WileyPLUS
Theorem 3.2.6 Abel’s Theorem) 4
3.2 Solutions of Linear Homogeneous Equations; the Wronskian 153
example is to make clear that a given differential equation has more than one fundamental set of solutions; indeed, it has infinitely many; see Problem 21. As a rule, you should choose the set that is most convenient.
Now let us examine further the properties of the Wronskian of two solutions of a second order linear homogeneous differential equation. The following theorem, perhaps surprisingly, gives a simple explicit formula for the Wronskian of any two solutions of any such equation, even if the solutions themselves are not known.
154 Chapter 3. Second Order Linear Equations
Equation (26) can be solved immediately since it is both a first order linear equation (Section 2.1) and a separable equation (Section 2.2). Thus
W(t)=cexp
−
p(t)dt
, (27)
wherecis a constant. The value ofcdepends on which pair of solutions of Eq. (21) is involved. However, since the exponential function is never zero, W(t) is not zero unlessc=0, in which caseW(t)is zero for allt, which completes the proof of Theorem 3.2.6.
Note that the Wronskians of any two fundamental sets of solutions of the same differential equation can differ only by a multiplicative constant, and that the Wron- skian of any fundamental set of solutions can be determined, up to a multiplicative constant, without solving the differential equation. Further, since under the condi- tions of Theorem 3.2.6 the WronskianWis either always zero or never zero, you can determine which case actually occurs by evaluatingWat any single convenient value oft.
E X A M P L E
7
In Example 5 we verified thaty1(t)=t1/2andy2(t)=t−1are solutions of the equation
2t2y+3ty−y=0, t>0. (28)
Verify that the Wronskian ofy1andy2is given by Eq. (22).
From the example just cited we know thatW(y1,y2)(t)= −(3/2)t−3/2. To use Eq. (22), we must write the differential equation (28) in the standard form with the coefficient ofyequal to 1. Thus we obtain
y+ 3 2ty− 1
2t2y=0, sop(t)=3/2t. Hence
W(y1,y2)(t)=cexp
− 3
2tdt
=cexp
−3 2lnt
=c t−3/2. (29)
Equation (29) gives the Wronskian of any pair of solutions of Eq. (28). For the particular solutions given in this example we must choosec= −3/2.
Summary. We can summarize the discussion in this section as follows: to find the general solution of the differential equation
y+p(t)y+q(t)y=0, α <t< β,
we must first find two functionsy1 andy2 that satisfy the differential equation in α <t< β. Then we must make sure that there is a point in the interval where the WronskianW ofy1andy2 is nonzero. Under these circumstancesy1andy2form a fundamental set of solutions and the general solution is
y=c1y1(t)+c2y2(t),
wherec1andc2are arbitrary constants. If initial conditions are prescribed at a point inα <t< β, thenc1andc2can be chosen so as to satisfy these conditions.
3.2 Solutions of Linear Homogeneous Equations; the Wronskian 155
PROBLEMS
In each of Problems 1 through 6 find the Wronskian of the given pair of functions.1. e2t, e−3t/2 2. cost, sint
3. e−2t, te−2t 4. x, xex
5. etsint, etcost 6. cos2θ, 1+cos 2θ
In each of Problems 7 through 12 determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.
7. ty+3y=t, y(1)=1, y(1)=2
8. (t−1)y−3ty+4y=sint, y(−2)=2, y(−2)=1 9. t(t−4)y+3ty+4y=2, y(3)=0, y(3)= −1 10. y+(cost)y+3(ln|t|)y=0, y(2)=3, y(2)=1 11. (x−3)y+xy+(ln|x|)y=0, y(1)=0, y(1)=1 12. (x−2)y+y+(x−2)(tanx)y=0, y(3)=1, y(3)=2
13. Verify that y1(t)=t2 and y2(t)=t−1 are two solutions of the differential equation t2y−2y=0 fort>0. Then show thaty=c1t2+c2t−1 is also a solution of this equa- tion for anyc1andc2.
14. Verify thaty1(t)=1 andy2(t)=t1/2are solutions of the differential equation
yy+(y)2=0 fort>0. Then show thaty=c1+c2t1/2 is not, in general, a solution of this equation. Explain why this result does not contradict Theorem 3.2.2.
