Then 6/21/2012 Part II

2nd Order Differential Equations The Reduction of order technique Example: Find the basis for the following D.E. Divide both sides by x2 Then 6/21/2012 Part II

2nd Order Differential Equations Homogeneous equations 2nd Order Differential Equations The differential equation: Is a second-order, constant coefficient, linear, homogeneous differential equation. Its solution is found from the solutions to the auxiliary equation: These are: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and different roots Real and equal roots Complex roots 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and different roots If the auxiliary equation: with solution: where: then the solution to: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and different roots Example: Find the general solution for the following D.E. The auxiliary equation is Thus the general solution is P.S. To find a particular solution we must be given enough information. 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and different roots Example: Find the general solution for the following D.E. The auxiliary equation is Thus the general solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and different roots Cont. Example: Find the general solution for the following D.E. Using the initial conditions Now Solving gives Thus the particular solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and equal roots If the auxiliary equation: with solution: where: then the solution to: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and equal roots Example: Find the general solution for the following D.E. The auxiliary equation is Thus the general solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Real and equal roots Example: Find the general solution for the following D.E. The auxiliary equation is  Thus the general solution is Using the initial conditions Now Thus the particular solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots If the auxiliary equation: with solution: where: Then the solutions to the auxiliary equation are complex conjugates. That is: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots Complex roots to the auxiliary equation: means that the solution of the differential equation: is of the form: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots Since: then: The solution to the differential equation whose auxiliary equation has complex roots can be written as:: 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots Example: Find the general solution for the following D.E. The auxiliary equation is Thus the general solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots Example: Find the general solution for the following D.E. The auxiliary equation is Thus the general solution is 6/21/2012 Part II

2nd Order Differential Equations The auxiliary equation Complex roots Cont. Example: Find the general solution for the following D.E. Using the initial conditions  Thus the particular solution is 6/21/2012 Part II

2nd Order Differential Equations Summary Differential equations of the form: Auxiliary equation: Roots real and different: Solution Roots real and the same: Solution Roots complex (  j): Solution 6/21/2012 Part II

2nd Order Differential Equations Differential Operators (Method of inverse operators) Sometimes, it is convenient to refer to the symbol “D” as the differential operator: But, 6/21/2012 Part II

2nd Order Differential Equations Differential Operators (Method of inverse operators) The differential operator D can be treated as an ordinary algebraic quantity with certain limitations. (1) The distribution law: A(B+C) = AB + AC which applies to the differential operator D (2) The commutative law: AB = BA which does not in general apply to the differential operator D Dxy  xDy (D+1)(D+2)y = (D+2)(D+1)y 6/21/2012 Part II

2nd Order Differential Equations Differential Operators (Method of inverse operators) The differential operator D can be treated as an ordinary algebraic quantity with certain limitations. (3) The associative law: (AB)C = A(BC) which does not in general apply to the differential operator D D(Dy) = (DD)y D(xy) = (Dx)y + x(Dy) The basic laws of algebra thus apply to the pure operators, but the relative order of operators and variables must be maintained. 6/21/2012 Part II

2nd Order Differential Equations Differential Operators to exponentials 6/21/2012 Part II

2nd Order Differential Equations Differential Operators to trigonometric functions where “Im” represents the imaginary part of the function which follows it. 6/21/2012 Part II

2nd Order Differential Equations Differential Operators The operator D signifies differentiation, i.e. D-1 is the “inverse operator” and is an “intergrating” operator. It can be treated as an algebraic quantity in exactly the same manner as D 6/21/2012 Part II

2nd Order Differential Equations Differential Operators Example: Find the general solution for the following D.E. Using the Differential Operators method The Differential Operators Then the general solution is 6/21/2012 Part II

2nd Order Differential Equations Euler-Cauchy Equation It is a differential Equation in the following form The basic method of solution Characteristic equation 1.two distinct, real values for r (when (A-1)2-4B > 0) 2.Only one real value for r (when (A-1)2-4B = 0) 3.Two distinct, complex values for r (when (A-1)2-4B < 0) 6/21/2012 Part II

Euler - Cauchy equation Case 1 : Two distinct, real values for r (when (A-1)2-4B > 0) The general solution : Where c1 and c2 are arbitrary constants 6/21/2012 Part II

2nd Order Differential Equations Euler-Cauchy Equation Example: Solve the following D.E. Solution Then the general solution is 6/21/2012 Part II

Euler - Cauchy equation Case 2 : Only one real value for r (when (A-1)2-4B = 0) To find y2(x), we apply a technique called reduction of order Try to produce u(x) such that u(x) y1(x) is a solution. Substituted into The general solution : Where c1 and c2 are arbitrary constants 6/21/2012 Part II

