Part 6 Euler Cauchy Equation-Advanced Engineering Mathematics-Solution Manual, Exercises of Engineering Mathematics

Download Part 6 Euler Cauchy Equation-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! 8. Characteristic equation (  2)2  0, ODE y
 4y  4y  0, Wronskian W  j j  e
4x. 10. Auxiliary equation (m  3)2  m2  6m  9  m(m  1)  7m  9  0. Hence the ODE (an Euler–Cauchy equation) is x2y
 7xy  9y  0. The Wronskian is W  j j  x
7. 12. The characteristic equation is (  2)2  2  2  4  4  2  0. Hence the ODE is y
 4y  (4  2)y  0. The Wronskian is W  j j  e
4x where e  e
2x, c  cos x, and s  sin x. 14. The auxiliary equation is (m  1)2  1  m2  2m  2  m(m  1)  3m  2  0. Hence the Euler–Cauchy equation is x2y
 3xy  2y  0. The Wronskian is W  j j  x
3 where x
1 in the second row results from the chain rule and (ln x)  1/x. Here, c  cos (ln x), s  sin (ln x). 16. The characteristic equation is (  k)2  2  0. This gives the ODE y
 2ky  (k2  2)y  0. The Wronskian is W  j j  e
2kx where e  e
kx, c  cos x, s  sin x. 18. Team Project. (A) c1e x  c2e
x  c*1 cosh x  c*2 sinh x. Expressing cosh and sinh in terms of exponential functions [see (17) in App. 3.1], we have 1_ 2(c*1  c*2)e x  1_2(c*1  c*2)e
x; es e(ks  c ) ec e(kc  s ) x
1 s x
2s  x
1cx
1 x
1 c x
2c  x
1sx
1 es e(2s  c) ec e(2c  s) x
3 ln x 3x
4 ln x  x
4 x
3 3x
4 xe
2x (1  2x)e
2x e
2x 2e
2x Instructor’s Manual 45 im02.qxd 9/21/05 10:57 AM Page 45 docsity.com hence c1  1_ 2(c*1  c*2), c2  1_ 2(c*1  c*2). The student should become aware that for second-order ODEs there are several possibilities for choosing a basis and making up a general solution. For this reason we say “a general solution,” whereas for first-order ODEs we said “the general solution.” (B) If two solutions are 0 at the same point x0, their Wronskian is 0 at x0, so that these solutions are linearly dependent by Theorem 2. (C) The derivatives would be 0 at that point. Hence so would be their Wronskian W, implying linear dependence. (D) By the quotient rule of differentiation and by the definition of the Wronskian, (y2 /y1)  (y2y1  y2y1) /y1 2  W/y1 2. In Prob. 10 we have (y2 /y1)y1 2  (ln x)x
6  x
7. y2 /y1 is constant in the case of linear dependence; hence the derivative of this quotient is 0, whereas in the case of linear independence this is not the case. This makes it likely that such a formula should exist. (E) The first two derivatives of y1 and y2 are continuous at x  0 (the only x at which something could happen). Hence these functions qualify as solutions of a second-order ODE. y1 and y2 are linearly dependent for x  0 as well as for x  0 because in each of these two intervals, one of the functions is identically 0. On 1  x  1 they are linearly independent because c1y1  c2y2  0 gives c1  0 when x  0, and c2  0 when x  0. The Wronskian is W  y1y2  y2y1  { }  0 if { . The Euler–Cauchy equation satisfied by these functions has the auxiliary equation (m  3)m  m(m  1)  2m  0. Hence the ODE is xy
 2y  0. Indeed, xy
1  2y1  x
6x  2
3x 2  0 if x  0, and 0  0 for x  0. Similarly for y2. Now comes the point. In the present case the standard form, as we use it in all our present theorems, is y
 y  0 and shows that p(x) is not continuous at 0, as required in Theorem 2. Thus there is no contradiction. This illustrates why the continuity assumption for the two coefficients is quite important. (F) According to the hint given in the enunciation, the first step is to write the ODE (1) for y1 and then again for y2. That is, y
1  py1  qy1  0 y
2  py2  qy2  0 where p and q are variable. The hint then suggests eliminating q from these two ODEs. Multiply the first equation by y2, the second by y1, and add: (y1y
2  y
1y2)  p(y1y2  y1y2)  W  pW  0 2  x x  0 x  0 0
3x2  x3
0 x3
0  0
3x2 46 Instructor’s Manual im02.qxd 9/21/05 10:57 AM Page 46 docsity.com New Features Resonance (11) y  At sin 0 t in the undamped case Beats (12) y  B(cos t  cos 0 t) if input frequency is close to natural Large amplitude if (15*) 2  0 2  c2/(2m2) (Fig. 56) Phase lag between input and output SOLUTIONS TO PROBLEM SET 2.8, page 90 2. yp  0.6 cos 1.5t  0.2 sin 1.5t. Note that a general solution of the homogeneous ODE is yh  e
t(A cos 1.5t  B sin 1.5t), and the student should perhaps be reminded that this is not resonance, of course. 4. yp  0.25 cos t. Note that yh  e