15. Show that ify=φ(t)is a solution of the differential equationy+p(t)y+q(t)y=g(t), whereg(t)is not always zero, theny=cφ(t), wherecis any constant other than 1, is not a solution. Explain why this result does not contradict the remark following Theorem 3.2.2.
16. Cany=sin(t2)be a solution on an interval containingt=0 of an equation y+p(t)y+q(t)y=0 with continuous coefficients? Explain your answer.
17. If the WronskianWoff andgis 3e4t, and iff(t)=e2t, findg(t).
18. If the WronskianWoff andgist2et, and iff(t)=t, findg(t).
19. IfW(f,g)is the Wronskian off andg, and ifu=2f−g,v=f +2g, find the Wronskian W(u,v)ofuandvin terms ofW(f,g).
20. If the Wronskian offandgistcost−sint, and ifu=f +3g,v=f−g, find the Wronskian ofuandv.
21. Assume thaty1andy2are a fundamental set of solutions ofy+p(t)y+q(t)y=0 and lety3=a1y1+a2y2andy4=b1y1+b2y2, wherea1,a2,b1, andb2are any constants. Show that
W(y3,y4)=(a1b2−a2b1)W(y1,y2).
Arey3andy4also a fundamental set of solutions? Why or why not?
In each of Problems 22 and 23 find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point.
22. y+y−2y=0, t0=0 23. y+4y+3y=0, t0=1
In each of Problems 24 through 27 verify that the functionsy1andy2are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
24. y+4y=0; y1(t)=cos 2t, y2(t)=sin 2t 25. y−2y+y=0; y1(t)=et, y2(t)=tet
156 Chapter 3. Second Order Linear Equations
26. x2y−x(x+2)y+(x+2)y=0, x>0; y1(x)=x, y2(x)=xex 27. (1−xcotx)y−xy+y=0, 0<x< π; y1(x)=x, y2(x)=sinx 28. Consider the equationy−y−2y=0.
(a) Show thaty1(t)=e−tandy2(t)=e2tform a fundamental set of solutions.
(b) Lety3(t)= −2e2t,y4(t)=y1(t)+2y2(t), andy5(t)=2y1(t)−2y3(t). Arey3(t),y4(t), andy5(t)also solutions of the given differential equation?
(c) Determine whether each of the following pairs forms a fundamental set of solutions:
[y1(t),y3(t)]; [y2(t),y3(t)]; [y1(t),y4(t)]; [y4(t),y5(t)].
In each of Problems 29 through 32 find the Wronskian of two solutions of the given differential equation without solving the equation.
29. t2y−t(t+2)y+(t+2)y=0 30. (cost)y+(sint)y−ty=0 31. x2y+xy+(x2−ν2)y=0, Bessel’s equation
32. (1−x2)y−2xy+α(α+1)y=0, Legendre’s equation
33. Show that ifpis differentiable andp(t) >0, then the WronskianW(t)of two solutions of[p(t)y]+q(t)y=0 isW(t)=c/p(t), wherecis a constant.
34. If y1 and y2 are a fundamental set of solutions of ty+2y+tety=0 and if W(y1,y2)(1)=2, find the value ofW(y1,y2)(5).
35. If y1 andy2 are a fundamental set of solutions of t2y−2y+(3+t)y=0 and if W(y1,y2)(2)=3, find the value ofW(y1,y2)(4).
36. If the Wronskian of any two solutions ofy+p(t)y+q(t)y=0 is constant, what does this imply about the coefficientspandq?
37. Iff,g, andhare differentiable functions, show thatW(fg,fh)=f2W(g,h).
In Problems 38 through 40 assume thatpandqare continuous and that the functionsy1and y2are solutions of the differential equationy+p(t)y+q(t)y=0 on an open intervalI.
38. Prove that ify1andy2are zero at the same point inI, then they cannot be a fundamental set of solutions on that interval.
39. Prove that ify1andy2have maxima or minima at the same point inI, then they cannot be a fundamental set of solutions on that interval.
40. Prove that ify1andy2have a common point of inflectiont0inI, then they cannot be a fundamental set of solutions onIunless bothpandqare zero att0.
41. Exact Equations. The equationP(x)y+Q(x)y+R(x)y=0 is said to be exact if it can be written in the form[P(x)y]+ [f(x)y]=0, wheref(x)is to be determined in terms of P(x),Q(x), andR(x). The latter equation can be integrated once immediately, resulting in a first order linear equation forythat can be solved as in Section 2.1. By equating the coefficients of the preceding equations and then eliminatingf(x), show that a necessary condition for exactness isP(x)−Q(x)+R(x)=0. It can be shown that this is also a sufficient condition.