2nd Order Differential Equations Euler-Cauchy Equation Example: Solve the following D.E. Solution Then the general solution is 6/21/2012 Part II

Where c1 and c2 are arbitrary constants Euler - Cauchy equation Case 3 : Two distinct, complex values for r (when (A-1)2-4B < 0) Any linear combination of solutions is also a solution ! The general solution : Where c1 and c2 are arbitrary constants 6/21/2012 Part II

2nd Order Differential Equations Euler-Cauchy Equation Example: Solve the following D.E. Solution Then the general solution is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations The second-order, constant coefficient, linear, inhomogeneous differential equation is an equation of the type: The solution is in two parts y1 + y2: part 1, y1 is the solution to the homogeneous equation and is called the complementary function which is the solution to the homogeneous equation part 2, y2 is called the particular integral. 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Particular integrals The general form assumed for the particular integral depends upon the form of the right-hand side of the inhomogeneous equation. The following table can be used as a guide: 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Example: Find the general solution for the following D.E. The auxiliary equation is Thus the CF is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Cont. Example: Find the general solution for the following D.E. Finding the PI: Substituting into the original equation Thus the PI is Hence the general solution is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Complementary function Example, to solve: Complementary function Auxiliary equation: m2 – 5m + 6 = 0 solution m = 2, 3 Complementary function y1 = Ae2x + Be3x where: 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Particular integral (b) Particular integral Assume a form for y2 as y2 = Cx2 + Dx + E then substitution in: gives: yielding: so that: 6/21/2012 Part II

complementary function + particular integral 2nd Order Differential Equations Inhomogeneous equations Complete solution (c) The complete solution to: consists of: complementary function + particular integral That is: 6/21/2012 Part II

Thus we should try another y(x) 2nd Order Differential Equations Inhomogeneous equations Example: Find the general solution for the following D.E. The auxiliary equation is Thus the CF is Finding the PI: But this is in the same form of one of the CF’s solutions. Thus we should try another y(x) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Cont. Example: Find the general solution for the following D.E. Thus, PI should be: Substituting into the original equation Equating Coefficients  Hence the general solution is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Example: Solve the following D.E.: Solution: The auxiliary equation is Thus the CF is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Example: Solve the following D.E.: Solution: Finding the PI: (Back to slide 45) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Cont. Example: Solve the following D.E.: Solution: Finding the PI: (Back to slide 45) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Cont. Example: Solve the following D.E.: Solution: Finding the PI: (Back to slide 45) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Cont. Example: Solve the following D.E.: Solution: Finding the PI: (Back to slide 45) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by undetermined Coefficients Cont. Example: Solve the following D.E.: Solution: Apply the conditions: 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Example: Find the general solution for the following D.E. Using the Differential Operators method The Differential Operators 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Cont. Example: Find the general solution for the following D.E. Using the Differential Operators method Or, The Differential Operators binomial expansion 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Cont. Example: Find the general solution for the following D.E. Using the Differential Operators method =2 6/21/2012 Part II

2nd Order Differential Equations Conclusion If f(p) = 0, Use dimensional analysis Non-zero 6/21/2012 Part II

2nd Order Differential Equations Conclusion y = 1, p = 0, Upcoming(D-p) replaced byD integration 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations: Solution by undetermined Coefficients or or or 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by Variation of Parameters The Linear inhomogeneous 2nd order D.E. will be in the following form: here: p(x) and q(x) need not to be constant Let y1(x) and y2(x) are solutions of y’’+ p(x)y’ + q(x)y = 0 on the interval J. Let yp(x) = u(x)y1(x) + v(x)y2(x)  But 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by Variation of Parameters Thus  Then Substitute them in the general equation 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations Solution by Variation of Parameters Thus  yp(x) = u(x)y1(x) + v(x)y2(x) 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations (Solution by Variation of Parameters) Example: Find the general solution for the following D.E. Solution: The auxiliary equation is Thus the CF is The Wronskian factor is 6/21/2012 Part II

2nd Order Differential Equations Inhomogeneous equations (Solution by Variation of Parameters) Cont. Example: Find the general solution for the following D.E. Solution: 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex1: Consider the following interconnected fluid tanks: Suppose both tanks, each holding 24 liters of a brine solution, are interconnected by pipes as shown. Fresh water flows into tank A at a rate of 6 L/min, and fluid is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex1: Consider the following interconnected fluid tanks: The liquids inside each tank are kept well stirred, so that each mixture is homogeneous. If initially tank A contains a kg of salt and tank B contains b kg of salt, determine the mass of salt in each tanks at any time t > 0. 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex1: Consider the following interconnected fluid tanks: For tank A, we have: For tank B, we have: 6/21/2012 Part II

