with Co

(1)

46

F F

Many important applications in mechanical and eJectrical

2 .4，立 8 ，and L9, are modeled linear可 differential

second order. Their is r可resentative of all linear ODEs as is seen when

to lin皂arODEs of third and higher ord凹 Howeve九 the solution fonnulas for second-order linear ODEs are than those so it is a natural

to ODEs of second order first in this chapter and then of order in

Although ordinary differential equations can be into linear and nonlinear ODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which many beautiful standard methods exist.

Chapter 2 includes the derivation of and so!utions^‘ the latter in connection with initial value

For those interested in solution methods for equations consult 5 and for Sturm-Liouville

是:0 輛輔臣關 T ， Numerics for second-order ODEs can be studied this See Sec. 21.3. which is of other sections in

Prerequisite: 1, in p征ticular， Sec. 1.5.

Sections that may be omitted in a shorter course: 2.10.

References and Answers to Problems: App. 1 Part and 2.

Ho

og會 F F

We have already considered first-order linear ODEs discuss linear ODEs of second order. These

and shall now define and have

applications, especially in connection with mechanical and eJectrical vibrations

2.8, 2.9) as well as in wave heat and other of ^‘ as we

shall see in Chap. 12.

A second-order ODE i8 called Ii:near if it can be written (1)

and :nonlinear if it cannot be written in this form.

The distinctive feature of this equatìon i8 that it is linear in y and its derivatives\wh巴reas

the functions p, q, and r on the right may be any given functions of x. If the begins with, say, , then divide 的 havethe standard fo.rm (1) with first term.

學習二階線性ODE 的重要性！

1.可應用於許多工程問題 2.其理論是高階ODE的基礎

y"的係數必為1(化簡成標準型式是解題的第一步驟) 學習重點：

1. 認識二階線性ODE 2. 齊次與非齊次ODE

注意標準型式

二階線性ODE 的應用領域！

(2)

SEC. 2.1 Homogeneous Linear ODEs of Second Order 尋7

The definitions of homogeneous and nonhomogenous second^<order linear ODEs are very similar to those of first-order ODEs discussed in Sec. 1.5. Indeed, if == 0 (that is, r(x) = 0 for all x considered; read "r(x) is identically then (1) reduces to (2)

and is called homogeneons. If r(抄手 0， then (1) is called nonhomogeneons. This is similar to Sec. 1.5.

An example of a nonhomogeneous linear ODE is y" + 25y = cosx, and a homogeneous linear ODE is

xy" + y' + xy = 0, written in standard form UJ + ^Vν ^-- ^AU + l-x

WJ

Finally, an example of a nonlinear ODE is

y"y + y'²= O.

The functions p and q in (1) and (2) are called the coefficients of the ODEs.

Solutions are defined similarly as for first-order ODEs in Chap. L A function y = h(x)

is called a solution of a (Iinear or nonlinear) second-order ODE on some open intervall if h is defined and twice differentiable throughout that interval and is such that the ODE becomes an identity if we replace the unknown y by h, the derivative y' h ', and the second derivative y" by 固 Examplesare given below.

的發言lr

Sections 2.1-2.6 will b巴 devoted to homogeneous linear ODEs (2) and the remaining sections of the chapter to nonhomogeneous linear ODEs.

Linear ODEs have a rich solution structure. For the homogeneous equation the backbone of this structure is the superposition principle or linearity principle^,which says that we can obtain further solutions from given ones by adding them or by multiplying them with any constants. Of course, this is a great advantage of homogen巴ouslinear ODEs. Let us first discuss an example.

Homog體的嘩。 us lin會arODEs: Sup曹rpositiCli1of 501臼tiOI'i!i

The functions ^y= cos ^xand ^y= sin ^xare solutions of the homogeneous linear ODE

y" + Y = 0

for all x. We verify this by differentiation and SUl了 stitution.We obtain (cos -cos x; henc巴 )，FF 十)'= (cos + cosx = -cosx + cosx = O.

