2ndrder ch03 1 pr a - marmara...
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![Page 1: 2ndrder ch03 1 pr a - Marmara Üniversitesimimoza.marmara.edu.tr/~msakalli/DE255Fall2013/2ndrder_ch03_1_p… · 1 DE 255 Fall 2013 DE-2013 Dr. M. Sakalli DE 255 Fall 2013 Ch 3.1:](https://reader035.vdocuments.net/reader035/viewer/2022070812/5f0af14e7e708231d42e19a3/html5/thumbnails/1.jpg)
1
DE 255 Fall 2013
DE-2013
Dr. M. Sakalli
DE 255 Fall 2013
Ch 3.1: Second Order Linear Homogeneous Equations with Constant Coefficients
A second order ordinary differential equation has the general form
where f is some given function.This equation is said to be linear if f is linear on y and y':
Give emphasis on superposition of linear parts. If G(t) = 0 for all t, then the equation is called homogeneous. Otherwise the equation is nonhomogeneous.IVP
),,( yytfy ′=′′
ytqytptgy )()()( −′−=′′ )()()()( tGytRytQytP =+′+′′
0=+′+′′ cyybya ( ) ( ) 0000 , ytyyty ′=′=
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DE 255 Fall 2013
Example: Infinitely Many Solutions Linear DE, two solutions, and combinations of these two in any numbers, and
Characteristic Eq. three possible cases.
,02 =++ cbrar
0=−′′ yy tt etyety −== )(,)( 21
tttt eetyetyety −− +=== 53)(,5)(,3)( 543
tt ececty −+= 21)(
1)0(,3)0(,0 =′==−′′ yyyy 1,21)0(3)0(
2121
21 ==⇒⎭⎬⎫
=−=′=+=
ccccyccy
,0=+′+′′ cyybya 02 =++ rtrtrt cebreear
aacbbr
242 −±−
=
DE 255 Fall 2013
2/323)( tety −−=
204.2;1542.6/77602118)( 3232
≈≈==>=
===>=+−= −−−−
yteeeeety
t
ttset
tt
tt eety 34
71
71)( +
−= −
1)0(,0)0(,012 =′==−′+′′ yyyyy
( )( ) 034012)( 2 =−+⇔=−+⇒= rrrrety rt
( ) 032032)( 2 =+⇔=+⇒= rrrrety rt
( ) ( ) 30,20,065 =′==+′+′′ yyyyy
( ) ( ) 30,10,032 =′==′+′′ yyyy
( )( ) 032065)( 2 =++⇔=++⇒= rrrrety rt
2/323)( tety −−=
tt eety 32 79)( −− −=
Check these examples for the case where roots are different.
this question asks t when y(t) reaching to zero..
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DE 255 Fall 2013
Ch 3.2: Fundamental Solutions of Linear Homogeneous Equations
p, q are continuous funtns on an interval I = (α, β), can be [∞]. For a function y twice differentiable on I, differential operator L by
ie in pieces,
L[y](t) = 0, Linear homogeneous equation, along with TWO initial conditions, and if so, are they unique.
[ ] )()()()()()( tytqtytptytyL +′+′′=
( )[ ] )sin(2)cos()sin()(
2,0),sin()(,)(,)(22
22
tettttyLIttyetqttp
t
t
++−=
==== π
1000 )(,)( ytyyty =′=
DE 255 Fall 2013
Theorem 3.2.1 Work out the example below yourself.Consider the initial value problem
where p, q, and g are continuous on an open interval I that contains t0. Then there exists a unique solution y = φ(t) on I.
Note: While this theorem says that a solution to the ivp above exists, it is often not possible to write down a useful expression for the solution which is a major difference between first and second order linear equations. Determine the longest interval on which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.
First put differential equation into standard form:
The longest interval containing the point t = 0 on which the coefficient functions are continuous is (-1, ∞)!!!!!.It follows from Theorem 3.2.1 that the longest interval on which this initial value problem is certain to have a twice differentiable solution is also (-1, ∞).
