Exam (elaborations) TEST BANK FOR Advanced Engineering Mathematics [Volume 1] By Herbert Kreyszig and Erwin Kreyszig (Student Solutions Manual and Study Guide)

Sec. 1.1 Basic Concepts. Modeling To get a good start into this chapter and this section, quickly review your basic calculus. Take a look at the front matter of the textbook and see a review of the main differentiation and integration formulas. Also, Appendix 3, pp. A63–A66, has useful formulas for such functions as exponential function, logarithm, sine and cosine, etc. The beauty of ordinary differential equations is that the subject is quite systematic and has different methods for different types of ordinary differential equations, as you shall learn. Let us discuss some Examples of Sec. 1.1, pp. 4–7. Example 2, p. 5. Solution by Calculus. Solution Curves. To solve the first-order ordinary differential equation (ODE) y
= cos x means that we are looking for a function whose derivative is cos x. Your first answer might be that the desired function is sin x, because (sin x)
= cos x. But your answer would be incomplete because also (sin x+2)
=cos x, since the derivative of 2 and of any constant is 0. Hence the complete answer is y= cos x+c, where c is an arbitrary constant. As you vary the constants you get an infinite family of solutions. Some of these solutions are shown in Fig. 3. The lesson here is that you should never forget your constants! Example 4, pp. 6–7. Initial Value Problem. In an initial value problem (IVP) for a first-order ODE we are given an ODE, here y
=3y, and an initial value condition y(0)=5.7. For such a problem, the first step is to solve the ODE. Here we obtain y(x)=ce3x as shown in Example 3, p. 5. Since we also have an initial condition, we must substitute that condition into our solution and get y(0)=ce3·0 = ce0 =c · 1= c=5.7. Hence the complete solution is y(x)=5.7e3x. The lesson here is that for an initial value problem you get a unique solution, also known as a particular solution. 2 Ordinary Differential Equations (ODEs) Part A Modeling means that you interpret a physical problem, set up an appropriate mathematical model, and then try to solve the mathematical formula. Finally, you have to interpret your answer. Examples 3 (exponential growth, exponential decay) and 5 (radioactivity) are examples of modeling problems. Take a close look at Example 5, p. 7, because it outlines all the steps of modeling. Problem Set 1.1. Page 8 3. Calculus. From Example 3, replacing the independent variable t by x we know that y
=0.2y has a solution y=0.2ce0.2x. Thus by analogy, y
=y has a solution 1 · ce1·x = cex, where c is an arbitrary constant. Another approach (to be discussed in details in Sec. 1.3) is to write the ODE as dy dx = y, and then by algebra obtain dy = y dx, so that 1 y dy = dx. Integrate both sides, and then apply exponential functions on both sides to obtain the same solution as above
1 y dy =
dx, ln|y| = x + c, eln |y| = ex+c, y = ex · ec = c∗ex, (where c∗ = ec is a constant). The technique used is called separation of variables because we separated the variables, so that y appeared on one side of the equation and x on the other side before we integrated. 7. Solve by integration. Integrating y
=cosh 5.13x we obtain (chain rule!) y=  cosh 5.13x dx = 1 5.13 (sinh 5.13x)+c. Check: Differentiate your answer:  1 5.13 (sinh 5.13x) + c 
= 1 5.13 (cosh 5.13x) · 5.13 = cosh 5.13x, which is correct. 11. Initial value problem (IVP). (a) Differentiation of y=(x+c)ex by product rule and definition of y gives y
= ex + (x + c)ex = ex + y. But this looks precisely like the given ODE y
=ex +y. Hence we have shown that indeed y=(x + c)ex is a solution of the given ODE. (b) Substitute the initial value condition into the solution to give y(0)=(0+c)e0 =c · 1= 12 . Hence c= 12 so that the answer to the IVP is y = (x + 12 )ex. (c) The graph intersects the x-axis at x=0.5 and shoots exponentially upward. Chap. 1 First-Order ODEs 3 19. Modeling: Free Fall. y

=g =const is the model of the problem, an ODE of second order. Integrate on both sides of the ODE with respect to t and obtain the velocity v=y
=gt +c1 (c1 arbitrary). Integrate once more to obtain the distance fallen y= 12 gt2 +c1t +c2 (c2 arbitrary). To do these steps, we used calculus. From the last equation we obtain y= 12 gt2 by imposing the initial conditions y(0)=0 and y
(0)=0, arising from the stone starting at rest at our choice of origin, that is the initial position is y=0 with initial velocity 0. From this we have y(0)=c2 =0 and v(0)= y
(0) = c1 = 0. Sec. 1.2 Geometric Meaning of y
=f (x, y). Direction Fields, Euler’s Method Problem Set 1.2. Page 11 1. Direction field, verification of solution. You may verify by differentiation that the general solution is y=tan(x+c) and the particular solution satisfying y(14 π)=1 is y=tan x. Indeed, for the particular solution you obtain y
= 1 cos2x = sin2x + cos2x cos2x = 1 + tan2x = 1 + y2 and for the general solution the corresponding formula with x replaced by x+c. 1 –1 –2 2 –1 –0.5 0 0.5 1 y x y(x) Sec. 1.2 Prob. 1. Direction Field 15. Initial value problem. Parachutist. In this section the usual notation is (1), that is, y
=f (x, y), and the direction field lies in the xy-plane. In Prob. 15 the ODE is v =f (t, v)=g −bv2/m, where v suggests velocity. Hence the direction field lies in the tv-plane.With m=1 and b=1 the ODE becomes v
= g −v2. To find the limiting velocity we find the velocity for which the acceleration equals zero. This occurs when g −v2 = 9.80−v2 =0 or v=3.13 (approximately). For v3.13 you have v
0 (increasing curves) and for v3.13 you have v
0 (decreasing curves). Note that the isoclines are the horizontal parallel straight lines g −v2 =const, thus v=const.