齊次的

非齊次的

重疊原理代入(1)式中滿足，即

y=h(x)為(1)式的解

y'=h'(x) y"=h"(x)

線性原理

(必須先滿足的條件！)

(解法以後會教)

學習重點：

認識重疊原理及其適用的條件(齊次線性ODE)

(3)

48 CHAP. 2 Second-Order Linear ODEs

Simil訂lyfor y = sin x (verify!). We can go an important step furth巴T.We multiply cos x by any constant, for

lllstanc巴，4.7, and sin x by, say, -2, and take tbe sum of the results, claiming tbat it is a solution. Indeed, differentiation and substitution giv的

(4.7 cosx - 2 sinx)" + (4.7 cosx - 2 sinx) = -4.7 cosx + 2sinx + 4.7 cosx - 2 sinx = O. 11

In this example we have obtained from Yl (= cos x) and Y2 (= sin x) a func位onof the form (3) Y = cIYl 十 C2Y2 (Cl, c2 arbitrary constants).

This is ca11ed a linear combination of Yl and Y2. In terms of this concept we can now formulate the result suggested by our example, often called the superposition principle or linearity principle.

Fundamental Theorem for the Homogeneous Linear ODE (2)

For a homogeneous linear ODE (2), any linear combination oftwo solutions on an open interval 1 is again a solution of (2) on l. In particular，戶rsuch an equation, sums and constant multiples of solutions are again solutions.

P R 00 F Let Yl and Y2 be solutions of (2) on 1. Then by substituting Y = CIYl + ^{C2月 and}

its derivatives into (2), and using the familiar rule (CIYl + ^C2Y2)' = CIY~ 十 C2泊， etc., we get

y" + PY' + qy = ^(CIYl+ C2Y2)" + P(CIYl + C2Y2)' 十 q(CIYl + C2Y2)

= cIyr + C2Y~ 十 P(CIY~ 十 C2Y~) + ^q(CIYl+ ^C2Y2)

= Cl(yr + PY~ + qYl) + C2(Y~ + Py~ + qY2) = 0,

since in the last line, (. . .) = 0 because Yl and Y2 征巴 solutions ，by assumption. This shows

that Y is a solution of (2) on 1. 11

CAUTION! Don't forget that this highly important theorem holds for homogeneous linear ODEs only but does not hold for nonhomogeneous linear or nonlinear ODEs, as the following two examples illustrate.

A Nonhomogeneous Linear ODE

Verify by substitution 血的 tbefunctions y = 1 + cos x and y = 1 + sin x are solutions of tbe nonhomogeneous linear ODE

y" + Y = 1,

but their sum is not a solution. Neither 時，for instanc巴，2(1 + cos x) or 5(1 + sin x). ^圖

A Nonlinear ODE

Verify by substitution that the functions y = x²and y = 1 are solutions of 自己 nonlinearODE

y叮y 一均" = 0,

but tbeir sum is not a solution. Neither is -x², so you cannot even multiply by -1!

•

y1與y2為解

亦為解(通解)

一定要滿足

0 0

(4)

SEC. 2.1 Homogeneous Linear ODEs of Second Order ^峰會

的

Recall from Chap. 1 that for a first-order ODE, an initial value problem consists of the ODE and one initial condition y(xo) = Yo. The initial condition is used to determine the arbitrary constant c in the general solution of the ODE. This results in a unique solution, as we need it in most applications. That solution is called a particular solution of the ODE. These ideas extend to second-order ODEs as follows 固

For a second-order homogeneous linear ODE (2) an initial valu監 problemconsists of (2) and two initial conditions

(4)

These conditions prescribe given values Ko and K 1 of the solution and its first derivative (the slope of its at the same given x = Xo in the open interval considered.

The conditions (4) are used to determine the two arbitrarγconstants Cl and C2 in a general solution

(5)

of the ODE; here, Yl and Y2 are suitable solutions of the ODE, with "suitable" to be explained after the next example. This results in a unique solution, through the point (xo, Ko) with as the tangent direction (the slope) at that point That solution is called a particular soluti.on of the ODE

Initi晶I Valu曹 Probl體間

Solve the initial value problem

YFF 十 y= 0, y(O) = 3.0, y' (0) = -0.5

Solution, Step 1. General SOllltiO^Il.The functÌons cos x and sin x are solutions of the ODE (by Example 1), 組dwe take

y32

y = C1 COS x + C2 sm x

This will turn out to be a general solution as defined below

Step 2. p，αrticlllar SOllltiO^Il.We need th巳 derivativey' -C1 sin x + ('2 COS x. From this and th巴 01423

一一

initial values we obtain, sinc巴 cos 0 二 I and sin 0 = 0, y(O) = C1 = 3.0 and y' (0) 二 C2= -0.5

This gives as the solution of our initial value problem the particular solution

立會 Particu!arsolution and initial tangent in

Example 4

y ⁼ 3.0 cos - 0.5 日nx.