0000 )(,)()()()(
ytyytytgytqytpy
′=′==+′+′′
( ) ( ) ( ) 00,10,13)(cos1 =′==+′−′′+ yyyytyt
( ) ( ) 00,10,1
11
31
cos=′=
+=
++′
+−′′ yy
ty
ty
tty
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DE 255 Fall 2013
Theorem 3.2.2 (Principle of Superposition)If y1=f and y2=g are solutions to the 2nd order LDE, then the linear combination c1y1 + c2y2 is also a solution, for all constants c1 and c2. To prove, substitute c1y1 + y2c2 and check…!!!! Please remember the notes given before (Proving linearity if always an exam question, in terms of additivity and scalability). Wronskian determinant: Can all solutions can be written this way, or do some solutions have a different form altogether? Suppose y1 and y2 are solutions to L[y] = y'' + p(t)y'=g(t), with initial values of y(t0)=y0 and y'(t0)= y0', from Thr 3.2.2, y = c1y1 + c2 y2 is a solution to this equation such that y = c1y1 + c2 y2 satisfies the given initial conditions ****
0022011
0022011
)()()()(
ytyctycytyctyc′=′+′
=+
)()()()()()(
)()()()()()(
02010201
0100102
02010201
0200201
tytytytytyytyyc
tytytytytyytyyc
′−′′+′−
=
′−′′−′
=
DE 255 Fall 2013
The Wronskian DeterminantArbitrary coefficients, in terms of determinants Wronskiandeterminant, If solution exists then, the determinant W cannot be zero.
)()()()()()(
)()()()()()(
02010201
0100102
02010201
0200201
tytytytytyytyyc
tytytytytyytyyc
′−′′+′−
=
′−′′−′
=
)()()()(
)()(
,
)()()()(
)()(
0201
0201
001
001
2
0201
0201
020
020
1
tytytyty
ytyyty
c
tytytyty
tyytyy
c
′′
′′=
′′
′′=
)()()()()()()()(
020102010201
0201 tytytytytytytyty
W ′−′=′′
=
Wytyyty
cW
tyytyy
c 001
001
2020
020
1
)()(
,)()(
′′=
′′=
( )( )021, tyyW
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DE 255 Fall 2013
Theorem 3.2.3The opposite of theorems given before, suppose y1 and y2 are solutions to the equation at a given t0, if the Wronskian
W= y1y'2 + y'1 y2 ≠ 0, then there is a choice of constants c1, c2for any of which y = c1y1 + c2 y2 is a solution to the DE for given initial conditions y(t0)=y0 and y'(t0)= y0'. Recall the following initial value problem and its solution:
The two functions that are are solutions: The Wronskian of y1 and y2 is
Since W ≠ 0 for all t, linear combinations of y1 and y2 can be used to construct solutions of the IVP for any initial value t0. y = c1y1 + c2 y2
tt eyey −== 21 ,
22 02121
21
21 −=−=−−=′−′=′′
= −− eeeeeyyyyyyyy
W tttt
( ) ( ) tt eetyyyyy −+=⇒=′==−′′ 2)(10,30,0
DE 255 Fall 2013
Theorem 3.2.4 (Fundamental Solutions)
Suppose y1 and y2 are solutions to the equation L[y]=0,If there is a point t0 such that W(y1,y2)(t0) ≠ 0, then the family of solutions y = c1y1 + c2 y2 with arbitrary coefficients c1, c2includes every solution to the differential equation. The expression y = c1y1 + c2 y2 is called the general solutionof the differential equation above, and in this case y1 and y2are said to form a fundamental set of solutions to the differential equation.
In previous example… 2121 ,, 21 rreyey trtr ≠==
( ) ( ) . allfor 021
21
21
122121
21 terrerer
eeyyyy
W trrtrtr
trtr
≠−==′′
= +
trtr ececy 2121 +=
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DE 255 Fall 2013
Example: Show that the functions below are fundamental solutions:
To show this, first substitute y1 into the equation, similar for y2:
To show both solutions form fundamental set of solutions.
