Math 251 Differential Equation I Course Outline PDF

Course Title: DIFFERENTIAL EQUATION I Code: MATH 251 Credit hours: 4 Tutor: Rhydal Esi Eghan (Math Department) Course Prerequisite Calculus I and II Differentiation and Integration Mode OF Delivery Course Objective ▶ Expose students to the basis of Differ...

Course Title: DIFFERENTIAL EQUATION I Code: MATH 251 Credit hours: 4 Tutor: Rhydal Esi Eghan (Math Department) Course Prerequisite Calculus I and II Differentiation and Integration Mode OF Delivery Course Objective ▶ Expose students to the basis of Differential Equations (DE) combined with some ideas from Linear Algebra : 1. Classify DE in terms of Types,Order,Degree,Linearity 2. Apply certain techniques in solving First-Order-Linear differential problems in ones field of study 3. Be able to solve DE of higher order with constant coefficient using the appropriate technique 4. Reduce higher order linear differential equations to system of first order differential equations and solve. Recommended Course Materials 1. Earl D. Rainville and Philip E. Bedient: Elementary Differential Equations, The Macmillan Company 2. Richard Bronson and Gabriel Costa (2006):Differential Equation (Third Edition), McGraw – Hill Inc. 3. https://tutorial.math.lamar.edu/pauldawkins Course Outline 1. Basic Concepts of Differential Equation (DE) 2. First-Order Ordinary Differential Equations 3. Higher-Order Linear ODE 4. Systems of Linear ODE 5. Laplace Transforms Assessment System (ONLINE) 1. Class Quizzes (5) 2. Mid-Semester Exams (1) 3. End of Semester Exams (1) Why Differential Equations for Engineering Mathematical models are used to understand, predict and optimise engineering systems. Many of these systems are deterministic and are modelled using differential equations Solve these models numerically on the computer: Numerical Methods/Analysis Outline 1: Basic Concepts of Modeling Definition: DE is an Equation involving an Unknown function and its Derivatives Application of Differential Equations Basic Concepts of Differential Equations ▶ Classify DE : Type,Order, Degree, Linearity, homogeneity ▶ Solution of a DE Classification of DE Ordinary DE vs Partial DE ODE PDE y = f (x) y = f (x, y , z,...) Single Independent variable Several Independent variables dy ∂y Ordinary derivatives. Partial derivatives. dx ∂x  ODE PDE d 2y dy ∂y ∂y 2 +3 + 2y = 0 + = e x−t dx dx ∂x ∂t  ODE PDE d 2y dy ∂y ∂y 2 +3 + 2y = 0 + = e x−t dx dx ∂x ∂t du ∂ 2u ∂ 2u = 3x −7 =0 dx ∂x 2 ∂x∂t  ODE PDE d 2y dy ∂y ∂y 2 +3 + 2y = 0 + = e x−t dx dx ∂x ∂t du ∂ 2u ∂ 2u = 3x −7 =0 dx ∂x 2 ∂x∂t ∂z ∂z y ′′ + 9y = e −2x =z +x ∂x ∂y Order vs Degree Order: the highest-ordered derivative Degree: power/exponent/ index that its highest-ordered derivative is raised, rationalized or cleared of fractions DE type order degree d 3y y 3 + =b ODE 3 1 dx x ∂ 2z ∂z + +z =0 PDE 2 1 ∂x 2 ∂x !3 d 2y d 3y x + =b dx 2 dx 3 Order vs Degree Order: the highest-ordered derivative Degree: power/exponent/ index that its highest-ordered derivative is raised, rationalized or cleared of fractions DE type order degree d 3y y 3 + =b ODE 3 1 dx x ∂ 2z ∂z + +z =0 PDE 2 1 ∂x 2 ∂x !3 d 2y d 3y x + =b ODE 3 1 dx 2 dx 3 xy ′′ + x(y ′ )2 − yy ′ = 0 Order vs Degree Order: the highest-ordered derivative