Figure 29 shows that at x = 0 it has the value 3.0 and the slope -0.5, 50 that its tangent inters凹的

the x

Obs啦。vati.Ol1. Our choice of Yl and Y2 was general enough to satisfy both initial conditions. Now let us take instead two proportional solutions Yl = COS x and Y2 = k CoS x, so that Yl/Y2 = l/k = const. Then we can write Y = CIYl + C2Y2 in the form

Y 二 ClCOS X + c2(k COS x) = C CoS x where C = C1 + C2k.

基底 y=c1y1+c²y2

代入

求出c1與c2

y1與y2即為 basis(基底) 兩個初始條件

，求出兩個c值代表x0點的斜率或切線方向

通解

特解 y(0)=3

y'(0)=-0.5(斜率)

兩個初始條件，但只有一個任意常數，矛盾！

y1與y2成比例關係

重要結論：y1與y2不可成比例關係，

否則重疊後的解不成立。

學習重點：

1.以初始條件(求出c值)，通解→特解 2.基底(basis)的概念，滿足基底的條件 3.如何以基底形成通解(以重疊原理)

(5)

CHAP. 2 Second-Order Linear ODEs

Hence we are no longer able to satisfy two initial conditions with only one arbitrary constant C. Consequent句，in defining the concept of a general solutìon, we must exclude proportionality. And we see at the same time why the concept of a general solution is of importance in connection with initial value problems.

A pal."ticular solution of on 1 is obtained if we assign specific values to Cl

UMM(5)

For the definition of an interval see Sec. 1.1. Furthermor皂，as usual^,Yl and Y2 are called proportional on 1 if for alJ X on 1,

(a) Yl = or Y2 二

where k and 1 are numbers, zero or not. that implies if and if k 手

Actually, we can reformulate our definition of a basis using a concept of g巴neral

importance. two functions Yl and Y2 are call巴d independent on an intervall where they are defined if

on 1 implies k1 = 0 and k2 = O.

=0

十

And Yl and Y2 are called linearly dependent on 1 if also holds for 80me constants k1, k2 not both zero. Then, if k1 ^守在 o or k2 芋 0, we can divide and see that Yl and Y2 are proportional,

的/AVV

。ρ-1i

k-K

一一吋A

VJ or Y2 = ---;--Yl.

"2

In contrast, in the case of linear可 independencethese functions are not proportional because then we cannot divide in This gives the following

If the coefficients p and q of (2) are continuous on 80me open interval 1, then (2) has a general solution. It yields the unique solution of any initial value problem (2), (4). It includes aU solutions of (2) on 1; hence has no singular solutions (solutions not obtainable from of a general solution; see also Problem Set 1.1). All this will be shown in Sec. 2.6.

通解y=c1y1+c²y2

general solution 特解

particular solution

利用初始條件求出c1與c2值

y1與y2為basis(基底)的條件：

y1與y2為線性獨立(即y1與y2不 可有比例關係)

線性獨立

線性相依

(6)

SEC. 2.1 51

區晶si車， G當時酬，聶 i P晶的ic臼 i晶『墨。i臼tiOi1

cos x and sin x in Example 4 form a basis of solutions of the ODE y" + Y = 0 for all x because their quo^tJent IS cot ^{x 芋 const}(or tan x 芋 const). H巳nc巴 y= Cl COS X + C2 sm x IS a ^g巴neral solution. Th巳 solution

y = 3.0 cos x - 0.5 sin X 01" the initial value problem is a particular solution 鰻轟轟5i盞， G告間體『晶l P晶rtk叫晶r 501闋的。關

Verify by substitution that Yl = eX and .1'2 = e -x are solutions of the ODE y" - Y = O. Th巴nsolve the initial value problem

y" - Y = 0, )'(0) = 6, /(0) = -2.

