Module 6 - Differential Equations 1 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 6

DIFFERENTIAL EQUATIONS I

Module Topics 1. Classification of differential equations 2. First order separable ordinary differential equations 3. Second order linear homogeneous ordinary differential equations with constant coefficients

A:

Work Scheme based on JAMES (THIRD EDITION)

1. Read section 10.1, the Introduction, on p.671. 2. Read quickly through section 10.2 on engineering examples. Four particular examples are discussed to illustrate the importance of differential equations in solving engineering problems. You are not expected to study these examples - they simply indicate the wide variety of engineering situations in which differential equations can be used. In this module, and two further modules on differential equations which you will study in the next semester, attention is concentrated on the mathematical methods used to solve differential equations. The applications will come in your engineering units throughout your degree course at Southampton. 3. Study the first paragraph of section 10.3 on the classification of differential equations, and then study section 10.3.1 on defining ordinary differential equations (ODEs) and partial differential equations (PDEs). Note that in the mathematics units this year you will only be asked to solve ODEs, because the solution of PDEs is much more difficult! Look carefully at Example 10.1. 4. Independent and dependent variables are defined next. Study section 10.3.2 and the various equations considered in Example 10.2. 5. Study the short section 10.3.3 and Example 10.3. 6. The important distinction between linear and nonlinear equations is discussed in section 10.3.4. Study the latter and Example 10.4. 7. Study section 10.3.5. Most people use inhomogeneous instead of the word nonhomogeneous introduced by J. Note also that on the eleventh line in the top paragraph on p.680 the word independent should be replaced by dependent. Study Examples 10.5 and 10.6. ***Do Exercises 2(a),(b),(c),(d),(j),(k) on p.681*** 8. After classifying a differential equation the most important stage is to obtain the solution. Read the introduction to section 10.4 at the top of p.682. 9. Study section 10.4.1 and Examples 10.7 and 10.8, and then study the related section 10.4.2. –1–

10. Study section 10.4.3 including Examples 10.9, 10.10 and 10.11. Note that the use of boundary conditions and initial conditions in J. is not standard. Boundary condition is normally used when the independent variable denotes distance (not time). When conditions are applied at two different boundaries then the problem is usually called a two-point boundary-value problem. The phrase initial conditions is normally used when time t is an independent variable and conditions are applied at the beginning of the process, which is often t = 0. If all the conditions were applied some time after the start of the process, when t = 10 s say, then it would not normally be given the special name initial-value problem as suggested in J. 11. Read section 10.4.4. ***Do Exercises 3(a),(e), 4(a),(b) on p.686*** 12. Read the introductory paragraph in section 10.5. 13. Study the beginning of section 10.5.3 on p.690 and the top of p.691, noting that it is usual to go direct from equation (10.13) to equation (10.15). Work through Example 10.13. Continue studying at the top of p.692, stopping just above equation (10.16) before the paragraph beginning “Some differential equations...”. ***Do Exercises 11(b), 15(a), 16(a),(d), 12(b) on p.693*** 14. You will study more first order ODEs next semester, but to complete this module some simple second order ODEs will be considered. Read the three lines introducing section 10.8 at the bottom of p.714, and then turn to p.723. 15. Study section 10.9.1. Equations of the type investigated in this section occur very frequently in engineering applications. One important example, for instance, is the equation governing simple harmonic motion. You should understand the theory on p.724 but in practice you would usually go straight from the given ODE (10.37) to the corresponding equation (10.38), solve the latter to find m1 and m2 and then write down the appropriate solution for x as in the summary at the top of p.725. These results do not appear on the Formula Sheet. Study Example 10.33, noting that the roots of the quadratic equation should be m1 = 8.275 and m2 = 0.725, with solution x(t) = A e8.275t + B e0.725t . Study Examples 10.34 and 10.35. ***Do Exercises 53(a), 55(b),(d), 57(a), 54(a), 57(b) on p.727*** 16. Almost all the ODEs in J. use x and t as the dependent and independent variables respectively. In practice, you must be able to solve equations involving different variables. Note also that the Data Book uses y and x as the dependent and independent variables.

B:

Work Scheme based on STROUD (FIFTH EDITION)

Although most of the mathematics in this module can be found in S., not all the terminology is discussed. You are advised, therefore, to work first through items 1-11 of the Work Scheme based on J. After completing the latter turn to S. and work through Programme 24, First-Order Differential Equations, frames 1-21 omitting Example 5 in frame 17. Finally, work through S. Programme 25, Second-Order Differential Equations, frames 1-16. –2–

Specimen Test 6 1.

Classify the following differential equations as ordinary/partial, linear/nonlinear, homogeneous/ inhomogeneous and also state the equation’s order and the dependent and independent variables: (i)

2.

dx 1 + x = 1, dt t

(ii)

∂2y ∂2y − 2 = 0, 2 ∂x ∂t

(iii)

d2 y dy + y = 0. + x2 dx2 dx

State how many arbitrary constants you expect to find in the general solution of the following differential d2 x = t + e2t . equation, and then find the general solution: dt2

3. (i) Determine the general solution of the equation (ii) Find the solution of the differential equation

4.

xt

dx = 1. dt

dx t+1 = dt x+1

given that x(0) = 1 .

Find the general solution of the following equations: (i)

5.

x

d2 x dx + 13 x = 0, −4 dt2 dt

(ii)

d2 y dy + y = 0. −2 dx2 dx

Find the solution of the differential equation

d2 x dx +2 x = 0 +3 dt2 dt

–3–

given that x(0) = 1,

dx (0) = −1 . dt