Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients


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1 Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous DE Paricular soluion o nonhomogeneous DE Mehod of undeermined coefficiens
2 A general second order linear DE has he form y p() y q() y g() where coefficien funcions p(), q(), and righ side g() are coninuous in some inerval I. If all he coefficiens are consan funcions, we say we have a second order linear DE wih consan coefficiens which we wrie as ay by cy g() Two imporan pieces of erminology: 1. If g() is idenically zero, hen DE is called HOMOGENEOUS. 2. If g() is no idenically zero, hen DE is called NONHOMOGENEOUS. Examples: y'' + 2 y' + 5y = 0 is a homogeneous DE y ''  4y'+ y = cos() is a nonhomogeneous DE ( wih consan coefficiens)
3 OPERATOR NOTATION & PROPERTIES An operaor is a funcion whose domain is a se of funcions and he range consiss of (oher) funcions. For example he derivaive is an operaor. Le D(y) y'(x). The domain of operaor D is he se of funcions whose derivaive exiss, usually on a specified inerval. The oupu is anoher funcion. For second order linear DEs i is convenien o define an operaor L given by he expression L[y] y'' +py' + qy, where p and q are some funcions of. If we recall ha a second derivaive is he derivaive of he firs derivaive, hen y'' = D(D(y)). I is convenien o wrie his as y '' = D 2 (y) where we mus keep in mind ha he D 2 means compue a pair of successive derivaives. Using his Dnoaion, he operaor L can be expressed as L[y] = D 2 (y) +p()d(y) + q()y or for consan coefficiens we wrie L[y] = = (ad 2 + bd  c)(y). Examples: If p() = 2 and q() = cos(), hen we have L[y] (D D + cos())(y) = y'' + 2 y' + cos()y. If a = 3, b = 2, and c = 4, hen L[y] (ad 2 + bd  c)(y) = 3y'' + 2 y'  4y. Noe ha he operaor noaion is basically a shor hand way of expressing he lef side of he DE.
4 Recall ha The derivaive of a sum is he sum of he individual derivaives. and ha The derivaive of a consan imes a funcion is he consan imes he derivaive of he funcion. Since our operaor L is jus a sum of muliples of derivaives i follows ha for any pair of consans c 1 and c 2 and any pair of funcions y 1 () and y 2 () ha L[c 1 y 1 + c 2 y 2 ] = c 1 L[y 1 ] + c 2 L[y 2 ]. An operaor ha saisfies his propery is said o be a LINEAR OPERATOR. We say ha a linear operaor preserves linear combinaions. (Recall L has a special form L[y] y +py + qy. No every operaor involving derivaives saisfies his propery. When his propery is no saisfied he operaor is called nonlinear.
5 Imporan Propery Every second order linear homogeneous DE L[y] y +py + qy = 0 saisfies he following propery. If c 1 and c 2 are any consans and y 1 () and y 2 () any soluions of he DE, hen linear combinaion c 1 y 1 () + c 2 y 2 () is also a soluion of he DE. (Principle of Superposiion) This implies ha from a few soluions o he linear homogeneous DE we can build los of ohers by forming linear combinaions. Example: Consider he second order linear homogeneous DE given by L[y] = D 2 (y)  3D(y) + 2y = 0. Verify ha y 1 (x) = e x and y 2 (x) = e 2x are soluions. Tha means L[y 1 ] = L[e x ] = 0 and L[y 2 ] = L[e 2x ] = 0. Since L is a linear operaor we know ha L[c 1 y 1 + c 2 y 2 ] = c 1 L[y 1 ] + c 2 L[y 2 ]. So his fac says ha since L[y 1 ] = 0 and L[y 2 ] = 0, hen so mus L[c 1 y 1 + c 2 y 2 ] = 0 for any consans c 1 and c 2. Thus every linear combinaion c 1 e x + c 2 e 2x is a soluion of he DE. Hence 26e x + 999e 2x and 45e x e 2x are soluions of his DE.