Solutùm. = 0 and (e- X)" - e- x = 0 show that eX an位 e-x are solutions. Th月! are not proportional, eXle -x = e^2x手const.Hence eX. e- x form a basis for allx. We now write down the con巳sponding g巳neralsolution and its derivative and equate th巳irvalues at 0 to the given initial conditions,

v + / = Clex C2e一弋 )'(0) = Cl + C2 = 6, = Cl - C2 = - 2。

By addition and subtraction, Cl 立， C2 = 4^‘50 that the a凡Hveris Y = 2e x + 4e -x. This is the particular solution

satisfying the two initial conditions. 喜喜

It often that one solution can be found 01' in some other way.

Then a second solution can be obtained a first-order ODE.

This Ìs called the method of 間ductiol:l.of ord創\1We first show how thi8 method works in an examDle and then in

民曹dudio詞。fOrd嘩rif 盈 Soh且tio官司 IsKnowl1. B轟51!>

Find a basis of SOlUtiOllS of the ODE

λ0" + Y = O.

Solution. lnspection shows that )'1 = x is a solution becaL間)'~ = 1 and y~ = 0, so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute

y = UYl 二 UX) y' = u'x + μ， YFF 二 u"x+ 2u' into the ODE. This gives

- x)(u 心 + 2u') 十 u) 十 UX= O.

ux and -xu cancel and we are lefl with the following ODE, which w巴 divid巴 byx, order, and simpli旬，

(文 2 一十立 -x²u' = 0, + (x - 2)u' = O.

This ODE is offirst order in v = u', namely, (x 2 - + (x - 2)v = O. Separation ofvariables and integration glves

dv x - 2 2\

一一=一一百一一-dλ=1 一一一-- -Id丸， Ix 一 l

lnlvl =lnlx 一 11 - 21n Ixl = ln~一

X

lCredited to the great mathematician JOSEPH LOUIS LAGRANGE (1736-1813), who was bom in Turin呵

。fFrench extractio且，got his first professorship when he was 19 (at th巳 MilitaryAcademy of Turin), became director of the m乳thematical section of the Berlin Academy in1766, and moved to Paris in 1787. His important

m呵。r work was in the ca1culus of variations, cclestial mechanics, g巴neralmechanics (且lécaniqueanalytique,

Paris, 1788)^‘ differentia! equations, approxin1ation theory 司 a!gebra，and number theory

階數化簡(二階ODE → 一階ODE)

注意解題步驟！

代入

化簡

一階ODE，求v 學習重點：

1. y1已知 → 如何求出y2? 2. 階數化簡(二階ODE→一 階ODE)

(原理推導在下頁)

二階ODE降階步驟：

1. y1已求得；

2. 令y2=uy1代入化簡；

3. 令u'=U代入得一階 ODE；

4. 分離變數法求出U；

5. u= ∫Udx；

6. u代入求y2=uy1。

(7)

豆豆 CHAP. 2 Second-Order Linear ODEs

We need no constant of integration because we want to obtain a particular solution; simi!arly in the next integration. Taking exponents and integrating again, we obtain

:( - 1

v 寸= 予 u = I v dλ= ln Ixl + -'-, hence Y2 二附 =x In Ixl + 1

x x r x

Sinc己 Yl= x and Y2 = x In Ixl + 1 are linearly indep巴ndent(their quotient is not constant), we have obtuined

a basis of soiutio肘，vaìid for aìl positive x﹒體

In this example we applied reduction of order to a homogeneous linear ODE [see (2月

十+ = O.

Note that we now take the ODE in standard form, with ， not 炒作出isis essential in applying our formulas. We assume a solution Yl of on an open interval

!, to be known and want to find a basis. For this we need a second Jinem勾 independent

solution Y2 of on 1. To get Y2, we substitute

Y = Y2 = uYl, Y 二 Y2 = ⁺ ^一一

斤。vd

一一 + 2u'y~ 十

into This gives

(8) + ^{F 一←} ^{+十 uY~)} + qUYl = 0

Collecting terms in , u', and u, we have

十 +十 +Pyi + = O.

