Mat 275 Modern Differential Equations



Mat 275 Modern Differential Equations

Lecture week 4

Test 2 Review

Dr. Firozzaman

Department of Mathematics and Statistics

Arizona State University

Section 1.6 Substitution Method and Exact Solution (Page # 58)

This review is giving just an idea. Do not forget to check all of your homework problems along with quizzes.

Example 1. Solve the DE [pic]. Let us make the substitution[pic]. So the transformed DE is

[pic].

Example 2. Solve the homogenous differential equation (HDE) [pic]. Let us make the substitution [pic]. The HDE [pic] yields [pic]

Example 3. Solve the exact DE [pic]. Observe that [pic] and [pic] with [pic]. The solution of the EDE is [pic].

3. [pic] use homogeneous system method. Rewrite the problem as [pic]. Now substitute [pic]

You will find [pic]

9. [pic]. Applying same method as in 3, you find [pic]

15. [pic], homogeneous

[pic]

[pic]

20. [pic] which is of the form [pic]. Substitute [pic]. We have [pic]. Now solve the problem by integrating factor. You will find [pic]

Then [pic]

29. [pic]. Substitute [pic]

We have [pic]. Now solve using integrating factor.

31. [pic]. Solve using the method of exact solution.

37. [pic]. Solve using the method of exact solution.

43. [pic]. To solve use the substitution [pic]

[pic]

59. [pic]. [pic]

We also have [pic] and [pic]

Now [pic]. Substitute [pic] and solve considering homogeneous system..

Home work Problems: (page # 71) 3, 9, 15, 20, 29, 31, 37, 43, 59

Page # 71.

Section 2.1 Population Models (Separable Differential Equations) (Page # 77)

Home work Problems: (page # 86) 1, 3, 7, 33a

Section 2.2 Equilibrium Solutions and Stability (Page # 90)

Homework problems: Page # 96: 1, 5, 9, 13, 17, 29

Section 2.3 Acceleration – Velocity Models (Page # 98)

Example 1. Suppose that a crossbow bolt is shot straight upward from the ground

(y(0) = 0) with initial velocity v(0) = 49 m/s, then g = 9.8 m/sq. s gives [pic].

Solving the equation we find [pic]. The bolt’s height y(t) is given by

[pic] since y(0) = 0. The bolt reaches its maximum height when v = 0, i.e., when t = 5, and [pic]. The bolt returned to the ground when y(t) = 0 implies t = 10 seconds.

Home work: (Page # 106) 2, 4

Section 2.4 Numerical Approximation: Euler’s Method (Page # 110)

To find numerical solution to the initial value problem [pic]using Euler’s method we have the following consideration:

[pic]

Example 1. Apply Euler’s method to approximate the solution of the initial value problem [pic]

a) With step size h = 1 on the interval [0, 5]

b) With step size h = 0.2 on the interval [0, 1]

c) Graph the actual solution curve and the estimated solution curve on the same viewing window.

d) Solve the initial value problem by hand.

Solution:

a) We have [pic], [pic]. Then by the iterative process we have the following results:

[pic]

b) We have [pic], [pic]. Then by the iterative process we have the following results:

[pic]

c) Actual solution:

[pic].

Home work: (Page # 119) 1, 3, 9, 15

Section 2.6 Numerical Approximation: The Runge-Kutta Method (Page # 132)

To find numerical solution to the initial value problem [pic]using Runge-Kutta method we have the following consideration (This method gives more accurate result compared to Euler’s method):

[pic]

Example 2. Apply Runge-Kutta method to approximate the solution of the initial value problem [pic] with step size h = 0.5

Solution:

[pic]Similarly we find [pic].

Home work: (Page # 139) 1, 5

Section 3.1 Introduction: Second Order Linear Equations (Page # 144)

A second order differential equation in the (unknown) function [pic]is one of the form [pic]. This differential equation is said to be linear provided that G is linear in the dependent variable y and its derivatives [pic].

The DE [pic]is a second order linear DE. On the other hand [pic] is not a second order linear FE because of the non linear term [pic]. The general second order linear DE has the form

[pic]………..(a)

Homogeneous Second Order Linear Equations: Consider the general second order linear equation [pic], where the coefficient functions A, B, C and F are continuous on the open interval I, and if the function F on the right hand side vanishes identically on I then the equation (a) is said to be homogeneous linear equation; otherwise it is non-homogeneous.

The second order linear equation [pic]is non-homogeneous; its associated homogeneous equation is [pic]. Note that the non-homogeneous term F(x,y) frequently corresponds to some external influence on the system.

Example 1. Verify that the functions [pic] and [pic]are solutions of the differential equation [pic] and then find a solution satisfying the initial conditions [pic] and [pic].

Solution: You can easily verify that [pic] and [pic]are solutions of the differential equation [pic]. Now by the superposition principle (Page# 146, Theorem 1) we know that the general solution is [pic]. We also have now [pic].

The following results are the found:

[pic]

Definition: Linear Independence of two functions: Two functions defined on an open interval I is said to linearly independent on I provided that neither is a constant multiple of the other.

Definition: Given two functions p and q, the Wronskian W(p, q) of p and q is the determinant

[pic]

Theorem: Wronskians of solutions: Suppose that [pic] are two solutions of the homogeneous second order linear equation [pic]on a open interval I, on which p and q are continuous.

a) If [pic] are linearly dependent, then the Wronskian [pic]on I.

b) If [pic] are linearly independent, then the Wronskian [pic]at each point of I.

Home work problems: (Page # 155) 1, 3, 6, 8, 10, 16, 17, 32, 51

Section 3.2 General Solutions of Linear Equations (Page # 158)

Definition: Linear Dependence of Functions: The n functions [pic] are said to be linearly dependent on I provided that there exists constants [pic]not all zero such that [pic] on I for all x.

Example 1. Show that the functions [pic]are linearly dependent by using Wronskian.

Example 2. Show that the functions [pic]are linearly independent.

Example 3. Show that [pic]is a solution of [pic].

Home work problems: (Page # 167) 1, 3, 5, 8, 10, 13, 17, 20, 39

Section 3.3 Homogeneous Equations with Constant Coefficients (Page # 170)

Characteristic equation for finding general solution: Suppose [pic]be a solution of the homogeneous equation [pic], with constant coefficients [pic], then [pic]is called the characteristic equation or auxiliary equation of the DE. The solution of DE is reduces to a solution of a purely algebraic equation.

Example 1. Solve the initial value problem [pic]

Solution: Let [pic]be the solution of the initial value problem. The characteristic equation of this DE is [pic]. By the above theorem the general solution is [pic]. Using initial condition one can find that [pic]. The particular solution is then [pic].

Home work problems: (Page # 180) 1, 3, 5, 8, 10, 13, 17, 25, 39

Section 3.5 Nonhomogeneous Equations and Undetermined Coefficient (Page # 195)

Example 1. Find the particular solution [pic]of [pic]. Our DE is nonhomogeneous of the form (b) where [pic] is polynomial of degree 1, so our guess is [pic], then [pic] will satisfy the DE provided that [pic]. We have the particular solution [pic].

Example 2. Find the general solution of [pic]

Example 3. Find the general solution of [pic]

Example 4. Find the general solution of [pic]

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