6 For now we will focus on second order nonhomogeneous DEs wih consan coefficiens. ay by cy g() We have seen ha for second order homogeneous DEs wih consan coefficiens we can find a fundamenal se of soluions y 1 () and y 2 () so ha he general soluion of ay + by +cy = 0 is given by he linear combinaion c 1 y 1 () + c 2 y 2 () where c 1 and c 2 are arbirary consans. Our sraegy o solve ay'' +by' + cy = g() is a hreesep procedure: Sep #1. Find he general soluion y h () of he associaed homogeneous equaion ay'' +by' + cy = 0. Sep #2. Find a paricular soluion y p () of he nonhomogeneous DE ay'' +by' + cy = g(). Sep #3. Then he general soluion of he nonhomogeneous DE ay'' +by' + cy = g() is y() = y h () + y p (). To verify ha his works we do he following.
7 Le () be any soluion of L(y) ay by cy g() Le y p () be a paricular soluion of his nonhomogeneous DE. (no arbirary consans in i) Le he general soluion of he associaed homogeneous equaion be: y h (x) = C 1 y 1 () + C 2 y 2 () We have L[()] = g() and L[y p ()] = g(). Since L is a linear operaor we have L[()  y p ()] = L[()]  L[y p ()] = g()  g() = 0 This says ha ()  y p () is a soluion of he associaed homogeneous equaion. Thus we mus be able o express ()  y p () in erms of he y 1 () and y 2 (). Tha is, here exis some consans C 1 and C 2 so ha ()  y p () = C 1 y 1 () + C 2 y 2 (). Rearranging his expression we have ha () which is any soluion of he nonhomogeneous DE can be expressed as () = C 1 y 1 () + C 2 y 2 () + y p () = y h (x) + y p () Summary: Every soluion of he nonhomogeneous DE is obained as a sum of a paricular soluion + he general soluion of he homogeneous DE; y() = y h () + y p ().
8 Examples: 1. For DE y'' y = 2 2, i is known ha a paricular soluion is y p () = 2. Find he general soluion. Characerisic equaion r 2 1 = 0 r = 1, 1 y h () = C 1 e + C 2 e  General soluion of nonhomogeneous DE is y() = y h () + y p () = C 1 e + C 2 e For DE y'' y' + y = sin(), i is known ha a paricular soluion is y p () = cos(). Find he general soluion. Characerisic equaion r 2 r + 1 = 0 General soluion of nonhomogeneous DE is 1 1 y() = y h () + y p () = i 3 r y 2 2 h() C1e cos 3 / 2 ) C2e sin 3 / 2 ) C e cos / ) C e sin / ) cos() We know how o find y h (x). We need o learn how o find y p (x).
9 Our firs sep is o find he soluion of he associaed homogeneous for he nonhomogeneous DE ay + by + cy = g(); his is, solve ay + by + cy = 0. The wo soluions ha make up he general soluion of he associaed homogeneous equaion may involve he following ypes of funcions: Polynomials 1,, Exponenial funcions e r, Trig funcions cos(k), sin(k), and Producs e r, e r cos(k), e r sin(k) One echnique for consrucing a paricular soluion y p () of ay + by + cy = g() is he mehod of undeermined coefficiens. This mehod requires us o consruc a rial expression for y p () having unknown coefficiens, subsiue i ino he DE, and solve for he unknown coefficiens involved. The major limiaion of his mehod is ha i is useful primarily for equaions for which we can easily wrie down he correc form of he paricular soluion in advance. For his reason, his mehod is usually used only for problems in which he homogeneous equaion has consan coefficiens and he nonhomogeneous erm is resriced o a relaively small class of funcions. In paricular, we consider only nonhomogeneous erms ha consis of polynomials, exponenial funcions, sines, and cosines. Despie his limiaion, he mehod of undeermined coefficiens is useful for solving many problems ha have imporan applicaions. This means ha he g() expression involves he same ype of funcion ha appears in he soluion of he associaed homogeneous DE.