Now comes the main Since Yl is a soJution of the expression in the last parentheses is zero. Hence u is gone，日lndwe are left with an ODE in u' and u^/l.We divide this remaining ODE Yl and set u' u" 二 u'，

+ PYl

+u 一一一一一一一= O. thus

Yl 十(羿+← O

This is the desired first-order OD且 the reduced ODE. Separation of variabJes and

mt巴gratlOnglVes

才= -(守十你別叫= ^-21nI九1- fp 此

By taking we finally obtain

(9)

Here U = u'^,⁵⁰th前 u = f U dx. Hence the de吋巴dsecond solution is Y2 = YI^U ^二 Yl

f

^{U dx}

The quotient = u = f U dx cannot be constant (since U > so that Yl and Y2 form a basis of solutions.

由v積分求u 求y2=uy1=ux

二階ODE降階步驟：

1. y1已求得；

5. u= ∫Udx；

6. u代入求y2=uy1。

0

降階

分離變數法求Ｕ

y1已知

代入標準型式下的P(x)

(8)

SEC. 2.2 Homogeneous linear ODEs with Constant Coefficients 53

巨星星DUC1鈍。關 OF ORDE良心 important b巳caus巴 It

gives a simpler ODE. A general second-order ODE F伏，y, y' , Y ") = 0, linear or not, can be reduced to first order if y does not occur 巴xplicitìy(Prob. ì) or if x does not occur explicitly (Prob. 2) or if the ODE is homogeneous linear and we know a solution (see the text)

1. Reduction. Show that F 紋， yly 叮= 0 can be reduced to first order in z = y' (from which y follows by integration). Give two examples of your own.

2. Redudion. Show that F(y, y', y叮=0 can be reduced to a first-order ODE with y as the independent variable and = (dzldy)乏， where z = y'; derive this by the chain rule. Give two examples.

[3=10J 臨之間101'1 OF ORD臨

Reduce to first order and solv巴，showing each step in detail.

3.YFF-yF 二 O

4. 旬，

5.

聶+ +λy = 0, Yl = (cos x)lx 7. + y'3 cos Y = 0

8. = 1 + y'²

9. + = 0, Yl = x² 10. 十(1 + lly)y'2 = 0

111-141 A押岫'TlONS OF 蠅。UCi屬U:ODlEs 11. Cm:ve. Find the curve through the origin in the

which satisfies and whose tangent at the origin has slope 1.

12. Hanging cable. It can be shown that the curve y(x) of an inextensible f1exible homogeneous cable hanging between two fixed points is obtained by solving

YFF=KU工芋， wh巴削heco削叫“叩endso叫le weight. This curve is called catenary (from Latin catena = the chain). Find 組dgraph y(x), assuming that k = I and those fixed points are (-1 多 0)and (1,0) in a vertìcal 刀一plane.

13. 闊的ion. 匠 in the motion of a small body on a

5虹 aight lin臣，the sum of velocity and acceleration equal s a positive constant, how will the distance y(t) depend on the initial velocity and position?

目。 Motion. In a straight-lin巴 motion，let the velocity be the reciprocal of the acceJeration. Find the distance y(t) for arbitrary initial position and velocity.

115-1~ G酬監MlSOlUT酬.11'11胡亂 VAlUE

PRO鷗lE輔

(More in the next set.) (a) Verify that the functions are linearly independent and form a basis of solutions of the given ODE. (0) Solve the IVP. Graph or sketch the solution.

15. + = 0, y(O) = 2; y' (0) = --1, cos 3x, sin 3x I盾+ +y 三 0， y(O) = 2, 1,

E txEX

17. 4x:ly" - 3y = 0, y(l) = -3, = 0, X- 1/ 2

18.λ?F 十 y = 0, = 1; ) = 2, x,xln (x)

1民 yll + + = 0, y(O) = 0, = 15, e -x COS x, e -x sin x

20. CAS PROJECT. Line晶宮 Independel!í且ce. Write a program for testing Iinear independenc扭扭d depen- dence. Try it out on some of the problems in this and the next problem set and on examples of your own.