In each of Problems 42 through 45 use the result of Problem 41 to determine whether the given equation is exact. If so, solve the equation.
42. y+xy+y=0 43. y+3x2y+xy=0 44. xy−(cosx)y+(sinx)y=0, x>0 45. x2y+xy−y=0, x>0
46. The Adjoint Equation. If a second order linear homogeneous equation is not exact, it can be made exact by multiplying by an appropriate integrating factorμ(x). Thus we require thatμ(x)be such thatμ(x)P(x)y+μ(x)Q(x)y+μ(x)R(x)y=0 can be written
3.3 Complex Roots of the Characteristic Equation 157
in the form[μ(x)P(x)y]+ [f(x)y]=0. By equating coefficients in these two equations and eliminatingf(x), show that the functionμmust satisfy
Pμ+(2P−Q)μ+(P−Q+R)μ=0.
This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general, the problem of solving the ad- joint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second order equation.
In each of Problems 47 through 49 use the result of Problem 46 to find the adjoint of the given differential equation.
47. x2y+xy+(x2−ν2)y=0, Bessel’s equation
48. (1−x2)y−2xy+α(α+1)y=0, Legendre’s equation 49. y−xy=0, Airy’s equation
50. For the second order linear equationP(x)y+Q(x)y+R(x)y=0, show that the adjoint of the adjoint equation is the original equation.
51. A second order linear equationP(x)y+Q(x)y+R(x)y=0 is said to be self-adjoint if its adjoint is the same as the original equation. Show that a necessary condition for this equation to be self-adjoint is thatP(x)=Q(x). Determine whether each of the equations in Problems 47 through 49 is self-adjoint.
3.3 Complex Roots of the Characteristic Equation
We continue our discussion of the equation
ay+by+cy=0, (1)
wherea,b, andcare given real numbers. In Section 3.1 we found that if we seek solutions of the formy=ert, thenrmust be a root of the characteristic equation
ar2+br+c=0. (2)
If the rootsr1andr2are real and different, which occurs whenever the discriminant b2−4acis positive, then the general solution of Eq. (1) is
y=c1er1t+c2er2t. (3) Suppose now thatb2−4acis negative. Then the roots of Eq. (2) are conjugate complexnumbers; we denote them by
r1=λ+iμ, r2=λ−iμ, (4)
whereλandμare real. The corresponding expressions foryare
y1(t)=exp[(λ+iμ)t], y2(t)=exp[(λ−iμ)t]. (5) Our first task is to explore what is meant by these expressions, which involve evaluat- ing the exponential function for a complex exponent. For example, if λ= −1,μ=2, andt=3, then from Eq. (5),
y1(3)=e−3+6i. (6)
158 Chapter 3. Second Order Linear Equations
What does it mean to raise the numbereto a complex power? The answer is provided by an important relation known as Euler’s formula.
Euler’s Formula. To assign a meaning to the expressions in Eqs. (5), we need to give a definition of the complex exponential function. Of course, we want the definition to reduce to the familiar real exponential function when the exponent is real. There are several ways to discover how this extension of the exponential function should be defined. Here we use a method based on infinite series; an alternative is outlined in Problem 28.
Recall from calculus that the Taylor series foretaboutt=0 is et=
∞ n=0
tn
n!, −∞<t<∞. (7)
If we now assume that we can substituteitfortin Eq. (7), then we have eit =∞
n=0
(it)n n!
=∞
n=0
(−1)nt2n (2n)! +i
∞ n=1
(−1)n−1t2n−1
(2n−1)! , (8)
where we have separated the sum into its real and imaginary parts, making use of the fact thati2= −1,i3= −i,i4=1, and so forth. The first series in Eq. (8) is precisely the Taylor series for costabout t=0, and the second is the Taylor series for sin t aboutt=0. Thus we have
eit=cost+isint. (9)
Equation (9) is known as Euler’s formula and is an extremely important mathe- matical relationship. Although our derivation of Eq. (9) is based on the unverified assumption that the series (7) can be used for complex as well as real values of the independent variable, our intention is to use this derivation only to make Eq. (9) seem plausible. We now put matters on a firm foundation by adopting Eq. (9) as the definitionofeit. In other words, whenever we writeeit, we mean the expression on the right side of Eq. (9).
