WileyPLUS
Theorem 3.2.2 Principle of Superposition)
Theorem 3.2.2
(Principle of Superposition)Ify1andy2are two solutions of the differential equation (2), L[y] =y′′+p(t)y′+q(t)y=0,
then the linear combination c1y1+c2y2 is also a solution for any values of the constantsc1andc2.
2A proof of Theorem 3.2.1 can be found, for example, in Chapter 6, Section 8 of the book by Coddington listed in the references at the end of this chapter.
148 Chapter 3. Second Order Linear Equations
A special case of Theorem 3.2.2 occurs if eitherc1orc2is zero. Then we conclude that any constant multiple of a solution of Eq. (2) is also a solution.To prove Theorem 3.2.2, we need only substitute
y=c1y1(t)+c2y2(t) (7) foryin Eq. (2). By calculating the indicated derivatives and rearranging terms, we obtain
L[c1y1+c2y2] = [c1y1+c2y2]′′+p[c1y1+c2y2]′+q[c1y1+c2y2]
=c1y′′1+c2y′′2+c1py′1+c2py′2+c1qy1+c2qy2
=c1[y′′1+py′1+qy1] +c2[y′′2+py′2+qy2]
=c1L[y1] +c2L[y2].
Since L[y1] =0 and L[y2] =0, it follows that L[c1y1+c2y2] =0 also. Therefore, regardless of the values ofc1andc2,yas given by Eq. (7) satisfies the differential equation (2), and the proof of Theorem 3.2.2 is complete.
Theorem 3.2.2 states that, beginning with only two solutions of Eq. (2), we can construct an infinite family of solutions by means of Eq. (7). The next question is whether all solutions of Eq. (2) are included in Eq. (7) or whether there may be other solutions of a different form. We begin to address this question by examining whether the constantsc1andc2in Eq. (7) can be chosen so as to satisfy the initial conditions (3). These initial conditions requirec1andc2to satisfy the equations
c1y1(t0)+c2y2(t0)=y0, c1y′1(t0)+c2y′2(t0)=y′0. (8) The determinant of coefficients of the system (8) is
W = 33 33 3
y1(t0) y2(t0) y′1(t0) y′2(t0) 33 33
3=y1(t0)y′2(t0)−y′1(t0)y2(t0). (9) IfW ̸=0, then Eqs. (8) have a unique solution(c1,c2)regardless of the values of y0andy′0. This solution is given by
c1= y0y′2(t0)−y′0y2(t0)
y1(t0)y′2(t0)−y′1(t0)y2(t0), c2= −y0y′1(t0)+y′0y1(t0)
y1(t0)y′2(t0)−y′1(t0)y2(t0), (10) or, in terms of determinants,
c1= 33 33 3
y0 y2(t0) y′0 y′2(t0) 33 33 3 3
33 33
y1(t0) y2(t0) y′1(t0) y′2(t0) 33 33 3
, c2= 33 33 3
y1(t0) y0 y′1(t0) y′0 33 33 3 3
33 33
y1(t0) y2(t0) y′1(t0) y′2(t0) 33 33 3
. (11)
With these values for c1 andc2, the linear combination y=c1y1(t)+c2y2(t) satis- fies the initial conditions (3) as well as the differential equation (2). Note that the denominator in the expressions forc1andc2is the nonzero determinantW.
On the other hand, if W =0, then the denominators appearing in Eqs. (10) and (11) are zero. In this case Eqs. (8) have no solution unlessy0andy′0have values that also make the numerators in Eqs. (10) and (11) equal to zero.Thus,ifW =0,there are many initial conditions that cannot be satisfied no matter howc1andc2are chosen.
3.2 Solutions of Linear Homogeneous Equations; the Wronskian 149
The determinantWis called theWronskian3determinant,or simply theWronskian, of the solutionsy1andy2. Sometimes we use the more extended notationW(y1,y2)(t0) to stand for the expression on the right side of Eq. (9), thereby emphasizing that the Wronskian depends on the functionsy1andy2, and that it is evaluated at the point t0. The preceding argument establishes the following result.Theorem 3.2.3
Suppose thaty1andy2are two solutions of Eq. (2) L[y] =y′′+p(t)y′+q(t)y=0, and that the initial conditions (3)y(t0)=y0, y′(t0)=y′0
are assigned. Then it is always possible to choose the constantsc1,c2so that y=c1y1(t)+c2y2(t)
satisfies the differential equation (2) and the initial conditions (3) if and only if the Wronskian
W =y1y′2−y′1y2 is not zero att0.
E X A M P L E
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In Example 2 of Section 3.1 we found thaty1(t)=e−2t andy2(t)=e−3t are solutions of the differential equation
y′′+5y′+6y=0.
Find the Wronskian ofy1andy2.
The Wronskian of these two functions is W =
33
33 e−2t e−3t
−2e−2t −3e−3t 33
33=−e−5t.
SinceWis nonzero for all values oft, the functionsy1andy2can be used to construct solutions of the given differential equation, together with initial conditions prescribed at any value oft.
One such initial value problem was solved in Example 3 of Section 3.1.