Since W ≠ 0 for t > 0, y1, y2 form a fundamental set of solutions for the differential equation
12
2/11 , −== tyty
0,032 2 >=−′+′′ tyytyt
0123
21
23
42 2/12/1
2/12/32 =⎟
⎠⎞
⎜⎝⎛ −+−=−⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛ − −−
tttttt ( ) ( ) ( ) 0134322 11232 =−−=−−+ −−−− tttttt
3
2/32/32/322/1
12/1
21
21
23
23
21
21
tttttt
tt
yyyy
W −=−=−−=−=′′
= −−−−−
−
0,032 2 >=−′+′′ tyytyt
DE 255 Fall 2013
SummaryTo find a general solution of the differential equation
we first find two solutions y1 and y2.Then make sure there is a point t0 in the interval such that W(y1, y2)(t0) ≠ 0.It follows that y1 and y2 form a fundamental set of solutions to the equation, with general solution y = c1y1 + c2 y2.If initial conditions are prescribed at a point t0 in the interval where W ≠ 0, then c1 and c2 can be chosen to satisfy those conditions. Exact and adjoint and self adjoint functions? Page 126, questions 26 and32
βα <<=+′+′′ tytqytpy ,0)()(
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DE 255 Fall 2013
Ch 3.3: Linear Independence and the WronskianTwo functions f and g are linearly dependent if they are multiples of each other. If the only solution to this equation is c1 = c2 = 0, then f and gare linearly independent. For example, f(x) = sin2x and g(x) = sinxcosx, and their linear combinationThis is satisfied if we choose c1 = 1, c2 = -2, and hence f and gare linearly dependent. Note that if a = b = 0, then the only solution to this system of equations is c1 = c2 = 0, provided D ≠ 0.
0)()( 21 =+ tgctfc
0cossin2sin 21 =+ xxcxc
bycycaxcxc
=+=+
2211
2211
21
2111
2121
112
22
2121
221
where,
,
yyxx
DD
bxayxyyx
bxayc
Dbxay
xyyxbxayc
=+−
=−+−
=
−=
−−
=
DE 255 Fall 2013
Example 1: Linear Independence Show that the following two functions are linearly independent on any interval:
Suppose for all t in an arbitrary interval (α, β).
We want to show the equation holds only for c1 = c2 = 0 for all t in (α, β), where t0 ≠ t1, except t0 = t1. Then
D ≠ 0, and therefore f and g are linearly independent.
tt etgetf −== )(,)(
0)()( 21 =+ tgctfc
0
011
00
21
21
=+
=+−
−
tt
tt
ecec
ecec
01101010
11
00tttttttt
tt
tt
eeeeeeeeee
D −−−−−
−
−=−==
10 0 0110 tteeD tttt =⇔=⇔= −−
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DE 255 Fall 2013
Theorem 3.3.1 from Wronskian p o view If f and g are a) differentiable functions on an open interval I and b) if W(f, g)(t0) ≠ 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(t) = 0 for all t in I.
Proof (outline): Let c1 and c2 be scalars, and suppose c1f(t)+c2g(t)=0for all t in I. In particular, when t = t0 we have
Since W(f, g)(t0) ≠ 0, it follows that c1f(t)+c2f(t)=0 only at c1 = c2 = 0, and hence f and g are linearly independent.
0)()(0)()(
0201
0201
=′+′=+
tgctfctgctfc
DE 255 Fall 2013
Theorem 3.3.2 (Abel’s Theorem)
Suppose y1 and y2 are solutions to the equation
where p and q are continuous on some open interval I. Then W(y1,y2)(t) is given by
where c is a constant that depends on y1 and y2 but not on t.
Note that W(y1,y2)(t) is either zero for all t in I only if c = 0 or else is never zero in I (if c ≠ 0).
