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Unformatted text preview: Instructor’s Solution Manual
Elementary Differential Equations, 8th Edition
° & Elementary Differential Equations w/Boundary Value Problems
Eighth Edition Boyce & DiPrima Pizhlishtzm Slum 186? Note: This manual is currently undergoing an accuracy check. This is NOT the ﬁnal
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diverge away from the equilibrium solution y(t) 3. CHAPTER 1. ———— For g > —— 1/2 , the slopes are positive, and hence the solutions increase. For
(2; < ,— 1/2, the slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution y(t) =  1/2.
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slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution y(t) = — 2. 8. For all solutions to approach the equilibrium solution y(t) = 2/3, we must have
y’ < 0 for y > 2/3 , and y’ > 0 for y < 2/3 . The required rates are satisfied by the differential equation y’ = 2 — 3y. 9. For solutions other than y(t) = 2 to diverge from y =2 2 , y(t) must be an increasing
function for y > 2, and a decreasing function f“r < 2 . The simplest differential equation
whose solutions satisfy these criteria is y’ :2 y — 2. 10. For solutions other than y(t) = 1/3 to diverge from y = 1/3 , we must have 31’ < O
for y < 1/3, and y’ > 0 for y > 1/3. The required rates are satisfied by the differential equation y’ 2 3y —— 1. 12. 6 I} {I‘M IN! fiff ff
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0 1 213 4 5 Note that y’ 2 O for y = 0 and y z: 5. The two equilibrium solutions are y(t) = O and
y(t) = 5 . Based on the direction field, y’ > 0 for y > 5 ; thus solutions with initial page 2 5
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l CHAPTER 1. —— values greater than 5 diverge from the solution y(t) 2 5. For 0 < y < 5, the slopes are
negative, and hence solutions with initial values between 0 and 5 all decrease toward the solution y(t) 2 0. For y < 0, the slopes are all positive; thus solutions with initial values ,
less than 0 approach the solution y(t) 2 0. 14. éiéé
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5 05 1 2 t Observe that y’ 2 0 for y 2 0 and y 2 2 . The two equilibrium solutions are y(t) 2 O
and y(t) 2 2. Based on the direction ﬁeld, y’ > 0 for y > 2 ; thus solutions with initial values greater than 2 diverge from y(t) 2 2. For 0 < y < 2, the slopes are also positive,
and hence solutions with initial values between 0 and 2 all increase toward the solution y(t) 2 2. For y < O, the slopes are all negative; thus solutions with initial
values less than 0 diverge from the solution y(t) 2 0. 15~ (3') y’=2~y
17 (9) 21’: 2y. 18. (b) y’=2+y. 20. (e) y’=y(y3) 22. (a) Let M (t) be the total amount of the drug (in milligrams) in the patient's body at any
given time t (hrs) . The drug is administered into the body at a constant rate of 500 mg/ hr.
The rate at which the drug leaves the bloodstream is given by 0.4M(t) . Hence the
accumulation rate of the drug is described by the differential equation (jg—Ag“ 2 500 —— 0.4M (mg/hr). (5) page 3 ,.,,.~__,..i__._wew.w,_T,MMW, CHAPTER 1. — 'xwmnmxxxx.w~u~w~wxmwt~.\xx
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equilibrium level of 1250 mg (within a: few hours). 23. The difference between the temperature of the object and the ambient temperature is
u — 70 , u in 0F . Since the object is cooling when u > 70 , and the rate constant is k = 0.05 min“, the
governing differential equation for the temperature of the object is do
3; — —— .05(u~70) 25. (a) Following the discussion in the text, the differential equation is mdv m m U2
or equivalently,
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pd? ~ g my ' (b) After along time, {3% m 0. Hence the object attains a terminal velocity given by (c) Using the relation 7 11020 : mg, the required drag coeﬂicient is 7 = 0.0408 leg/sec. (01) page 4 itindwnwinm_m.,..~__.H_¢__E_WW.“Minnowme— CHAPTER 1. m
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which is also a solution corresponding to the initial value y(0) 2 ~— 5/ 2.