There are some variations of Euler’s formula that are also worth noting. If we replacetby−tin Eq. (9) and recall that cos(−t)=costand sin(−t)= −sint, then we have
e−it=cost−isint. (10)
Further, ift is replaced by μt in Eq. (9), then we obtain a generalized version of Euler’s formula, namely,
eiμt=cosμt+isinμt. (11)
Next, we want to extend the definition of the exponential function to arbitrary com- plex exponents of the form(λ+iμ)t. Since we want the usual properties of the expo- nential function to hold for complex exponents, we certainly want exp[(λ+iμ)t]to satisfy
e(λ+iμ)t=eλteiμt. (12)
3.3 Complex Roots of the Characteristic Equation 159
Then, substituting foreiμtfrom Eq. (11), we obtain e(λ+iμ)t=eλt(cosμt+isinμt)
=eλtcosμt+ieλtsinμt. (13) We now take Eq. (13) as the definition of exp[(λ+iμ)t]. The value of the exponential function with a complex exponent is a complex number whose real and imaginary parts are given by the terms on the right side of Eq. (13). Observe that the real and imaginary parts of exp[(λ+iμ)t]are expressed entirely in terms of elementary real-valued functions. For example, the quantity in Eq. (6) has the value
e−3+6i=e−3cos 6+ie−3sin 6∼=0.0478041−0.0139113i.
With the definitions (9) and (13) it is straightforward to show that the usual laws of exponents are valid for the complex exponential function. You can also use Eq. (13) to verify that the differentiation formula
d
dt(ert)=rert (14)
holds for complex values ofr.
E X A M P L E
1
Find the general solution of the differential equation
y+y+9.25y=0, (15)
Also find the solution that satisfies the initial conditions
y(0)=2, y(0)=8, (16)
and draw its graph.
The characteristic equation for Eq. (15) is
r2+r+9.25=0 so its roots are
r1= −12 +3i, r2= −12−3i.
Therefore two solutions of Eq. (15) are
y1(t)=exp[(−12+3i)t] =e−t/2(cos 3t+isin 3t) (17) and
y2(t)=exp[(−12−3i)t] =e−t/2(cos 3t−isin 3t). (18) You can verify that the WronskianW(y1,y2)(t)= −6ie−t, which is not zero, so the general solution of Eq. (15) can be expressed as a linear combination ofy1(t)andy2(t)with arbitrary coefficients.
However, rather than using the complex-valued solutionsy1(t)andy2(t), let us seek instead a fundamental set of solutions of Eq. (15) that are real-valued. From Theorem 3.2.2 we know that any linear combination of two solutions is also a solution, so let us form the linear combinationsy1(t)+y2(t)andy1(t)−y2(t). In this way we obtain from Eqs. (17) and (18)
y1(t)+y2(t)=2e−t/2cos 3t, y1(t)−y2(t)=2ie−t/2sin 3t.
160 Chapter 3. Second Order Linear Equations
y
t 3
–1 2
1
2 4 6 8 10
FIGURE 3.3.1 Solution of the initial value problem y+y+9.25y=0, y(0)=2, y(0)=8
Dropping the multiplicative constants 2 and 2ifor convenience, we are left with
u(t)=e−t/2cos 3t, v(t)=e−t/2sin 3t (19) as real-valued solutions of Eq. (15). [If you are not completely sure thatu(t)andv(t)are solutions of the given differential equation, you should substitute these functions into Eq. (15) and confirm that they satisfy it.] On calculating the Wronskian ofu(t) andv(t), we find thatW(u,v)(t)=3e−t; thusu(t)andv(t)form a fundamental set of solutions and the general solution of Eq. (15) can be written as
y=c1u(t)+c2v(t)=e−t/2(c1cos 3t+c2sin 3t), (20) wherec1andc2are arbitrary constants.
To satisfy the initial conditions (16), we first substitutet=0 andy=2 in Eq. (20) with the result thatc1=2. Then by differentiating Eq. (20), settingt=0, andy=8, we obtain
−12c1+3c2=8, so thatc2=3. Thus the solution of the initial value problem (15), (16) is
y=e−t/2(2 cos 3t+3 sin 3t). (21)
The graph of this solution is shown in Figure 3.3.1.