0)()(][ =+′+′′= ytqytpyyL
∫=− dttp
cetyyW)(
21 ))(,(
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DE 255 Fall 2013
Example 2: Wronskian and Abel’s TheoremRecall the following equation and two of its solutions:
The Wronskian of y1and y2 is
Thus y1 and y2 are linearly independent on any interval I, by Theorem 3.3.1. Now compare W with Abel’s Theorem:
Choosing c = -2, we get the same W as above.
tt eyeyyy −===−′′ 21 ,,0
. allfor 022 0
21
21 teeeeeyyyy
W tttt ≠−=−=−−=′′
= −−
ccecetyyWdtdttp=∫=∫=
−− 0)(21 ))(,(
DE 255 Fall 2013
Theorem 3.3.3Suppose y1 and y2 are solutions to equation below, whose coefficients p and q are continuous on some open interval I:
Then y1 and y2 are linearly dependent on I iff W(y1, y2)(t) = 0 for all t in I. Also, y1 and y2 are linearly independent on I iff W(y1, y2)(t) ≠ 0 for all t in I.
0)()(][ =+′+′′= ytqytpyyL
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DE 255 Fall 2013
SummaryLet y1 and y2 be solutions of
where p and q are continuous on an open interval I. Then the following statements are equivalent:
The functions y1 and y2 form a fundamental set of solutions on I.The functions y1 and y2 are linearly independent on I.W(y1,y2)(t0) ≠ 0 for some t0 in I.W(y1,y2)(t) ≠ 0 for all t in I.
0)()( =+′+′′ ytqytpy
DE 255 Fall 2013
Linear Algebra NoteLet V be the set
Then V is a vector space of dimension two, whose bases are given by any fundamental set of solutions y1 and y2. For example, the solution space V to the differential equation
has bases
with { } { }ttSeeS tt sinh,cosh,, 21 == −
( ){ }βα ,,0)()(: ∈=+′+′′= tytqytpyyV
0=−′′ yy
21 SpanSpan SSV ==
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DE 255 Fall 2013
Unit Circle, Taylor and Mc Laurin series.
z=x+iy, [x, y]=rcos(Θ), r(sin(Θ));(r, Θ)={sqrt(x2+y2), arctang(x/y)}z=x+iy=|z|cos(Θ)+isin(Θ)=|z| eiΘ
f(a)+f'(a)(x-a)/1!+f''(a)(x-a)2/2!+f'''(a)(x-a)3/2!+….{1/(1-x)}=1+x+x2+x3+..Wikipedia..
DE 255 Fall 2013
Ch 3.4: Complex Roots of Characteristic Equation
Recall our discussion of the equation
where a, b and c are constants. Assuming an exponential soln leads to characteristic equation:
Quadratic formula (or factoring) yields two solutions, r1 & r2:
If b2 – 4ac < 0, then complex roots: r1 = λ + iµ, r2 = λ - iµThus
0=+′+′′ cyybya
0)( 2 =++⇒= cbrarety rt
aacbbr
242 −±−
=
( ) ( )titi etyety µλµλ −+ == )(,)( 21
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DE 255 Fall 2013
Euler’s Formula; Complex Valued SolutionsSubstituting it into Taylor series for et, we obtain Euler’s formula:
Generalizing Euler’s formula, we obtain
Then
Therefore
( ) ( ) titn
tin
tnite
n
nn
n
nn
n
nit sincos
!12)1(
!2)1(
!)(
1
121
0
2
0
+=−
−+
−== ∑∑∑
∞
=
−−∞
=
∞
=
tite ti µµµ sincos +=( ) [ ] tietetiteeee ttttitti µµµµ λλλµλµλ sincossincos +=+==+
( )
( ) )sin(cos)(
)sin(cos)(
2
1
titeety
titeetytti
tti
µµ
µµλµλ
λµλ
−==
+==−
+
DE 255 Fall 2013
Real Valued Solutions, The WronskianTo achieve this, recall that linear combinations of solutions are
themselves solutions: Ignoring constants
Checking the Wronskian
Thus y3 and y4 form a fundamental solution set, and the general solution
tietyty
tetytyt
t
µ
µλ
λ
sin2)()(
cos2)()(
21
21
=−
=+tety
tetyt
t
µ
µλ
λ
sin)(
,cos)(
4
3
=
=
( ) ( )0
cossinsincossincos
2 ≠=
+−=
t
tt
tt
e
ttettetete
W
λ
λλ
λλ
µ
µµµλµµµλµµ
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DE 255 Fall 2013
All Exam Quest: Example 1: y'' + y' +y =0
Therefore and thus the general solution is
Example 2: y'' +4y =0
( ) ( ))2/3sin2/3cos()( 212/ tctcety t += −
iir
rrety rt
23
21
231
2411
01)( 2
±−=±−
=−±−
=
=++⇒=
2/3,2/1 =−= µλ
irrety rt 204)( 2 ±=⇔=+⇒= 2,0 == µλ
( ) ( )tctcty 2sin2cos)( 21 +=
DE 255 Fall 2013
Example 3: y'' -2 y' +y =0
Cnt.ed fr Example 1: For the initial values y(0)=1, y' (0) =1, find (a) the solution u(t) and (b) the smallest time T for which |u(t)| ≤ 0.1
Find the smallest time T for which |u(t)| ≤ 0.1graphing calculator or computer algebra system, we find that T ≅ 2.79.
