Transcription of second order equations{Undetermined Coe - cients
1 September 29, 20139-19. Particular Solutions of Non-homogeneoussecond order equations undetermined Coeffi-cientsWe have seen that in order to find the general solution tothe second order differential equationy +p(t)y +q(t)y=g(t)(1)wherep,q,gare continuous functions in an intervalI,it suffices to find two linearly independent solutions tothe associated homogeneous equationy +p(t)y +q(t)y= 0(2)and one particular solution to (1).Here we will describe some methods for finding partic-ular 1: undetermined coefficientsThis method is useful when the the differential equa-tion hasconstant coefficientsand the functiong(t) hasa special form: some linear combination of functions ofthe formtn,e t,e tcos( t),e tsin( t).
2 (3)In each of the above cases, one can assume a particularform for the solutiony(t) involving a polynomialU(t)September 29, 20139-2with unknown iny(t) to the equation (1) then gives a systemof linear equations for the coefficients ofU(t) which canbe is important to know the correct form to be assumedfor the solutiony(t).We consider the four cases (t) denote a polynomial of degreenin the fol-lowing. That is,P(t) =n i=0aitiwherean6= (t) will be known fromg(t).In each of the following cases,U(t) will be an unknownpolynomialU(t) =n i=0 Aitiof the same degree asP(t), possibly with complex co-efficients, and the coefficientsAiare to be 1:g(t) =P(t)Write the characteristic polynomial asz(r) =ar2+br+ (0)6= 0, ( ,c6= 0) then one assumesSeptember 29, 20139-3y(t) =U(t) =n i=0 AitiwhereU(t) is a polynomial of the same degree asP(t).
3 Ifz(0) = 0 andz (0)6= 0 ( , 0 is a root of multiplicity1 ofz(r)), then one assumesy(t) =tU(t).Ifz(0) = 0 =z (0), butz (0)6= 0, ( , 0 is a root ofmultiplicty 2 ofz(r)), then one assumesy(t) =t2U(t).Case 2:g(t) =e tP(t)Ifz( )6= 0, then one assumesy(t) =U(t)e ( ) = 0, butz ( )6= 0, then one assumesy(t) =tU(t)e ( ) = 0,z ( ) = 0, butz ( )6= 0, then one as-sumesy(t) =t2U(t)e 3:g(t) =P(t)e tcos( t)Here we reduce to Case 2 using complex methods. Wecan writeg(t) as the real part of the complex functiongc(t) =P(t)exp(( +i )t)September 29, 20139-4We find a complex solutionyc(t) =tsUc(t)exp(( +i )t) wheresis the multiplicity of +i as a root ofz(r).
4 We then take the real part ofyc(t) to get our 4:g(t) =P(t)e tsin( t)This is just like Case 3 except that we replace the wordsreal partbyimaginary an alternative to using complex meth-ods in Cases 3 and 4, one may assume a solutiony(t) ofthe formy(t) =ts[U(t)e tcos( t) +V(t)e tsin( t)]wheresis the multiplicity of +i as a root ofz(r)andU(t) andV(t) are two real polynomials of the samedegree asP(t). Note that in this case one has to useboth the unknown polynomialsU(t),V(t) even in Case3 or Case we will see, the complex method is more efficientbecause we replace the 2nreal linear equations for theunknowns toncomplex linear us consider some + 2y 3y= 29, 20139-5 Let us try to find a solution of the formy=At2+Bt+ in, we get2A+ 2(2At+B) 3(At2+Bt+C)
5 = coefficients oft2,t,1 have to be equal on bothsides, so we get 3A= 2,4A 3B= 0,2A+ 2B 3C= these equations forA,B,Cwe getA= 23, B= 89, C= 49 + 2y 3y= we tryy= get9Ae3t+ 6Ae3t 3Ae3t= 4,A= 1 29, 20139-6It is worthwhile to notice something about this the general situation in which we havey +py +qy=Aebt(4)in whichbis not a root ofz(r) =r2+pr+q.(5)Let us assume a solutionyp(t) of the formyp(t) = into the equation (4), we get(Cebt) +p(Cebt) +qCebt=Aebtor(Cb2ebt) +p(Cbebt) +qCebt=AebtCebt(b2+pb+q) =AebtCebtz(b) =AebtorSeptember 29, 20139-7C=Az(b)Thus, under the assumption (5), we have a simpleformula for the particular solution:yp(t) =Az(b)ebtExample + 2y 3y= 2t2+ 4e3t.