10 Examples: We consider DE y 3y  4y = g() for differen selecions of g() The associaed homogeneous DE y 3y 4y = 0 has characerisic equaion r 2 3r 4 = (r 4)(r + 1) = 0. So r = 1, 4 and y h () = C 1 e  + C 2 e 4. Case g() = 3e 2 This suggess rial soluion Y() = Ae 2 for he paricular soluion. Taking derivaives we ge Y() Ae Y'() 2Ae, Y''() 4Ae Subsiuing hese derivaives ino he differenial equaion we ge Ae 6Ae 4Ae 3e Simplify o ge 6Ae 3e A 1/ 2 Thus a paricular soluion o he nonhomogeneous DE is 1 Y() y p() e 2 2 The general soluion of y 3y  4y = 3e 2 is y() = y h () + y p () = C 1 e  + C 2 e 4 (1/2)e 2.
11 y 3y  4y = g() y h () = C 1 e  + C 2 e 4 Case g() = 2sin() Because he derivaive of sin() will repea in successive differeniaions, his suggess rial soluion Y() = Asin() for he paricular soluion. Taking derivaives we ge Y() Asin Y'() Acos, Y''() Asin Subsiuing hese derivaives ino he differenial equaion we ge Asin 3Acos 4Asin 2sin Simplify o ge 2 5A sin 3Acos 0 Since sin() and cos() are no muliples of one anoher he individual coefficiens mus boh be zero; ha is, 2 + 5A = 0 and 3A = 0, which is impossible. (WHY?) Our nex aemp a finding a paricular soluion Y is o use rial soluion Y() = Asin() + Bcos(). Y() A sin For derivaives we have Bcos Y () Acos Bsin, Y () A sin Bcos Subsiuing hese derivaives ino he differenial equaion we ge simplifing we ge 5A 3B sin 3A 5B cos 2sin Equaing coefficiens of like erms on eiher side of he equal sign we ge A sin Bcos Acos Bsin A sin Bcos sin 5A 3B 2, 3A 5B 0 which gives A 5 / 17, B 3 / Thus a paricular soluion o he nonhomogeneous DE is Y() y () sin cos The general soluion of y 3y  4y = 2sin() is y() = y h () + y p () = C 1 e  + C 2 e 4 (5/17)sin() +(3/17)cos(). p
12 y 3y  4y = g() Case g() = 8e cos(2) Subsiuing in he DE we ge y h () = C 1 e  + C 2 e 4 Because he derivaives of cos(2) will involve sin(2), his suggess rial soluion Y() = Ae cos(2) + Be sin(2) for he paricular soluion. Taking derivaives we ge 2 2 Y() Ae cos Be sin Y '() Ae cos Ae sin Be sin Be cos A 2Be cos2 2A Be sin Y () A B e cos A B e sin A B e sin A Be cos A 4B e cos2 4A 3Be sin Y () A B e cos A B e sin 3Y () ( 3) A 2B e cos 2 ( 3) 2A B e sin 2 4Y() ( 4)Ae cos 2 ( 4)Be sin 2 8e cos 2 10A 2B e cos 2 2A 10B e sin 2 The general soluion of y 3y  4y = 8e cos(2) is y() = y h () + y p () = C 1 e  + C 2 e 4 + (10/13)e cos(2) + (2/13)e sin(2). Equaing coefficiens of like erms we ge 8 = 10A 2B 0 = 2A 10B Solving for A and B gives A = 10/13 and B = 2/ Y() e cos e sin
13 Summary: The general soluion of y 3y  4y = g() in he hree cases are displayed nex. 1 e 2 Case g() = 3e 2 y() = y 2 h () + Y()= C 1 e  + C 2 e 4 Case g() = 2sin() y() = y h () + Y()= C 1 e  + C 2 e 4 + sin cos Case g() = 8e 10 2 cos(2) y() = y h () + Y()= C 1 e  + C 2 e 4 + e cos2 e sin Consider he equaion 2 y 3y 4y 3e 2sin 8e cos2 Our equaions o solve individually are y 3y 4y 3e 2 y 3y 4y 2sin To ge he paricular soluion of he y 3y 4y 8e cos 2 DE jus add he paricular soluion of he individual cases Y() e cos sin e cos 2 e sin So he general soluion of y 3y 4y 3e 2sin 8e cos y() = y h () + Y()= C 1 e  + C 2 e e cos sin e cos2 e sin2 is