Degree: power/exponent/ index that its highest-ordered derivative is raised, rationalized or cleared of fractions DE type order degree d 3y y 3 + =b ODE 3 1 dx x ∂ 2z ∂z + +z =0 PDE 2 1 ∂x 2 ∂x !3 d 2y d 3y x + =b ODE 3 1 dx 2 dx 3 xy ′′ + x(y ′ )2 − yy ′ = 0 ODE 2 1 x(y ′′ )2 + y ′′′ = −1 Order vs Degree Order: the highest-ordered derivative Degree: power/exponent/ index that its highest-ordered derivative is raised, rationalized or cleared of fractions DE type order degree d 3y y 3 + =b ODE 3 1 dx x ∂ 2z ∂z + +z =0 PDE 2 1 ∂x 2 ∂x !3 d 2y d 3y x + =b ODE 3 1 dx 2 dx 3 xy ′′ + x(y ′ )2 − yy ′ = 0 ODE 2 1 x(y ′′ )2 + y ′′′ = −1 ODE 3 1 Order Vs Degree Find the order and degree of the given DE " 3 #3/2 d 3y dy = + ex (1) dx 3 dx Order Vs Degree Find the order and degree of the given DE " 3 #3/2 d 3y dy = + ex (1) dx 3 dx Multiply equation by 2: ie order=3, degree=2 Linear vs Non-linear Linear DE 1. The dependent variable and ALL its derivatives occur only in the first degree (or to the first power). 2. No product of the dependent variable, say y , and/or any of its derivatives present. 3. No transcendental function (trigonometric, logarithmic or exponential) of the dependent variable and/or its derivatives occurs. Else Non-linear Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) (1 − y )y ′ + 2y = e x Nonlin (1st term coeff. depends on y) Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) (1 − y )y ′ + 2y = e x Nonlin (1st term coeff. depends on y) !4 d 3y d 2y + =3 Nonlin (1st-term deg > 1) dx 3 dx 2 Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) (1 − y )y ′ + 2y = e x Nonlin (1st term coeff. depends on y) !4 d 3y d 2y + =3 Nonlin (1st-term deg > 1) dx 3 dx 2 e x y ′′ + sin(xy ) = cosx Non (transcends function sin(y)) y ′ + 3y = 0 Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) (1 − y )y ′ + 2y = e x Nonlin (1st term coeff. depends on y) !4 d 3y d 2y + =3 Nonlin (1st-term deg > 1) dx 3 dx 2 e x y ′′ + sin(xy ) = cosx Non (transcends function sin(y)) y ′ + 3y = 0 linear (all deg=1) z ′′ + (z ′′′ )2 = 0 Linear and Non-Linear DE linearity y ′′ + 3y ′ + 9y = 0 linear (all deg=1) (1 − y )y ′ + 2y = e x Nonlin (1st term coeff. depends on y) !4 d 3y d 2y + =3 Nonlin (1st-term deg > 1) dx 3 dx 2 e x y ′′ + sin(xy ) = cosx Non (transcends function sin(y)) y ′ + 3y = 0 linear (all deg=1) z ′′ + (z ′′′ )2 = 0 Nonlin(2nd term deg > 1) Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homo. (y ′ )2 + 2y ′ = sin(x) Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homo. (y ′ )2 + 2y ′ = sin(x) NonHomo.(non-diff-term =sin(x)) yy ′ + x = 0 Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homo. (y ′ )2 + 2y ′ = sin(x) NonHomo.(non-diff-term =sin(x)) yy ′ + x = 0 NonHomo.(non-diff-term = −x) x 2 (y ′′ + x) = xy ′ − y Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homo. (y ′ )2 + 2y ′ = sin(x) NonHomo.(non-diff-term =sin(x)) yy ′ + x = 0 NonHomo.(non-diff-term = −x) x 2 (y ′′ + x) = xy ′ − y NonHomo (non-diff-term =x 2 x) dy = ax dx Homogeneous & Non-Homogeneous Homogeneous: if it contains No Non-differential terms and are set to 0 Differential Equation Homogeneity y ′ + t 2y = 0 Homo. y ′′ + y ′ = 1 NonHomo (non-diff-term =1) u ′′ + 2u ′ + u = 0 Homo. (y ′ )2 + 2y ′ = sin(x) NonHomo.