irrrety rt
32
31
612420123)( 2 ±=
−±=⇔=+−⇒=
( ) ( ))3/2sin3/2cos()( 213/ tctcety t +=
( ) ( )2/3sin2/3cos)( 2/2
2/1 tectectu tt −− +=
33
3,11
23
21
1
2121
1
===⇒⎪⎭
⎪⎬
⎫
=+−
=cc
cc
c
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DE 255 Fall 2013
Ch 3.5: Repeated Roots; Reduction of OrderRecall our 2nd order linear homogeneous ODE
where a, b and c are constants. Assuming an exponential soln leads to characteristic equation:
Quadratic formula (or factoring) yields two solutions, r1 & r2:
When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives one solution:
0=+′+′′ cyybya
0)( 2 =++⇒= cbrarety rt
aacbbr
242 −±−
=
atbcety 2/1 )( −=
DE 255 Fall 2013
Exam QuestionSecond Solution: Multiplying Factor v(t)
We know that
Since y1 and y2 are linearly dependent, we generalize this approach and multiply by a function v, and determine conditions for which y2 is a solution:
Then ****
solution a )()(solution a )( 121 tcytyty =⇒
atbatb etvtyety 2/2
2/1 )()( try solution a )( −− =⇒=
atbatbatbatb
atbatb
atb
etva
betva
betva
betvty
etva
betvty
etvty
2/2
22/2/2/
2
2/2/2
2/2
)(4
)(2
)(2
)()(
)(2
)()(
)()(
−−−−
−−
−
+′−′−′′=′′
−′=′
=
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DE 255 Fall 2013
Finding Multiplying Factor v(t)Substituting derivatives into ODE, seek a formula for v: b2-4ac=0
43
2
222
22
22
2
22/
)()(0)(
0)(4
4)(
0)(44
4)(0)(
44
42
4)(
0)(24
)(
0)()(2
)()(4
)()(
0)()(2
)()(4
)()(
ktktvktvtv
tva
acbtva
tvaac
abtvatv
aac
ab
abtva
tvca
ba
btva
tcvtva
btvbtva
btvbtva
tcvtva
btvbtva
btvabtvae atb
+=⇒=′⇒=′′
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −−′′
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−+′′⇔=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−+′′
=⎟⎟⎠
⎞⎜⎜⎝
⎛+−+′′
=+−′++′−′′
=⎭⎬⎫
⎩⎨⎧
+⎥⎦⎤
⎢⎣⎡ −′+⎥
⎦
⎤⎢⎣
⎡+′−′′−
DE 255 Fall 2013
The General Solution
Thus every solution is a linear combination of
Wrnskian
Thus y1 and y2 form a fundamental solution set for equation
( )abtabt
abtabt
abtabt
tecec
ektkek
etvkekty
2/2
2/1
2/43
2/1
2/2
2/1 )()(
−−
−−
−−
+=
++=
+=
abtabt tececty 2/2
2/1)( −− +=
abtabt tetyety 2/2
2/1 )(,)( −− ==
te
abte
abtee
abte
ab
teetyyW
abt
abtabtabtabt
abtabt
allfor 0
221
21
2))(,(
/
//2/2/
2/2/
21
≠=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ −=
⎟⎠⎞
⎜⎝⎛ −−=
−
−−−−
−−