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slopes
eventually increase very rapidly in magnitude. l
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—«—..~.‘~«.x\\\ \. \ \ \ The direction field is rather complicated. Nevertheless, the collection of points at which
the slope field is zero, is given by the implicit equation y3 * 6y 2 2152 . The graph of
these points is shown below: page 6 CHAPTER 1. ——— The yintercepts of these curves are at y 2 0, d:\/_6_ . It follows that for solutions with
initial values y > \/6— , all solutions increase without bound. For solutions with initial values in the range y < — x/g and 0 < y < WG— , the slopes remain negative, and
hence these solutions decrease without bound. Solutions with initial conditions in the range
— \/5 < y < O initially increase. Once the solutions reach the'critical value, given by
the equation y3 — 6y 2 2t2 , the slopes become negative and remain negative. These
solutions eventually decrease without bound. page 7 CHAPTER 1. ——— Section 1.2 1(a) The differential equation can be rewritten as
dy
————~—— 2 dt.
5 — 2/ Integrating both sides of this equation results in  ln5 — yl 2 t + cl , or equivalently,
5 — y 2 c e4 . Applying the initial condition y(0) 2 yo results in the specification of
the constant as c 2 5 ~ yo. Hence the solution is y(t) = 5 + (310 — 5)e‘t . All solutions appear to converge to the equilibrium solution y(t) 2 5. 1(a). Rewrite the differential equation as dy __
10—2y—dt' Integrating both sides of this'equation results in — élnllO — 2y 2 t + 01 , or
equivalently, 5 —~ y 2 c €~2t . Applying the initial condition y(0) 2 yo results in the specification of
the constant as c 2 5 — yo . Hence the solution is y(t) 2 5 + (90 2 5)?” . U 2 41E» 810 All solutions appear to converge to the equilibrium solution y(t) 2 5, but at a faster rate
than in Problem la . 2(a). The differential equation can be rewritten as page 8 aM,._#Aﬂ_.*,__w_m,_m?wwwMWMWMWWMWW CHAPTER 1. ——, .11.?” 2 dt, 3/ —» 5 Integrating both sides of this equation results in lnly m 5 2 t + cl , or equivalently,
y — 5 2 c e’ . Applying the initial condition y(0) 2 yo results in the specification of
the constant as c 2 yo w 5. Hence the solution is y(t) = 5 + (yo — 5)? . All solutions appear to diverge from the equilibrium solution y(t) 2 5 . 2(1)). Rewrite the differential equation as
d
ML 2 dt_
2y w 5 Integrating both sides of this equation results in %ln]2y —— 5} 2 t + C1 , or equivalently,
2y ~— 5 2 c 6% . Applying the initial condition y(0) 2 yo results in the specification of
the constant as c 2 2310 — 5. Hence the solution is y(t) 2 2.5 + (yo 2 2.5)62t. All solutions appear to diverge from the equilibrium solution y(t) 2 2.5. 2(0). The differential equation can be rewritten as dy 2dt.
2y—10 Integrating both sides of this equation results in %ln12y — 101 2 t + 61 , or equivalently,
y — 5 2 c e” . Applying the initial condition y(0) 2 yo results in the specification of
the constant as c 2 ya — 5. Hence the solution is y(t) 2 5 + (yo — 5)e2t . page 9 ,Wjﬂdmwmﬂﬂwwam. CHAPTER 1. ———— All solutions appear to diverge from the equilibrium solution y(t) 2 5. 3(a). Rewrite the differential equation as dy
2dt,
b—ay which is valid for y 2 b /a. Integrating both sides results in film!) —— ayl 2 75+ 01 , or equivalently, I) ~— ay 2 c 6““ . Hence the general solution is y(t) 2 (b ~ c 6““) /a.
Note that if y 2 Wu , then dy/dt 2 O, and y(t) 2 b/a is an equilibrium solution. (5) (9.5}. a=‘1.b=2) 2.5 W)?
