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:

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 ′=′=

2

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..

3

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

4

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

5

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 +=

6

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)()(

7

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 =⇔=⇔= −−

8

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 ))(,(

9

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

10

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 ==

11

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

12

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

13

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

14

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

)()(

)()(

−−−−

−−

−

+′−′−′′=′′

−′=′

=

15

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

≠=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −−=

−

−−−−

−−

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

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 ++=

18

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)(

−−++−=

+−+

+−++−+=′′

+−++=

++−=′

+=

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

=⇒

==⇒=−