![Page 16: 2ndrder ch03 1 pr a - Marmara Üniversitesimimoza.marmara.edu.tr/~msakalli/DE255Fall2013/2ndrder_ch03_1_p… · 1 DE 255 Fall 2013 DE-2013 Dr. M. Sakalli DE 255 Fall 2013 Ch 3.1:](https://reader035.vdocuments.net/reader035/viewer/2022070812/5f0af14e7e708231d42e19a3/html5/thumbnails/16.jpg)
16
DE 255 Fall 2013
Example 1: y'' + 2y' +y =0, y(0)=1, y' (0) =1,
Example 2: y'' - 2y' +0.25y =0, y(0)=2, y' (0) =1/2,
10)1(012)( 22 −=⇔=+⇔=++⇒= rrrrety rt
tt tececty −− += 21)( 2,111
2121
1 ==⇒⎭⎬⎫
=+−=
cccc
c
tt teety −− += 2)(
2/10)2/1(025.0)(
2
2
=⇔=−⇔
=+−⇒=
rrrrety rt
2/2
2/1)( tt tececty +=
21,2
21
21
221
21
1−==⇒
⎪⎭
⎪⎬⎫
=+
=cccc
c
2/2/
212)( tt teety −=
DE 255 Fall 2013
Reduction of Order also works for equations with nonconstant coefficients
That is, given that y1 is solution, and y2 = v(t)y1:
this last equation reduces to a first order equation in v′ :
0)()( =+′+′′ ytqytpy
)()()()(2)()()()()()()()(
)()()(
1112
112
12
tytvtytvtytvtytytvtytvty
tytvty
′′+′′+′′=′′′+′=′
=
( ) ( ) 02 111111 =+′+′′+′+′+′′ vqyypyvpyyvy
( ) 02 111 =′+′+′′ vpyyvy
![Page 17: 2ndrder ch03 1 pr a - Marmara Üniversitesimimoza.marmara.edu.tr/~msakalli/DE255Fall2013/2ndrder_ch03_1_p… · 1 DE 255 Fall 2013 DE-2013 Dr. M. Sakalli DE 255 Fall 2013 Ch 3.1:](https://reader035.vdocuments.net/reader035/viewer/2022070812/5f0af14e7e708231d42e19a3/html5/thumbnails/17.jpg)
17
DE 255 Fall 2013
Example of Reduction of Order Similar Exam questions
Given the variable coefficient equation and solution y1, use reduction of order method to find a second solution:
Substituting these into ODE and collecting terms,
ktctv += ln)(
,)(;0,03 11
2 −=>=+′+′′ ttytyytyt
3212
212
12
)(2 )(2 )()(
)( )()( )()(
−−−
−−
−
+′−′′=′′
−′=′
=
ttvttvttvty
ttvttvtyttvty ( ) ( )
)()( where,00
033220322
111
1213212
tvtuuutvvt
vtvtvvtvtvvtvttvtvttvtvt
′==+′⇔=′+′′⇔
=+−′++′−′′⇔
=+−′++′−′′−−−
−−−−−−
tcv =′
.0 since,
lnln0
11 >=⇔=⇔
+−=⇔=+
−− tctuetu
Ctuudtdut
C
( ) 1112 lnln)( −−− +=+= tktcttktcty .ln)( 1
2 ttty −=
ttctcty ln)( 12
11
−− +=
DE 255 Fall 2013
Ch 3.6: Nonhomogeneous Equations
Recall the nonhomogeneous equation p, q, g are continuous functions on an open interval I.
Theorem 3.6.1 (Exam question very potential)If Y1, Y2 are solutions of nonhomogeneous equationthen Y1 - Y2 is a solution of the homogeneous equationIf y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that
The general solution of nonhomogeneous equation, where Yis a specific solution to the nonhomogeneous equation.