Hor芳芳。審會問

with Co

臨的時會議 r

DEs

C晶體ffici 慰問ts

We shall now consider second-order homogeneous linear ODEs whose coefficients a and b are constant,

These equations have important applications in mechanical and 巳lectrical vibrations, as we shall see in S巳cs. 2.4, 2.8, and 2.9.

To solve (1), we recall from Sec. 1.5 that the solution of the first-order Iinear ODE with a constant coefficient k

y' + = 0

注意標準型式

1.分離變數法或

2.λ+k=0故λ=-k 學習重點：

1. 求解二階線性ODE (具常數係數) 2. 何謂特性方程式與其三種形態特性根 (兩相異實根、兩重根、共軛根)

3. 三種特性根所對應之三種通解格式

令代入

微分運算子(2.3節)：

1. Dy = y' = dy/dx, D²y = y"

2. (D²+aD+bI)y = y"+ay'+by

(9)

CHAP. 2 Second-Order Linear ODEs

is an exponential function Y 二 ce-^kx.This us the idea to try as a solution of (1) the function

(2)

一一 ρlν M VJ

Substituting and its derivatives

YF=λeA:J; and into our equation (1), we obtain

(λ2 +αλt 二 O.

Hence if À. is a solution of the characteristic (3)

then the exponential function is a solution of the ODE 因 Nowfrom

that the roots of this are

we recall

寸 (-G+ 必亡 ^=主 ^{-y宗三豆豆)}

and wiil be basic because our derivation shows that the functions Yl = ^{e' ‘主砂} and Y2 二

are solutions of 因 this

From algebra we further know that the roots, U\.， IA.lll" 只 onthe of th巴 discriminant

In this case, a basis of solutions of on any interval is Yl = e…1"" and Y2 =

becaus巴 Yl and Y2 are defined general solution is

is not constant. The for all and their

(1)式的解以y1與y2為basis(基底)，

形成通解y=c1y1+c2y2

求解λ1與λ2 先假設為(1)式的解

特性方程式輔助方程式

不可能為0

討論三種λ根的情況

當作基底(basis)

重要形式兩相異根λ₁與λ₂

重根λ₁= λ2 = -a/2 共軛根λ₁, λ2 = -a/2 ± iω

(10)

SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 55

G曹詞曹『晶I 501l.ltlol'l in tI嘻嘻 Ca串串 ofDistinc::t Re晶IRoots

We can now solve y" - y 二 o in Example 6 of Sec. 2.1 systematical旬. Th巴 characteristìc 巳quation is λ2 1 = O. Its roots ar巳 λ1= 1 and λ2 =一 1. Henc巴 abasis of solutìons is e^Xand e -x and gives the same

g巴n巴ral solutìo且 as b 巴 fore ，

y = cle'" 十 C2e ^山團

Initi晶 IV晶 11.1睡 Problemin the C晶晶睡 ofDistind R曹õilRoots

Solv巴 theinitìal value problem

y" + / - 2y = 0, y(O) = 4, /(0) 二 5.

Solution. Step 1. General solution. The characteristic equatìon is λ2+λ-2 = O.

Its roots ar 巳

Àl =主(-] +丸。) = I and λ2 寸 (-1 一向= -2 so that we obtain the gen巴ral solutìo且

y = Cl^ex ^十C2e-2x

Step 2. Particular solution. Sinc巴 v 引力 -2c2e -2X, we obtain from the general solution and the initial conditions

y(O) = Cl + ^C2= 4, / (0) = Cl - 2C2 = -5

Hence Cl = 1 and C2 = 3. This gives theαnswer y = eX + 3e -2x, Figur巴 30 shows that the curve begins at y = 4 with a negative slope (-5, but note that the axes have ditferent scales!), in agreement with the initial

condìtions 國

:L/

0." o 0.5 1.5 2 x 10. Solution in Example 2

@

i i

If the discriminant a2 - 4b is zero, we see directly from that we get only one root,

λ=λ1=λ2 = -a/2, hence only one solution,

一(日/2)x

Yl = ^e

To obtain a second independent solution Y2 (n巴ededfor a basis), we use the method of reduction of order discussed in the last section, setting Y2 = uYl' Substituting this and its derivatives Y; 二 U'^Y1+ uy~ and Y~ into (1), we fi叫 have

(U"y^l+ ^2u'A + uy1) 十 α(u'Yl + uy~) + bu九三 O.