1.5 U 8.2 114 t0.8 0.8 l (i) As a increases, the equilibrium solution gets closer to y(t) 2 O , from above.
Furthermore, the convergence rate of all solutions, that is, a , also increases. (ii) As I) increases, then the equilibrium solution y(t) 2 b / 0. also becomes larger. In
this case, the convergence rate remains the same. If a and I) both increase (but b/a 2 constant), then the equilibrium solution
y(t) 2 b/a remains the same, but the convergence rate of all solutions increases. 4. The equilibrium solution satisfies the differential equation dye 2 0.
dt
Setting aye —— b 2 0 , we obtain ye(t) 2 b/a.
6(a). Consider the simpler equation dyl /dt 2 —— ayl . As in the previous solutions, re— write the equation as page 10 CHAPTER 1. ——— as
y1 = ~adt. at Integrating both sides results in y1(t) :2 e e“ .
(b). New set y(t) = y1(t) + k , and substitute into the original differential equation. We findthat
~ay1+O= ~a(y1+k)+b. That is, — ah + b = O, and hence h = b/a. (e). The general solution of the differential equation is y(t) :2 c 6““ + b/a. This is
exactly the form given by Eq. (17) in the text. Invoking an initial condition y(0) : yo ,
the solution may also be expressed as y(t) = b / a + (yo  b/a)e‘“t . 6(a). The general solution is p(t) = 900 + c em, that is, p(t) = 900 + (Po ~ 900%“2 .
With 120 = 850 , the speciﬁc solution becomes p(t) = 900 — 506t/2. This solution is a
decreasing exponential, and hence the time of extinction is equal to the number of months
it takes, say it}: , for the population to reach zero. Solving 900 ~ 5069/2 2 O , we find that t, = amass/50) = 5.78 months.
(b) The solution, p(t) : 900 + (190 *’ 900)?” , is a decreasing exponential as long as
100 < 900. Hence 900 + (190 — 900)e’*f/‘q = 0 has only one root, given by 900
t :2 2l .
f ”(%0»n) (c). The answer in part (b) is a general equation relating time of extinction to the value of
the initial population. Setting tf = 12 months, the equation may be written as 900 _ 6
900290“ 9 which has solution p0 2 897.7691 . Since p0 is the initial population, the appropriate
answer is 190 :2 898 mice. 8(a). The general solution is phi) : pg 6” . Based on the discussion in the text, time t is
measured in months. Assuming 1 month = 30 days , the hypothesis can be expressed as
130 e?"1 2 2130 . Solving for the rate constant, T :2 ln(2) , with units of per month. rN/SO (b). N days = N / 30 months. The hypothesis is stated mathematically as pee = 2120 .
It follows that rN/3O : ln(2) , and hence the rate constant is given by r 2 3O ln(2) / N .
The units are understood to be per month. 9(a). Assuming no air resistance, with the positive direction taken as downward, Newton's
Second Law can be expressed as page 11 4...4.,_,_d_~,s.j_,_w.mw_wa_w, CHAPTER 1. —— in which 9 is the gravitational constant measured in appropriate units. The equation can be
written as dv/dt 2 g, with solution 21(t) 2 gt + no . The object is released with an initial
velocity v0 . Suppose that the object is released from a height of h units above the ground. Using the
fact that v 2 dac/dt, in which a: is the downward displacement of the object, we obtain the
differential equation for the displacement as dm/dt 2 gt + 720. With the origin placed at the point of release, direct integration results in $(i) 2 gt2 / 2 —l 710 t. Based on the chosen
coordinate system, the object reaches the ground when m(t) 2 it. Let t 2 T be the time that it takes the object to reach the ground. Then 9T2 / 2 + ’UQT 2 h . Using the quadratic
formula to solve for T , "' ’UQIl: 'U() + J‘:: The positive answer corresponds to the time it takes for the object to fall to the ground. The
negative answer represents a previous instant at which the object could have been launched
upward (with the same impact speed), only to ultimately fall downward with speed 110 , from a height of h units above the ground. (0). The impact speed is calculated by substituting t 2 T into v(t) in part (a). That is, U(T) 2 t/vo + 29h . 12(a,b). The general solution of the differential equation is 6205) 2 c e‘” . Given that
62(0) 2 100 mg , the value of the constant is given by c 2 100. Hence the amount of
thorium234 present at any time is given by C20?) 2 100 6"”. Furthermore, based on the
hypothesis, setting If 2 1 results in 82.04 2 100 6”. Solving for the rate constant, we
find that r 2 — ln(82.04/100) 2 .19796/week or r 2 .02828/day. (c). Let T be the time that it takes the isotope to decay to one—half of its original amount.