From the graph we see that the solution of this problem is a decaying oscillation. The sine and cosine factors control the oscillatory nature of the solution, while the negative exponential factor in each term causes the magnitude of the oscillations to diminish as time increases.
Complex Roots; The General Case. The functionsy1(t)andy2(t), given by Eqs. (5) and with the meaning expressed by Eq. (13), are solutions of Eq. (1) when the roots of the characteristic equation (2) are complex numbersλ±iμ. Unfortunately, the solutionsy1andy2are complex-valued functions, whereas in general we would prefer to have real-valued solutions, if possible, because the differential equation itself has real coefficients. We can proceed just as in Example 1 to find a fundamental set of
3.3 Complex Roots of the Characteristic Equation 161
real-valued solutions. In particular, let us form the sum and then the difference ofy1 andy2. We have
y1(t)+y2(t)=eλt(cosμt+isinμt)+eλt(cosμt−isinμt)
=2eλtcosμt and
y1(t)−y2(t)=eλt(cosμt+isinμt)−eλt(cosμt−isinμt)
=2ieλtsinμt.
Hence, neglecting the constant multipliers 2 and 2i, respectively, we have obtained a pair of real-valued solutions
u(t)=eλtcosμt, v(t)=eλtsinμt. (22) Observe thatuandvare simply the real and imaginary parts, respectively, ofy1.
By direct computation you can show that the Wronskian ofuandvis
W(u,v)(t)=μe2λt. (23)
Thus, as long asμ=0, the WronskianWis not zero, souandvform a fundamental set of solutions. (Of course, ifμ=0, then the roots are real and the discussion in this section is not applicable.) Consequently, if the roots of the characteristic equation are complex numbersλ±iμ, withμ=0, then the general solution of Eq. (1) is
y=c1eλtcosμt+c2eλtsinμt, (24) wherec1andc2 are arbitrary constants. Note that the solution (24) can be written down as soon as the values ofλandμare known. Let us now consider some further examples.
E X A M P L E
2
Find the solution of the initial value problem
16y−8y+145y=0, y(0)= −2, y(0)=1. (25) The characteristic equation is 16r2−8r+145=0 and its roots arer=1/4±3i. Thus the general solution of the differential equation is
y=c1et/4cos 3t+c2et/4sin 3t. (26) To apply the first initial condition, we sett=0 in Eq. (26); this gives
y(0)=c1= −2.
For the second initial condition we must differentiate Eq. (26) and then sett=0. In this way we find that
y(0)= 14c1+3c2=1,
from whichc2 =1/2. Using these values ofc1andc2in Eq. (26), we obtain
y= −2et/4cos 3t+12et/4sin 3t (27) as the solution of the initial value problem (25). The graph of this solution is shown in Fig- ure 3.3.2.
162 Chapter 3. Second Order Linear Equations
10
5
–5
–10
2 4 6 8
y
t y = –2et/4 cos 3t + e1 t/4 sin 3t
2
FIGURE 3.3.2 Solution of 16y−8y+145y=0, y(0)= −2, y(0)=1.
In this case we observe that the solution is a growing oscillation. Again the trigonometric factors in Eq. (27) determine the oscillatory part of the solution, while the exponential factor (with a positive exponent this time) causes the magnitude of the oscillation to increase with time.
E X A M P L E
3
Find the general solution of
y+9y=0. (28)
The characteristic equation isr2+9=0 with the rootsr= ±3i; thusλ=0 andμ=3. The general solution is
y=c1cos 3t+c2sin 3t; (29)
y
t 2
1
–1
–2
–3 3
6 4
2 8 10
FIGURE 3.3.3 Two typical solutions ofy+9y=0.
3.3 Complex Roots of the Characteristic Equation 163
note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution. Figure 3.3.3 shows the graph of two typical solutions of Eq. (28). In each case the solution is a pure oscillation whose amplitude is determined by the initial conditions.
Since there is no exponential factor in the solution (29), the amplitude of each oscillation remains constant in time.
PROBLEMS
In each of Problems 1 through 6 use Euler’s formula to write the given expression in the form a+ib.1. exp(1+2i) 2. exp(2−3i)
3. eiπ 4. e2−(π/2)i
5. 21−i 6. π−1+2i
In each of Problems 7 through 16 find the general solution of the given differential equation.