The next theorem justifies the term “general solution” that we introduced in Section 3.1 for the linear combinationc1y1+c2y2.
Theorem 3.2.4
Suppose thaty1andy2are two solutions of the differential equation (2), L[y] =y′′+p(t)y′+q(t)y=0.Then the family of solutions
y=c1y1(t)+c2y2(t)
with arbitrary coefficientsc1andc2includes every solution of Eq. (2) if and only if there is a pointt0where the Wronskian ofy1andy2is not zero.
3Wronskian determinants are named for Jósef Maria Hoëné-Wronski (1776–1853), who was born in Pol- and but spent most of his life in France. Wronski was a gifted but troubled man, and his life was marked by frequent heated disputes with other individuals and institutions.
150 Chapter 3. Second Order Linear Equations
Letφbe any solution of Eq. (2). To prove the theorem, we must determine whether φ is included in the linear combinations c1y1+c2y2. That is, we must determine whether there are values of the constants c1 andc2 that make the linear combi- nation the same asφ. Lett0be a point where the Wronskian ofy1andy2is nonzero.Then evaluateφandφ′at this point and call these valuesy0andy′0, respectively; thus y0=φ(t0), y′0=φ′(t0).
Next, consider the initial value problem
y′′+p(t)y′+q(t)y=0, y(t0)=y0, y′(t0)=y′0. (12) The functionφis certainly a solution of this initial value problem. Further, because we are assuming that W(y1,y2)(t0) is nonzero, it is possible (by Theorem 3.2.3) to choosec1 andc2 such that y=c1y1(t)+c2y2(t) is also a solution of the initial value problem (12). In fact, the proper values ofc1andc2are given by Eqs. (10) or (11). The uniqueness part of Theorem 3.2.1 guarantees that these two solutions of the same initial value problem are actually the same function; thus, for the proper choice ofc1andc2,
φ(t)=c1y1(t)+c2y2(t), (13) and thereforeφis included in the family of functionsc1y1+c2y2. Finally, sinceφis an arbitrarysolution of Eq. (2), it follows thateverysolution of this equation is included in this family.
Now suppose that there is no point t0 where the Wronskian is nonzero. Thus W(y1,y2)(t0)=0 no matter which pointt0is selected. Then (by Theorem 3.2.3) there are values ofy0andy′0such that the system (8) has no solution forc1andc2. Select a pair of such values and choose the solutionφ(t)of Eq. (2) that satisfies the initial condition (3). Observe that such a solution is guaranteed to exist by Theorem 3.2.l.
However, this solution is not included in the familyy=c1y1+c2y2. Thus this linear combination does not include all solutions of Eq. (2) ifW(y1,y2)=0. This completes the proof of Theorem 3.2.4.
Theorem 3.2.4 states that,if and only if theWronskian ofy1andy2is not everywhere zero, then the linear combinationc1y1+c2y2 contains all solutions of Eq. (2). It is therefore natural (and we have already done this in the preceding section) to call the expression
y=c1y1(t)+c2y2(t)
with arbitrary constant coefficients thegeneral solution of Eq. (2). The solutions y1andy2are said to form afundamental set of solutionsof Eq. (2) if and only if their Wronskian is nonzero.
We can restate the result of Theorem 3.2.4 in slightly different language: to find the general solution, and therefore all solutions, of an equation of the form (2), we need only find two solutions of the given equation whose Wronskian is nonzero.
We did precisely this in several examples in Section 3.1, although there we did not calculate the Wronskians. You should now go back and do that, thereby verifying that all the solutions we called “general solutions” in Section 3.1 do satisfy the nec- essary Wronskian condition. Alternatively, the following example includes all those mentioned in Section 3.1, as well as many other problems of a similar type.
3.2 Solutions of Linear Homogeneous Equations; the Wronskian 151
E X A M P L E
4
Suppose thaty1(t)=er1tandy2(t)=er2tare two solutions of an equation of the form (2). Show that they form a fundamental set of solutions ifr1̸=r2.
We calculate the Wronskian ofy1andy2: W =
33
33er1t er2t r1er1t r2er2t
33
33=(r2−r1)exp[(r1+r2)t].
Since the exponential function is never zero, and since we are assuming thatr2−r1̸=0, it follows thatW is nonzero for every value oft. Consequently,y1andy2form a fundamental set of solutions.
E X A M P L E
5
Show thaty1(t)=t1/2andy2(t)=t−1form a fundamental set of solutions of
2t2y′′+3ty′−y=0, t>0. (14) We will show how to solve Eq. (14) later (see Problem 34 in Section 3.3). However, at this stage we can verify by direct substitution thaty1andy2are solutions of the differential equation. Sincey′1(t)= 12t−1/2andy′′1(t)=−14t−3/2, we have
2t2(−14t−3/2)+3t(12t−1/2)−t1/2=(−12+32−1)t1/2=0.
Similarly,y′2(t)=−t−2andy′′2(t)=2t−3, so
2t2(2t−3)+3t(−t−2)−t−1=(4−3−1)t−1=0.