6 (6)Here we use thesuperposition we have the equationL(y) =g1(t) +g2(t)(7)andy1, y2are solutions toL(y) =g1, L(y) =g2,respectively, theny1+y2is a solution to (7).So, we get the particular solutiony(t) =13e3t 23t2 89t 49 1627to (6).September 29, 20139-8 Example + 2y 3y= 4e 3t.(8)We again tryy=Ae get9Ae 3t 6Ae 3t 3Ae 3t= 4e has no solution. The problem is that 3 is aroot of the characteristic polynomialr2+ 2r this case, we tryy=At(e 3t).We gety =Ae 3t 3 Ate 3ty = 3Ae 3t 3A(e 3t 3te 3t) + 2Ae 3t 6 Ate 3t 3 Ate 3t= 4e 3t. 4A= 4, A= , we get the solutiony= te is going on with the method?Here it helps to go to the abstract 29, 20139-9y +py +qy=Be tWe try to find a solution of the formy=Ae getA( 2+p +q)e t=Be is not a root ofz(r) =r2+pr+q, then we getAasA=Bz( ).
7 If is a root of multiplicity one, and we tryy=Ate t, we findAte t[ 2+p +q] +Ae t(2 +p) =Be is the same asAte tz( ) +Ae tz ( ) =Be , we getA=Bz ( ).September 29, 20139-10 Similarly, if is a root of multiplicity 2 toz(r), wegety=At2e twhereA=B2=Bz ( ).In general, we try a function of the formy=Atse twheresis the multiplicity of as a root ofz(r).Example 5. As we saw above, similar methods apply to righthand sides of the forme tsin( t),e tcos( t)making use of complex arithmetic methods and tak-ing real and imaginary instance, considery + 2y 3y= 3cos(2t)(9)We recognize that this equation is the real part ofthe differential equationy + 2y 3y= 3e2it.
8 (10)The characteristic polynomial isz(r) =r2+ 2r 29, 20139-11 Since 2iis not a solution ofz(r) = 0, we get a com-plex solution of the formy(t) =Ae2it. Its real partwill turn out to be solution to (9).Let us be more the method in the first part of Example 4(which also works for complex differential equations ), and the fact that 2iis not a root ofz(r), we canfind a particular complex solution to (10) asyc(t) =3z(2i)e2itSince we equation (9) was the real part of equation(10), we take the real part of the solution to (10) toget a particular solution to (9).Let us carry this haveyc(t) =3z(2i)e2it=34i 7e2it=3 7 + 4ie2it=3( 7 4i)72+ 42e2itSeptember 29, 20139-12= 21 12i65e2it= ( 2165 1265i)e2it= ( 2165 1265i)(cos(2t) +isin(2t))= 2165cos(2t) +1265sin(2t) +i( 1265cos(2t) 2165sin(2t))Let us useRe(w) andIm(w) to denote the real andimaginary parts of the complex , we see that the real part ofyc(t) isRe(yc(t)) = 2165cos(2t) +1265sin(2t)This is our particular solution to equation (9).
9 If we had started withy + 2y 3y= 3sin(2t),(11)then the same method applies. We simply take theimaginary part ofyc(t) to get our particular solutionto (11) asIm(yc(t)) = 1265cos(2t) 2165sin(2t)September 29, 20139-13As we saw in this example, it is usually necessaryto divide by complex numbers and then take real orimaginary instance, we may have the numbera+bic+di,and we have to take the real or imaginary in example 4 in Section 8, it is useful to transformthe division into a multiplation using the formula forthe inverse1c+di=c dic2+ givesa+bic+di= (a+bi)c dic2+d2=ac+bd+ (bc ad)ic2+d2,and we can read the real and imaginary parts offeasily as we did in the above method to solve equations (9) or (11)
10 , whichdoes not use complex methods is to assume a solutionSeptember 29, 20139-14of the formAcos(t) +Bsin(t), plug in and solve forAandB. As we mentioned above, the advantage ofcomplex arithmetic is that it allows us to keep trackof coefficients two at a time, and thus only do halfthe number of the 6. Let us consider another the IVPy + 5y + 2y= 3sin(4t),y(0) = 1,y (0) = 2 (12)Solution:Step the general solution to the associatedhomogeneous characteristic equation isz(r) =r2+ 5r+ 2which has the rootsr1= 5 + 172, r2= 5 172 The general solution to this homogeneous equationisy(t) =c1exp(r1t) +c2exp(r2t)(We do not need this here, but let us observe that thefirst fundamental solution of the homogeneous equa-September 29, 20139-15tion isy1(t) =exp(r1t), and the second fundamentalsolution isy2(t) =exp(r2t)).