From part (a), it follows that 50 2 100 e‘TT, in which 7" 2 .19796/week. Taking the
natural logarithm of both sides, we find that T 2 3.5014 weeks or T 2 24.51 days. 13. The general solution of the differential equation dQ/dt 2 —~ 1' Q is Q(t) 2 6208‘”, page 12 CHAPTER 1. —~— in which Q0 2 62(0) is the initial amount of the substance. Let r be the time that it takes
the substance to decay to onehalf of its original amount, Q0 . Setting t 2 7' in the solution,
we have 0.5 Q0 2 6206‘”. Taking the natural logarithm of both sides, it follows that — rr 2 ln(0.5) or rr 2 Z712. 14. The differential equation governing the amount of radium—226 is dQ/dt 2 — r Q ,
with solution Q(t) 2 Q(0)e“”. Using the result in Problem 11, and the fact that the
halflife 7' 2 1620 years, the decay rate is given by r 2 ln(2) / 1620 per year. The
amount of radium—226, after 75 years, is therefore Q(t) 2 Q(0)e“0‘00042786‘. Let T be
the time that it takes the isotope to decay to 3 / 4 of its original amount. Then setting t 2 T, and Q(T) 2 262(0) , we obtain 262(0) 2 Q(0)e“0‘00042786T. Solving for the decay
time, it follows that — 0.00042786T 2 ln(3/4) or T 2 672.36 years. 16. Based on Problem 15, the governing differential equation for the temperature in the
room is du
———~ 2 — .15 — 10 .
dt (u )
Setting t 2 0 at the instant that the heating system fail, the initial condition is
u(0) 2 70 °F . Using separation of variables, the general solution of the differential equation is
u(t) 2 10 + C’exp( — .15 t). Invoking the given initial condition, the temperature in the room is given by
u(t) 2 10 + 60 emp( — .15 t) . Setting Mt) 2 32, we obtain it 2 6.69 hr. 70
80 50 40
u 30
20 10 U 2 4,5 810 page 13 l
l
l
l
l
l CHAPTER 1. ——— 17. The solution of the differential equation, with 62(0) = 0, is Q(t) : C'V(l —— €_t/CR). As ta co , the exponential term vanishes, and hence the limiting value is 62,; 2 CV. 18(a). The accumulation rate of the chemical is (0.01)(300) grams per hour. At any
given time t, the concentration of the chemical in the pond is Q(t) /106 grams per gallon Consequently, the chemical leaves the pond at a rate of (3 x 10‘4)Q(t) grams per hour.
Hence, the rate of change of the chemical is given by fig? : 3 ~ 0.0003 Q(t) gm/hr. Since the pond is initially free of the chemical, 62(0) :2 0. (b). The differential equation can be rewritten as
V (£62
10000 — Q Integrating both sides of the equation results in —— ln10000 —— Q] = 0.0003t + C.
Taking the natural logarithm of both sides gives 10000 ~— Q = c 6‘0'00033. Since 62(0) 2 0, the
value of the constant is c = 10000. Hence the amount of chemical in the pond at any
time is Q(t) : 10000(1 ~— 600003“) grams. Note that 1 year 2 8760 hours. Setting t = 8760 , the amount of chemical present after one year is Q(87 60) = 9277.77 grams , that is, 9.27777 kilograms. = 0.0003 dt. (c). With the accumulation rate now equal to zero, the governing equation becomes
dQ/dt = — 0.0003 Q(t) gm/hr. Resetting the time variable, we now assign the new
initial value as 62(0) 2: 9277.77 grams. (d). The solution of the differential equation in Part (c) is Q(t) : 927 7.77 e"0'0003t.