This gives us a system of First Order Equations 2nd Order Differential Equations Applications: Ex1: This gives us a system of First Order Equations 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit The total charges at any time in these two circuits 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit The total charges at any time in these two circuits The voltage drop across R The voltage drop across L The voltage drop across C 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit The total charges at any time in these two circuits The total voltage drop is 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit The total charges at any time in these two circuits The total voltage drop is 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit Find the current I(t) in the given RLC circuit with R=100Ω, L=0.1 H, C=1mF, and E(t)=155sin(377t)V. Q(0)=I(0)=0. 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit Find the current I(t) in the given RLC circuit with R=100Ω, L=0.1 H, C=1mF, and E(t)=155sin(377t)V. Q(0)=I(0)=0. The auxiliary equation is The CF solution 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit Find the current I(t) in the given RLC circuit with R=100Ω, L=0.1 H, C=1mF, and E(t)=155sin(377t)V. Q(0)=I(0)=0. The Ip is 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit Find the current I(t) in the given RLC circuit with R=100Ω, L=0.1 H, C=1mF, and E(t)=155sin(377t)V. Q(0)=I(0)=0. Thus, I=Ih+ Ip Apply the initial conditions 6/21/2012 Part II

2nd Order Differential Equations Applications: Ex2: Two cases of RLC Circuit Find the current I(t) in the given RLC circuit with R=100Ω, L=0.1 H, C=1mF, and E(t)=155sin(377t)V. Q(0)=I(0)=0. Applying the initial conditions 6/21/2012 Part II

2nd Order Differential Equations Applications: 1. Free falling stone where s is distance or height and g is acceleration due to gravity. 2. Spring vertical displacement where y is displacement, m is mass and k is spring constant 3. RLC – circuit, Kirchoff ’s Second Law q is charge on capacitor, L is inductance, c is capacitance. R is resistance and E is voltage 6/21/2012 Part II

2nd Order Differential Equations Applications: 4. Newton’s Low of Cooling where is rate of cooling of the liquid, is temperature difference between the liquid ‘T’ and its surrounding Ts 5. Growth and Decay y is the quantity present at any time 6/21/2012 Part II

Higher Order Differential Equations The most general nth order linear differential equation is, Thus, the standard form could be written as In this type of D.Es. The roots/basis of the characteristics function becomes larger and the use of Wronskian plays a vital rule in the solution. In this case the GENERAL SOLUTION (GS)will be extended to be in the form of: 6/21/2012 Part II

Higher Order Differential Equations The basis of the GS are linearly independent if and only if the Wronskian (the n-th order determinant )is not equals to zero In this case the Wronskian is given by on some point in an open interval 6/21/2012 Part II

Higher Order Differential Equations As previously explained for n-th order D.E. r(x)=0 Homogenous r(x)0 Inhomogeneous 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients The solution will be based on the same solution method of the 2nd O.D.Es. (substitute ) Thus, there will be three different cases Real and Unequal Roots Complex Roots Repeated Roots 6/21/2012 Part II

Higher Order Differential Equations Real and Unequal Roots Homogenous Higher O.D.Es with Constant coefficients If roots of characteristic polynomial P(m) are real and unequal, then there are n distinct solutions of the differential equation: If these functions are linearly independent, then general solution of differential equation is The Wronskian can be used to determine linear independence of solutions. 6/21/2012 Part II

Higher Order Differential Equations Complex Roots Homogenous Higher O.D.Es with Constant coefficients If the characteristic polynomial P(r) has complex roots, then they must occur in conjugate pairs, Note that: not all the roots needed be complex. Thus, General Solution 6/21/2012 Part II

Higher Order Differential Equations Repeated Roots Homogenous Higher O.D.Es with Constant coefficients Suppose a root m of characteristic polynomial P(r) is a repeated root with multiplicity n. Then linearly independent solutions corresponding to this repeated root have the form 6/21/2012 Part II

Use Long Division (Based on the basis of 2) Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: The auxiliary equation is Use Long Division (Based on the basis of 2) 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Thus the CF is 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: The auxiliary equation is 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Thus the CF is 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: The auxiliary equation is Thus the CF is 6/21/2012 Part II

Higher Order Differential Equations Homogenous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: The auxiliary equation is 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients The Standard form is as follows The Constant-coefficient equation General Solution Constants 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: The auxiliary equation is Thus the CF is 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Finding the PI: (Back to slide 45) If we take it, we will see that its been contained in the homogenous solution 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Finding the PI: (Back to slide 45) 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Get the first and second derivatives of the general solution 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: Substitute the initial conditions 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es with Constant coefficients Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Method of variation of parameters. The standard form of the general solution is given by where 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: Euler-Cauchy Equation 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: 6/21/2012 Part II

Higher Order Differential Equations Inhomogeneous Higher O.D.Es Example: Solve the following D.E. Solution: 6/21/2012 Part II