7. y−2y+2y=0 8. y−2y+6y=0
9. y+2y−8y=0 10. y+2y+2y=0
11. y+6y+13y=0 12. 4y+9y=0
13. y+2y+1.25y=0 14. 9y+9y−4y=0 15. y+y+1.25y=0 16. y+4y+6.25y=0
In each of Problems 17 through 22 find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasingt.
17. y+4y=0, y(0)=0, y(0)=1 18. y+4y+5y=0, y(0)=1, y(0)=0 19. y−2y+5y=0, y(π/2)=0, y(π/2)=2 20. y+y=0, y(π/3)=2, y(π/3)= −4 21. y+y+1.25y=0, y(0)=3, y(0)=1 22. y+2y+2y=0, y(π/4)=2, y(π/4)= −2 23. Consider the initial value problem
3u−u+2u=0, u(0)=2, u(0)=0.
(a) Find the solutionu(t)of this problem.
(b) Fort>0 find the first time at which|u(t)| =10.
24. Consider the initial value problem
5u+2u+7u=0, u(0)=2, u(0)=1.
(a) Find the solutionu(t)of this problem.
(b) Find the smallestTsuch that|u(t)| ≤0.1 for allt>T.
25. Consider the initial value problem
y+2y+6y=0, y(0)=2, y(0)=α≥0.
(a) Find the solutiony(t)of this problem.
(b) Findαso thaty=0 whent=1.
(c) Find, as a function ofα, the smallest positive value oftfor whichy=0.
(d) Determine the limit of the expression found in part (c) asα→ ∞.
164 Chapter 3. Second Order Linear Equations
26. Consider the initial value problem
y+2ay+(a2+1)y=0, y(0)=1, y(0)=0.
(a) Find the solutiony(t)of this problem.
(b) Fora=1 find the smallestTsuch that|y(t)|<0.1 fort>T.
(c) Repeat part (b) fora=1/4, 1/2, and 2.
(d) Using the results of parts (b) and (c), plotTversusaand describe the relation between Tanda.
27. Show thatW(eλtcosμt,eλtsinμt)=μe2λt.
28. In this problem we outline a different derivation of Euler’s formula.
(a) Show that y1(t)=cost and y2(t)=sint are a fundamental set of solutions of y+y=0; that is, show that they are solutions and that their Wronskian is not zero.
(b) Show (formally) thaty=eitis also a solution ofy+y=0.Therefore
eit=c1cost+c2sint (i) for some constantsc1andc2. Why is this so?
(c) Sett=0 in Eq. (i) to show thatc1=1.
(d) Assuming that Eq. (14) is true, differentiate Eq. (i) and then sett=0 to conclude that c2=i. Use the values ofc1andc2in Eq. (i) to arrive at Euler’s formula.
29. Using Euler’s formula, show that
cost=(eit+e−it)/2, sint=(eit−e−it)/2i.
30. Ifertis given by Eq. (13), show thate(r1+r2)t =er1ter2tfor any complex numbersr1andr2. 31. Ifertis given by Eq. (13), show that
d
dtert=rert for any complex numberr.
32. Let the real-valued functions p and q be continuous on the open interval I, and let y=φ(t)=u(t)+iv(t)be a complex-valued solution of
y+p(t)y+q(t)y=0, (i)
whereuandvare real-valued functions. Show thatuandvare also solutions of Eq. (i).
Hint:Substitutey=φ(t)in Eq. (i) and separate into real and imaginary parts.
33. If the functionsy1 andy2 are a fundamental set of solutions ofy+p(t)y+q(t)y=0, show that between consecutive zeros ofy1 there is one and only one zero ofy2. Note that this result is illustrated by the solutionsy1(t)=costandy2(t)=sintof the equation y+y=0.
Hint:Suppose thatt1 andt2are two zeros ofy1between which there are no zeros ofy2. Apply Rolle’s theorem toy1/y2to reach a contradiction.
Change of Variables. Sometimes a differential equation with variable coefficients,
y+p(t)y+q(t)y=0, (i)
can be put in a more suitable form for finding a solution by making a change of the independent variable. We explore these ideas in Problems 34 through 46. In particular, in Problem 34 we show that a class of equations known as Euler equations can be transformed into equations with constant coefficients by a simple change of the independent variable. Problems 35 through