2t(A cos t  B sin t); of course, this is not resonance. Furthermore, it is interesting that whereas a single term on the right will generate two terms in the solution, here we have—by chance—the converse. 6. yp  _1 10 cos t  1_ 5 sin t  _1 90 cos 3t  _1 45 sin 3t 8. yp  2 cos 4t  1.5 sin 4t 10. y  (c1  c2t)e
2t  0.03 cos 4t  0.04 sin 4t 12. y  c1e
t  c2e
4t  0.1 cos 2t. Note that, ordinarily, yp will consist of two terms if r(x) consists of a single trigonometric term. 14. y  e
t(A cos 2t  B sin 2t)  0.2  0.1 cos t  0.2 sin t 16. y  _163 cos 8t  1_ 8 sin 8t  _1 63 cos t. From the graph one can see the effect of (cos t) /63. There is no corresponding sine term because there is no damping and hence no phase shift. 18. y  (33  31t)e
t  37.5 cos t  6 cos 2t  4.5 sin 2t  1.5 cos 3t  2 sin 3t 20. y  100 cos 4.9t  98 cos 5t 22. The form of solution curves varies continuously with c. Hence if you start from c  0 and let c increase, you will at first obtain curves similar to those in the case of c  0. For instance, consider y
 0.01y  25y  100 cos 4.9t  98 cos 5t. 24. CAS Experiment. The choice of needs experimentation, inspection of the curves obtained, and then changes on a trial-and-error basis. It is interesting to see how in the case of beats the period gets increasingly longer and the maximum amplitude gets increasingly larger as /(2 ) approaches the resonance frequency. 26. If 0  t  , then a particular solution yp  K0  K1t  K2t 2 gives y
p  2K2 and y
p  yp  K0  2K2  K1t  K2t 2  1  t2; thus, K2   , K1  0, K0  1  2K2  1  . Hence a general solution is y  A cos t  B sin t  1   t2. 1  2 2  2 2  2 1  2 1  2 Instructor’s Manual 49 im02.qxd 9/21/05 10:57 AM Page 49 docsity.com From this and the first initial condition, y(0)  A  1   0, A  (1  ) . The derivative is y  A sin t  B cos t  t and gives y(0)  B  0. Hence the solution is (I) y(t)  (1  2/ 2)(1  cos t)  t2/ 2 if 0  t  , and if t , then (II) y  y2  A2 cos t  B2 sin t with A2 and B2 to be determined from the continuity conditions y( )  y2( ), y( )  y2( ). So we need from (I) and (II) y( )  2(1  2/ 2)  1  1  4/ 2  y2( )  A2 and y(t)  (1  2/ 2) sin t  2t / 2 and from this and (II), y( )  2/  B cos  B2. This gives the solution y  (1  4/ 2) cos t  (2/ ) sin t if t . Answer: y  { . The function in the second line gives a harmonic oscillation because we disregarded damping. SECTION 2.9. Modeling: Electric Circuits, page 91 Purpose. To discuss the current in the RLC-circuit with sinusoidal input E0 sin t. ATTENTION! The right side in (1) is E0 cos t, because of differentiation. Main Content Modeling by Kirchhoff’s law KVL Electrical–mechanical strictly quantitative analogy (Table 2.2) Transient tending to harmonic steady-state current A popular complex method is discussed in Team Project 20. if 0  t  if t (1  2/ 2)(1  cos t)  t2/ 2 (1  4/ 2) cos t  (2/ ) sin t 2  2 2  2 2  2 50 Instructor’s Manual im02.qxd 9/21/05 10:57 AM Page 50 docsity.com SOLUTIONS TO PROBLEM SET 2.9, page 97 2. The occurring integral can be evaluated by integration by parts, as is shown (with other notations) in standard calculus texts. From (4) in Sec. 1.5 we obtain I  e
Rt/L [ eRt/L sin t dt  c]  ce
Rt/L  (R sin t  L cos t)  ce
Rt/L  sin ( t  ),   arctan . 4. This is another special case of a circuit that leads to an ODE of first order, RI  I/C  E  E0 cos t. Integration by parts gives the solution I(t)  e
t/(RC) [ et/(RC) cos t dt  c]  ce
t/(RC)  (cos t  RC sin t)  ce
t/(RC)  sin ( t  ), where tan   1/( RC). The first term decreases steadily as t increases, and the last term represents the steady-state current, which is sinusoidal. The graph of I(t) is similar to that in Fig. 62. 6. E  t2, E  2t, 0.5I
 (104/8)I  2t, I
 2500I  4t, I(0)  0 is given. I(0)  0 follows from LI(0)  Q(0) /C  E(0)  0. Answer: I  0.0016(t  0.02 sin 50t). 8. Write 1     and 2    , as in the text before Example 1. Here   R/(2L) 0, and  can be real or imaginary. If  is real, then   R/(2L) because R2  4L /C  R2. Hence 1  0 (and 2  0, of course). If  is imaginary, then Ih(t) represents a damped oscillation, which certainly goes to zero as t * . 10. E  200 cos 2t, 0.5I
 8I  10I  200 cos 2t, so that the steady-state solution is I  5 cos 2t  10 sin 2t A. 12. The ODE is I
 2I  20I  _1573 cos 3t. The steady-state solution is Ip  2 sin 3t  _11 3 cos 3t. Note that if you let C decrease, the sine term in the solution will become increasingly smaller, compared with the cosine term. 14. The ODE is 0.1I
 0.2I  0.5I  377 cos 0.5t. E0C 
1  ( RC)2 E0C  1  ( RC)2 E0  R L  R E0 
R2  2L2 E0  R2  2L2 E0  L Instructor’s Manual 51 im02.qxd 9/21/05 10:57 AM Page 51 docsity.com