Hence, one year after the source is removed, the amount of chemical in the pond is Q(8760) = 670.1 grams. (6). Letting t be the amount of time after the source is removed, we obtain the equation
10 =2 9277.77 6’0'0009“. Taking the natural logarithm of both sides, — 0.0003t 2' = ln(10/9277.77) or t = 22, 776 hours 2 2.6 years. (f) page 14 CHAPTER 1. —— 1 0000
8000
E3000
£1000 2000 U 2000 8000 10000 15000 18000 22000 28000 19(a). It is assumed that dye is no longer entering the pool. In fact, the rate at which the
dye leaves the pool is 200  [q(t) / 60000] kg/min 2 200(60/ 1000) [q(t) / 60] gm per hour.
Hence the equation that governs the amount of dye in the pool is dq
«~— 2 ~— . m/h .
dt 02a (g r) The initial amount of dye in the pool is q(0) 2 5000 grams. (0). The solution of the governing differential equation, with the specified initial value,
is q(t) 2 5000 8—“0'2t. (c). The amount of dye in the pool after four hours is obtained by setting 75 2 4. That is,
(1(4) 2 5000 6‘0'8 2 2246.64 grams. Since size of the pool is 60, 000 gallons, the concentration of the dye is 0.0374 grams/gallon. (d). Let T be the time that it takes to reduce the concentration level of the dye to 0.02 grams/gallon. At that time, the amount of dye in the pool is 1, 200 grams. Using
the answer in part (b), we have 5000 e’0'2T 2 1200 . Taking the natural logarithm of
both sides of the equation results in the required time T 2 7.14 hours. (6). Note that 0.2 2 200/1000. Consider the differential equation
dq 7‘ alt '“ 1000 q . Here the parameter 7" corresponds to the ﬂow rate, measured in gallons per minute.
Using the same initial value, the solution is given by q(t) 2 5000 reﬁt/1°00 . In order to determine the appropriate flow rate, set i 2 4 and q 2 1200. (Recall that 1200 gm of
dye has a concentration of 0.02 gm/gal ). We obtain the equation 1200 2 5000 (fr/250 .
Taking the natural logarithm of both sides of the equation results in the required flow rate
r 2 357 gallons per minute. page 15 L. ,, CHAPTER 1. —— Section 1.3 l. The differential equation is second order, since the highest derivative in the equation
is of order two. The equation is linear, since the left hand side is a linear function of y and
its derivatives. 3. The differential equation is fourth order, since the highest derivative of the function y
is of order four. The equation is also linear, since the terms containing the dependent
variable is linear in y and its derivatives. 4. The differential equation is ﬁrst order, since the only derivative is of order one. The
dependent variable is squared, hence the equation is nonlinear. 5. The differential equation is second order. Furthermore, the equation is nonlinear,
since the dependent variable y is an argument of the sine function, which is not a linear
function. 7. y1(t) 2 at 2> y1’(t) 2 y{’(t) 2 et. Hence yl” ~— yl 2 0».
Also, y2(t) 2 cosht 2> y1’(t) 2 sinht and y2”(t) 2 cosht. Thus 3/2” — y, 2 0. 9. y(t) 2 3t + t2 2 y’ (t) 2 3 + 2t. Substituting into the differential equation, we have
t(3 + 275) — (3t —— t2) 2 315 + 2152 ~ 3t — t2 2 t2 . Hence the given function is a solution. 10. yitt) = t/ 3 => y1/(t) = 1/ 3 and 3/1"“) = gill/(t) = 313"“) = 0 Clearly, ydt) is
a solution. Likewise, y2(t) 2 64 + t/3 2> 2 —— aft + 1/3 , y2”(t) 2 e“t , 112’” (t) 2  e‘t , yz’”’(t) 2 6” ..Substituting into the left hand side of the equation, we
find that e"t + 4( —— e“) + 3(e"t + t/3) 2 6‘75 w 4e‘t + 36”: + t 2 t. Hence both
functions are solutions of the differential equation. ll. 3/103) 2 “ti/2 2 y1’(t) 2 t‘1/2/2 and yﬂt) 2 — 23’3/2/4. Substituting into the left
hand side of the equation, we have 2t2( — 273/2/4) + 3t(t“1/2/2) 41/2 = — t1/2/2 + 3751/2/2  t”?