(non-diff-term =sin(x)) yy ′ + x = 0 NonHomo.(non-diff-term = −x) x 2 (y ′′ + x) = xy ′ − y NonHomo (non-diff-term =x 2 x) dy = ax NonHomo dx Trial Tabulate the classifications of the following Diff y ′ = 2x − y (2) y ′′ + ln(y ) = x 2 (3) d 2y dy +( )2 = sin(y ) (4) dx 2 dx ∂u ∂ 2u − =0 (5) ∂t ∂x 2 d 2y dy = + 12xe x (6) dx 2 dx Solution of Differential Equation 1. The condition under which we can be assured that a DE is solvable is the Existence Theorem 2. A solution of DE is any function/expression of a dependent variables say ′ ϕ′ satisfying the DE on a specified open interval ′I ′ F (x, y , y ′ ) = 0 =⇒ F (x, ϕ, ϕ′ ) = 0 Solutions of DE General Particular Has n arbitrary constants Assign definite values to arbitrary constants C1 ,..., Cn Solves at specific ’C’ Given IVP,BVP Rule 1. State the Solution 2. Differentiate to Highest derivative 3. Substitute into the DE Test of General Solution of DE: Example 1 Show that for any values of the arbitrary constants c1 , c2 , the function ϕ = c1 cosx + c2 sinx is the solution of the DE y ′′ + y = 0 Test of General Solution of DE: Example 1 Show that for any values of the arbitrary constants c1 , c2 , the function ϕ = c1 cosx + c2 sinx is the solution of the DE y ′′ + y = 0 soln: Given Solution : ϕ = c1 cosx + c2 sinx Differentiate terms : ϕ′ = −c1 sinx + c2 cosx ϕ′′ = −c1 cosx − c2 sinx DE Substn. : y ′′ + y = 0 =⇒ ϕ′′ + ϕ = (−c1 cosx − c2 sinx) + (c1 cosx + c2 sinx) =0 Test of General Solution of DE: Example 2 Verify that u = e 3x is a solution of the 2nd-order ODE u ′′ + 2u ′ − 15u = 0 Test of General Solution of DE: Example 2 Verify that u = e 3x is a solution of the 2nd-order ODE u ′′ + 2u ′ − 15u = 0 soln: Given Soln. : u = e 3x Differentiate terms : u ′ = 3e 3x u ′′ = 9e 3x DE Substn. : u ′′ + 2u ′ − 15u = 9e 3x + 6e 3x − 15e 3x =0 Test of Particular Solution of DE: Example 1 Initial Value Problem: Verify that y = 2e x + xe x is a solution of the 2nd-order ODE y ′′ − 2y ′ + y = 0 given y (0) = 2 and y ′ (0) = 3 Test of Particular Solution of DE: Example 1 Initial Value Problem: Verify that y = 2e x + xe x is a solution of the 2nd-order ODE y ′′ − 2y ′ + y = 0 given y (0) = 2 and y ′ (0) = 3 soln: Given Soln. : y = 2e x + xe x Differentiate : y ′ = 2e x + xe x + e x y ′′ = 4e x + xe x DE Substn. : y ′′ − 2y ′ + y =0 1st Condition test : y (0); 2e 0 + 0 = 2 2nd Condition test : y ′ (0); 2e 0 + 0 + e 0 = 3 Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x False y =x y ′′ − xy ′ + y = 0 Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x False y =x y ′′ − xy ′ + y = 0 True y = 2sin(2x) y ′′ + 4y = 0 Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x False y =x y ′′ − xy ′ + y = 0 True y = 2sin(2x) y ′′ + 4y = 0 True y = 2e x + xe x y ′′ − 2y ′ + y = 2 Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x False y =x y ′′ − xy ′ + y = 0 True y = 2sin(2x) y ′′ + 4y = 0 True y = 2e x + xe x y ′′ − 2y ′ + y = 2 False y = 2x + Ce x y ′ − y = 2(1 − x) Trial Verify by direct substitution that each given function is a solution of the given differential equation Solution DE Test (T/F) y =1 y ′′ + 2y ′ + y = x False y =x y ′′ − xy ′ + y = 0 True y = 2sin(2x) y ′′ + 4y = 0 True y = 2e x + xe x y ′′ − 2y ′ + y = 2 False y = 2x + Ce x y ′ − y = 2(1 − x) True Outline 2: First-Order DE (FODE Definition: A First Order Differential Equation is an equation containing only the first differential. 