Next we calculate the WronskianWofy1andy2: W=
33
33 t1/2 t−1
1
2t−1/2 −t−2 33
33=−32t−3/2. (15)
SinceW ̸=0 fort>0, we conclude thaty1andy2form a fundamental set of solutions there.
In several cases we have been able to find a fundamental set of solutions, and therefore the general solution, of a given differential equation. However, this is often a difficult task, and the question arises as to whether a differential equation of the form (2) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question.
Theorem 3.2.5
Consider the differential equation (2),L[y] =y′′+p(t)y′+q(t)y=0,
whose coefficientspandqare continuous on some open intervalI. Choose some pointt0inI. Lety1be the solution of Eq. (2) that also satisfies the initial conditions
y(t0)=1, y′(t0)=0,
and lety2be the solution of Eq. (2) that satisfies the initial conditions y(t0)=0, y′(t0)=1.
Theny1andy2form a fundamental set of solutions of Eq. (2).
152 Chapter 3. Second Order Linear Equations
First observe that theexistenceof the functionsy1andy2is ensured by the existence part of Theorem 3.2.1. To show that they form a fundamental set of solutions, we need only calculate their Wronskian att0:W(y1,y2)(t0)= 33 33 3
y1(t0) y2(t0) y′1(t0) y′2(t0) 33 33
3=
33 331 0
0 1 33 33=1.
Since their Wronskian is not zero at the pointt0, the functionsy1andy2do form a fundamental set of solutions, thus completing the proof of Theorem 3.2.5.
Note that the potentially difficult part of this proof, demonstrating the existence of a pair of solutions, is taken care of by reference to Theorem 3.2.1. Note also that Theorem 3.2.5 does not address the question of how to find the solutionsy1andy2
by solving the specified initial value problems. Nevertheless, it may be reassuring to know that a fundamental set of solutions always exists.
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Find the fundamental set of solutionsy1andy2specified by Theorem 3.2.5 for the differential equation
y′′−y=0, (16)
using the initial pointt0=0.
In Section 3.1 we noted that two solutions of Eq. (16) arey1(t)=et andy2(t)=e−t. The Wronskian of these solutions is W(y1,y2)(t)=−2̸=0, so they form a fundamental set of solutions. However,they are not the fundamental solutions indicated byTheorem 3.2.5 because they do not satisfy the initial conditions mentioned in that theorem at the pointt=0.
To find the fundamental solutions specified by the theorem, we need to find the solutions satisfying the proper initial conditions. Let us denote byy3(t)the solution of Eq. (16) that satisfies the initial conditions
y(0)=1, y′(0)=0. (17)
The general solution of Eq. (16) is
y=c1et+c2e−t, (18)
and the initial conditions (17) are satisfied ifc1=1/2 andc2=1/2. Thus y3(t)= 12et+12e−t=cosht.
Similarly, ify4(t)satisfies the initial conditions
y(0)=0, y′(0)=1, (19)
then
y4(t)= 12et−12e−t=sinht.
Since the Wronskian ofy3andy4is
W(y3,y4)(t)=cosh2t−sinh2t=1,
3.2 Solutions of Linear Homogeneous Equations; the Wronskian 153
these functions also form a fundamental set of solutions, as stated by Theorem 3.2.5. Therefore, the general solution of Eq. (16) can be written as
y=k1cosht+k2sinht, (20)
as well as in the form (18). We have usedk1andk2 for the arbitrary constants in Eq. (20) because they are not the same as the constants c1 andc2 in Eq. (18). One purpose of this example is to make it 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.
In the next section we will encounter equations that have complex-valued solu- tions. The following theorem is fundamental in dealing with such equations and their solutions.
Theorem 3.2.6
Consider again the equation (2),L[y] =y′′+p(t)y′+q(t)y=0,
wherepandqare continuous real-valued functions. Ify=u(t)+iv(t)is a complex- valued solution of Eq. (2), then its real part u and its imaginary part vare also solutions of this equation.
To prove this theorem we substituteu(t)+iv(t)foryinL[y], obtaining
L[y] =u′′(t)+iv′′(t)+p(t)[u′(t)+iv′(t)] +q(t)[u(t)+iv(t)]. (21) Then, by separating Eq. (21) into its real and imaginary parts (and this is where we need to know thatp(t)andq(t)are real-valued), we find that
L[y] =u′′(t)+p(t)u′(t)+q(t)u(t)+i[v′′(t)+p(t)v′(t)+q(t)v(t)]
=L[u](t)+iL[v](t).
Recall that a complex number is zero if and only if its real and imaginary parts are both zero. We know thatL[y] =0 becauseyis a solution of Eq. (2). Therefore, L[u](t)=0 andL[v](t)=0 also; consequently,uandvare also solutions of Eq. (2), so the theorem is established. We will see examples of the use of Theorem 3.2.6 in Section 3.3.
Incidentally, the complex conjugateyof a solution yis also a solution. This is a consequence of Theorem 3.2.2 sincey=u(t)−iv(t)is a linear combination of two solutions.
Now let us examine further the properties of the Wronskian of two solutions of a second order linear homogeneous differential equation. The following the- orem, 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.