= 0 Likewise, y2(t) 2 t‘1 22> y2’(t) 2  it”? and yﬂt) 2 2t"3. Substituting into the left
hand side of the differential equation, we have 2152(2 t‘3) + 3t( ~— 16”) —— t‘1 2 4 t”1 ~
— 3 t"1 w t‘1 2 0 . Hence both functions are solutions of the differential equation. 12. y1(t) 2 t“2 2 y1’(t) 2 m 2t‘3 and yl”(t) 2 6t‘4. Substituting into the left hand
side of the differential equation, we have t2(6 t‘4) —+— 5t(  275*) + 4f2 2 6 t—2 — —~ 10 t"? + lit“2 2 0. Likewise, y2(i) 2 t"2lnt 2> yz’Ct) 2 t‘3 —— Zt‘glnt and y,” (t) 2  5 75—4 + 6 t‘4ln t. Substituting into the left hand side of the equation, we have
t2( ~ 5 r4 + 6t”4lnt) + 5t(t“3 2 2152317; t) + 4(t—2lnt) = —— M4 + 6 ﬁlm + page 16 CHAPTER 1. ——— + 5 t“2 —— 10 t’zln t + 4 t“2lnt 2 0 . Hence both functions are solutions of the
differential equation. 13. y(t) 2 (cos t)ln cost + t sint 2> y’(t) 2 — (sin t)ln cost + tcost and y” (t) 2 — (cos t)ln cost —— t sint + sec t. Substituting into the left hand side of the
differential equation, we have ( — (cos t)ln cost  t sint + sect) + (cos t)ln cost +
+ taint 2  (costﬂncost —— tsint + sect + (cost)lncost+ tsint 2 sect.
Hence the function y(t) is a solution of the differential equation. 15. Let y(t) 2 6”. Then y” (t) 2 r26” 7, and substitution into the differential equation
results in We“ + 2 e” 2 0. Since 6” 79 0 , we obtain the algebraic equation r2 + 2 2 0. The roots of this equation are rm 2 2E . 17. y(t) 2 a” 2> y’(t) 2 remt and y”(t) 2 r26“. Substituting into the differential
equation, we have r26” + re” ~— 6 e” 2 0. Since 6” 2 0, we obtain the algebraic
equation r2 + r — 6 2 O, that is, (r — 2)(r+ 3) 2 0. Theroots are m; 2  3, 2. 18. Let y(t) 2 6”. Then y’(t) 2 re”, y”(t) 2 7°26” and y”’(t) 2 r36”. Substituting the derivatives into the differential equation, we have r38" — 3726” + fare” 2 0. Since
6" 2 O , we obtain the algebraic equation r3 —— 37’2 + 27' 2 O . By inspection, it follows
that r(r —— 1)(r  2) 2 0. Clearly, the roots are r, 2 0 , r2 2 1 and r3 2 2. 20. y(t) 2 75" => y’(t) 2 rtf‘l and y”(t) 2 r(r — 1)t"”2. Substituting the derivatives
into the differential equation, we have t2[r(r ~ Dim] — 471M“) + 4d 2 0. After
some algebra, it followsthat r(r — 1)t’" —~ 47" t’" + 4t?" 2 0. For t 75 0, we obtain the
algebraic equation r2 — 5r + 4 2 0 . The roots of this equation are 7', 2 1 and r2 2 4. 21. The order of the partial differential equation is two, since the highest derivative, in
fact each one of the derivatives, is of second order. The equation is linear, since the left
hand side is a linear function of the partial derivatives. 23. The partial differential equation is fourth order, since the highest derivative, and in
fact each of the derivatives, is of order four. The equation is linear, since the left hand
side is a linear function of the partial derivatives. 24. The partial differential equation is second order, since the highest derivative of the
function u(m, y) is of order two. The equation is nonlinear, due to the product a  am on
the left hand side of the equation. 25. u1(m,y) 2 cosmcoshy 2:» @5321} 2 — cosxcoshy and 595% 2 cosmcoshy. 2“
2 2 y
It is evident that %8%1 + 63?? 2 0. Likewise, given u2(ac, y) 2 [71(932 + 3/2), the second
derivatives are page 17 i
Z
a
l
i
l
l
i
E
r»
r
l
l
.