通解，以初始值求出c1與c2

特解

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5盛 CHAP. 2 Second國 OrderLinear ODEs

Collecting terms in u", u', and u，的 in the last section, we obtain u 叮1 + u'(2A + aYl) 十 μ(y'{ + ay~ + bYl) = o.

The expression in the last parentheses is zero, since Yl is a solution of (1). The expression in the first parentheses is zero, too, since

2y~ = -ae 一間12 = -aYl

執1e are thus left with u" Yl = O. Hence u" = O. By two integrations, u = clx + c2' To get a second independent solution Y2 = uYb we can simply choose cl = 1, C2 = 0 and take U = x. Then Y2 = XYl' Since these solutions are not proportional, they form a basis Hence in the case of a double root of (3) a basis of solutions of (1) on any interval is

。xl2 日x/2

xe

The corresponding general solution is

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WARNING! Ifλis a simple root of (4), then 十 C2x)eÀX with C2 弓亡。 1日not a solution of (1).

G鹽的體ral 501閣制制1in 1::11曹 C晶事曹 of 晶l)ol.lbl由民001::

The characteristic equation of the ODE y" + 6y' + 9y = 0 isλ2 + 6λ+9=(λ+ 3)2 = O. It has the double rootλ-3. Henc巴 abasis is e -3x and xe-3x. Th思 correspondinggeneral solution is y = (Cl + 鸚 i岡 itial V晶岫曹 Probl體mi的 th曹 C晶晶體 01'晶I)ol.lbl曹 Root

Solve the initial value problem

y" + y' + 0.25y = 0, y(O) = 3.0, y' (0) = -3.5

Solution. The characteristic equation i日 λ2+λ 十 0.25= (λ+ 0.5) 2 = O. It has the double rootλ= -0.5 。 This gives the general solution

一 O.5x y 二 (Cl + c2x)e We need its derivative

y' = C2e -O.5x - 0.5(Cl + C2λ)e-^O.^5X. From this and the initial conditions we ^obtai日

y(O) = Cl = 3.0, y' (0) = C2 - 0.5Cl = 3.5; hence C2 = -2.

Th巴 particularsolution of the initial value pτoblem is y = (3 - 2x)e -O.5x. See Fig. 31. ^國

γνquqrι

O

-1

31. Solution in Examp!e 4

0 0

因為

重要形式當作基底(basis) 二階ODE降階步驟：

1. y1已求得；

5. u= ∫Udx；

6. u代入求y2=uy1。

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51 Homogeneous Linear ODEs with Constant Coefficients

SEC. 2.2

-is

This case occurs if the discrimin且nta^{2 -} 4b of the characteristic equation (3) is negative.

In this case, the roots of (3) are the co即lexλ=iG±iωthat^{give the}complex叫ltlOns

of the ODE (1). However, we will show that we can obtain a basis of real solutions

-is ^十 P

e

j j

(ω > 0) where ω2 = b - !a^{2 .}^Itcan be verified by subst削lOn 出atthese are solutions in the present case. We shall derive them systematically aft凹. the two examples using the complex exponential function. They form a basis on any interval since their quotient cot ω'x is not constant. Hence a real general solution in Case 111 is

Y2 2 = = e e -axl2 ~....- smωx si

一 α xl2

Yl = e cos ω'x ，

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B arbitrary).

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Comple耳展。的囂.I l'i iti晶I Valu曹 Problem

Solv巳 th巴 initialvalue problem

y' (0) = 3

Solution. Step 1圖 Generalso[utiofl. The charact凹的tic 巴quatlOn 18λ2 + 0 .4λ+ 9.04 = O. It has the roots

0.2 土 3i ，Hence ÚJ = 3, and a general solution (9) is

y(O) 二 O.

y" + 0.4y' + 9.04y ⁼ 0,

y = e-^O.²:屯 (Acos 3x + ^Bsi且 3x).

Step 2. p，αrticular solution, The first initial condition gives y(O) = A 二 O. Th巴 remaining expression is y = Be -O.2x sin 衍.We need the 出口vative(chain rule!)