14 Special case: If any erm in he righ side g() in he nonhomogeneous DE ay + by + cy = g() is par of he soluion o he associaed homogeneous DE ay + by + cy = 0, hen we need o muliply he erms of rial soluion by or possibly a power of. Consider he nonhomogeneous equaion y'' 3y' 4y 2e So g() = 2e . Y The characerisic equaion of he associaed homogeneous DE is r 23r  4 = (r  4)(r + 1) =0, so r = 4, 1. Thus 4 y () C e C e h 1 2 We see ha a naural rial soluion is Y = Ae . However he differenial operaor will be fooled ino hinking ha he paricular soluion of his form is a soluion of he associaed homogeneous equaion. To have he differenial operaor no be deceived we adjus he rial soluion o be Y = Ae .(We muliplied by.) Now we proceed in he usual manner compuing derivaives of he rial soluion and subsiuing hem ino he nonhomogeneous DE. Equae coefficiens of like Ae 4Y 4Ae erms on from he sides of he 3Y' 3Ae 3Ae equal sign; 2 = 5A, A = 2/5. Y' Ae Ae Y'' Ae Ae Ae 2Ae Ae Y'' 2Ae Ae ADD 2e 5A e 4A 3A A e zero Paricular soln is Y() = y p () = (2/5)e  General soluion is 2 Y() y h() C e C e e
15 Always compue he soluion of he associaed homogeneous DE firs! How o consruc he "rial" paricular soluion. If in any group of erms generaed by differeniaing g() has a funcion form ha is he same as one of he funcions in he soluion of he associaed homogeneous equaion, hen we mus muliply each erm of he group by s where s is he smalles posiive power so ha all he erms of he rial soluion are disinc from he members of he soluion of he homogeneous DE.
16 Pracice wih rial soluion: DE ay + by + cy = g() has he soluion o he associaed homogeneous DE given by y h () = C 1 e 2 + C 2 e. If g() = 4cos() wha should your rial soluion be? y() = Acos() + Bsin() If g() = 2e  sin() wha should your rial soluion be? y() = Ae  cos() + Be  sin() If g() = wha should your rial soluion be? y() = A + B If g() = 3e 2 wha should your rial soluion be? y() = Ae 2 If g() = 5cos() wha should your rial soluion be? y() = Acos() + Bsin() + Ccos() +Dsin()
17 Solving an iniial value problem of he form ay + by + cy = g(), y( 0 ) = y 0, y'( 0 ) = y 0 '. Firs find y h () he general soluion o ay + by + cy = 0; y h () = C 1 y 1 () + C 2 y 2 () Nex find he paricular soluion y p () for ay + by + cy = g(). So he general soluion of ay + by + cy = g() is y() = y h () + y p () Finally use he iniial condiions o deermine C 1 and C 2. Subsiue he general soluion y() = y h () + y p () ino he DE and equae coefficiens of liked powered erms o provided a se of equaions o solve for C 1 and C 2.
18 Example: Solve he IVP y'' y' 2sin(), y( 0) 1, y'( 0) 1 The characerisic polynomial of y''+ y' = 0 is r 2 + r = 0 r = 0, 1 y h () = C 1 + C 2 e . Applying he mehod of undeermined coefficiens we have rial soluion Y() = A sin() + B cos(). Differeniaing we have Y() A sin() Bcos() Y'() Acos() Bsin() Y''() A sin() Bcos() Subsiuing ino y'' + y' = 2sin() we have A sin() Bcos() A cos() B sin() 2sin() (A B)cos() ( A B) sin() 2 sin() collecing erms Equaing coefficiens of like erms A  B = 0  A  B = 2 Solving for A and B A =  1, B =  1 So he paricular soluion is y p () =  sin()  cos() Thus he general soluion is y() y h() y p() C1C2e sin() cos() Applying he iniial condiions: y(0) = 1 1 = C 1 + C 21 and y'(0) = 11 = C 21. Solving for C 1 and C 2 we have C 1 = 2 and C 2 = 0. So he soluion of he IVP is y() y () y () 2 sin() cos() h p
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