2 Representation: 1. Standard Form dy y ′ = f (x, y ) ≡ = f (x, y ) (7) dx 2. Differential Form M(x, y )dy + N(x, y )dx = 0 (8) Standard and Differential Representations 1. Write the general equation xy ′ − y 2 = 0 in a standard form ′ y2 solve for y’ : y = x Standard and Differential Representations 1. Write the general equation xy ′ − y 2 = 0 in a standard form ′ y2 solve for y’ : y = x 2. Write the differential equation y (yy ′ − 1) = x in a Differential form x +y solve for y’ : y ′ = y2 dy x +y =⇒ : = dx y2 y 2 dy = (x + y )dx Diff. form : y 2 dy −(x + y ) dx = 0 |{z} | {z } M(x,y) N(x,y) Standard and Differential Forms Write the following differential equations in their respective Standard or Differential forms Standard Differential y y′ = x Standard and Differential Forms Write the following differential equations in their respective Standard or Differential forms Standard Differential y y′ = xdy − ydy = 0 x Standard and Differential Forms Write the following differential equations in their respective Standard or Differential forms Standard Differential y y′ = xdy − ydy = 0 x Differential Standard (xy + 3)dx + (2x − y 2 + 1)dy = 0 Standard and Differential Forms Write the following differential equations in their respective Standard or Differential forms Standard Differential y y′ = xdy − ydy = 0 x Differential Standard dy −(xy + 3) (xy + 3)dx + (2x − y 2 + 1)dy = 0 = dx 2x − y 2 + 1 Solutions to FODE The most general first order differential equation can be written as y ′ = f (y , t) (9) Equation (9) has No General Formula for it Solution : Special Cases 1. Linear Equation 2. Homogeneous Equation 3. Separable Equation 4. Exact Equation 5. Bernoulli & Ricatti Linear First Order DE Standard Form: dy + p(t)y = q(t) (10) dt ▶ Where both p(t) and q(t) are continuous functions ▶ Solve using Integrating Factor Solving Steps Given dy + p(t)y = q(t) dt 1. Write DE in Standard Form: y ′ + p(t)y = q(t) R 2. Find the Integrating Factor :I (t) = e p(x)dx 3. Multiply Equation by the integrating factor. d 4. Find derivative computed by the Product Rule: [I (t)y ] dt 5. Integrate both sides 6. Solve for unkwown (solution) TEST THE FINAL SOLUTION Linear FODE Example: 1 Find the solution of y ′ = −4y Soln: Standard Form : y ′ + 4y = 0 R 4dx Integrating Factor : e = e 4x Multiplication : e 4x y ′ + 4e 4x y = 0 d   Product Rule : e 4x y =0 dx Integrate : e 4x y = C Solution : y = Ce −4x Linear FODE Example: 2 Find the solution of v ′ − 2v = x Soln: Standard Form : v ′ − 2v = x R −2dx Integrating Factor : e = e −2x Multiplication : e −2x v ′ − 2e −2x v = xe −2x d   Product Rule : e −2x v = xe −2x dx integration by part 1 1 Integrate : e −2x v = xe −2x − xe −2x + C 2 4 1 1 Solution : v = x − + Ce −2x 2 4 Linear FODE Example: 3 1 2 Find the solution of u′ − u = xcosx x x2 Soln: Linear FODE Example: 3 1 2 Find the solution of u′ − u = xcosx x x2 Soln: Multiply through by x to obtain standard form 2 Standard Form : u ′ − u = x 2 cosx x R 2 − dx Integrating Factor : e = e −2lnx = x −2 x 2 Multiplication : x −2 (u ′ − u) = cosx x d  −2  Product Rule : x u = cosx dx Integrate : x −2 u = sinx + C Solution : u = x 2 sinx + Cx 2