l
l
l
l
. CHAPTER I. —— 821% m 2 4302 5932 "‘ $2 + y2 ($2 + 92)? z
62714 2 411/2 i ll Adding the partial derivatives, 82% + 62% __ 2 42:2 + 2 __ 4y2
3:32 ayz " $2 + yz ($2 + y2)2 x2 + 312 (x2 + y2)2
4 4(932 + 712) Hence nix, y) is also a solution of the differential equation. 27. Let u1(ac, t) = sin Am sin Aat. Then the second derivatives are 2:121 2 — AZSi‘n Am sin Aat
8;? == — A2a2sin Ax sin Aat It is easy to see that awgﬁl = 995% . Likewise, given u2(a:, t) = sin(:c ~ at) , we have 86:22 2 ~— sin(m —— at) l
(22:32 2  a28in(w — at) Clearly, 1L2 (x, t) is also a solution of the partial differential equation. 28. Given the function Mm, t) : x/W/t e”‘”2/4“2t , the partial derivatives are W 6—102/4a2t W$28—x2/4a2t Um : _ 2a2t + Zia/it2 u m __ V 7ft swig/401% + ﬁx2e”$2/4O‘2t t —" 2t2 4a2t2\/t_ 12 _ 2 —$2/4012t
It follows that a2 um = W 2 ~ W . Hence Mac, t) is a solution of the partial differential equation. page 18 CHAPTER 1. — Viimg (b). The path of the particle is a circle, therefore polar coordinates are intrinsic to the
problem. The variable r is radial distance and the angle 6 is measured from the vertical. Newton‘s Second Law states that E: F 2 ma. In the tangential direction, the equation of motion may be expressed as 2 F9 2 m a9 , in which the tangential acceleration, that is, the linear acceleration along the path is a9 = L d219/dt2 . (a9 is positive in the direction
of increasing 6 Since the only force acting in the tangential direction is the component
of weight, the equation of motion is
(120
v m sin (9 2 mL—— . 9 . di2
(Note that the equation of motion in the radial direction will include the tension in the
rod). ' I (c). Rearranging the terms results in the differential equation d26 ,
E; + ism 6 = 0.
30. (a) The kinetic energy of a particle of mass m is given by T 2 % m 112, in which i) is its speed. A particle in motion on a circle of radius L has speed L (dd/alt), where 0
is its angular position and dﬁ/dt is its angular speed. (17) Gravitational potential energy is given by V :: mgh, where h is the height above a
certain datum. Choosing the lowest point of the swing as the datum (V 2 0), it follows
from trigonometry that h 2 1 — cos 6. (c) From parts (a) and (b), page 19 TMWMMMWMWWW” l
E
t
i
i
i
i
t l
l
i CHAPTER 1. —— 1 d6 2
E = — 2 —— 1 ~
2mL (alt) +mgL( cos 0)
Applying the Chain Rule for Diﬂerentiation,
dE d6 d29 d0
u 2 L2 M —— L ' a —.
dt m dt dt2 +mg 3m dt
Setting dE/dt = O and dividing both sides of the equation by dQ/dt results in
(126
2 . *
mL (—125 +mgLsm6 — 0 , which leads to equation (12). 31. Angular Momentum is the moment (about a certain point) of linear momentum,
which is given by d6
2 L—.
mv m dt
Taking a moment about the pivot point, the angular momentum is
d6
MP 2 ng—Et— . The moment of the gravitational force (about the same point) is Mg 2 mg  Lsz‘n 6. page 20 ...
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 Spring '08
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