)()(0)()()21()2()1( 2211 tyctyctgtgyyLyLyL +==−=−=−
)()()( tgytqytpy =+′+′′
)()()()( 2211 tYtyctycty ++=
![Page 18: 2ndrder ch03 1 pr a - Marmara Üniversitesimimoza.marmara.edu.tr/~msakalli/DE255Fall2013/2ndrder_ch03_1_p… · 1 DE 255 Fall 2013 DE-2013 Dr. M. Sakalli DE 255 Fall 2013 Ch 3.1:](https://reader035.vdocuments.net/reader035/viewer/2022070812/5f0af14e7e708231d42e19a3/html5/thumbnails/18.jpg)
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DE 255 Fall 2013
Method of Undetermined Coefficients: g(t) g(t) is expSince exponentials replicate through differentiation, a good guess for Y is:
g(t)=sine
Since sin(x) and cos(x) are linearly independent (they are not multiples of each other), we must have c1= c2 = 0, and hence 2 + 5A = 3A = 0, which is impossible Our next attempt at finding a Y is
( )0cossin
0cos3sin52sin2sin4cos3sin
21 =+⇔=++⇔=−−−
tctctAtA
ttAtAtA
teyyy 2343 =−′−′′
ttt AetYAetYAetY 222 4)(,2)()( =′′=′⇒=
tyyy sin243 =−′−′′tAtYtAtYtAtY sin)(,cos)(sin)( −=′′=′⇒=
tBtAtYtBtAtYtBtAtY
cossin)(,sincos)(cossin)(
−−=′′−=′⇒+=
DE 255 Fall 2013
Polynomial g(t):
Product g(t)
teyyy t 2cos843 −=−′−′′
1443 2 −=−′−′′ tyyy
AtYBAttYCBtAttY 2)(,2)()( 2 =′′+=′⇒++=
811
23)( 2 −+−= tttY
( ) ( )( ) ( ) ( )
( )( ) ( ) teBAteBA
teBAteBAteBAteBAtY
teBAteBAtBetBetAetAetY
tBetAetY
tt
t
ttt
tt
tttt
tt
2sin342cos432cos22
2sin22sin222cos2)(2sin22cos2
2cos22sin2sin22cos)(2sin2cos)(
−−++−=
+−+
+−++−+=′′
+−++=
++−=′
+=
![Page 19: 2ndrder ch03 1 pr a - Marmara Üniversitesimimoza.marmara.edu.tr/~msakalli/DE255Fall2013/2ndrder_ch03_1_p… · 1 DE 255 Fall 2013 DE-2013 Dr. M. Sakalli DE 255 Fall 2013 Ch 3.1:](https://reader035.vdocuments.net/reader035/viewer/2022070812/5f0af14e7e708231d42e19a3/html5/thumbnails/19.jpg)
19
DE 255 Fall 2013
Sum g(t) is sum of functions
If Y1, Y2 are solutions of
respectively, then Y1 + Y2 is a solution of the nonhomogeneous equation above.
)()()( 21 tgtgtg +=
)()()()()()(
2
1
tgytqytpytgytqytpy
=+′+′′=+′+′′
teteyyy tt 2cos8sin2343 2 −+=−′−′′
tetettetY ttt 2sin1322cos
1310sin
175cos
173
21)( 2 ++−+−=
teyyytyyy
eyyy
t
t
2cos843sin243
343 2
−=−′−′′
=−′−′′=−′−′′
DE 255 Fall 2013
Pay Attension
Failure: Substituting these derivatives into ODE:
Thus no particular solution exists of the form
tyy 2cos34 =+′′
tBtAtYtBtAtYtBtAtY
2cos42sin4)(,2sin22cos2)(2cos2sin)(
−−=′′−=′⇒+=
( ) ( )( ) ( )
tttBBtAAttBtAtBtA
2cos302cos32cos442sin442cos32cos2sin42cos42sin4
==+−++−=++−−
tBtAtY 2cos2sin)( +=
tBttAttBtAtYtBttBtAttAtY
tBttAttY
2cos42sin42sin42cos4)(2sin22cos2cos22sin)(
2cos2sin)(
−−−=′′−++=′
+=
tttY
BAttBtA
2sin43)(
0,4/32cos32sin42cos4
=⇒
==⇒=−