JUST THE MATHS SLIDES NUMBER 15.7 ORDINARY DIFFERENTIAL EQUATIONS - PDF document

“JUST THE MATHS” SLIDES NUMBER 15.7 ORDINARY DIFFERENTIAL EQUATIONS 7 (Second order equations (D)) by A.J.Hobson 15.7.1 Problematic cases of particular integrals

UNIT 15.7 - ORDINARY DIFFERENTIAL EQUATIONS 7 SECOND ORDER EQUATIONS (D) 15.7.1 PROBLEMATIC CASES OF PARTICULAR INTEGRALS Difficulties can arise if all or part of any trial solution would already be included in the complementary function We illustrate with examples: EXAMPLES 1. Determine the complementary function and a partic- ular integral for the differential equation d 2 y d x 2 − 3d y d x + 2 y = e 2 x . Solution The auxiliary equation is m 2 − 3 m + 2 = 0, with solutions m = 1 and m = 2. Hence, the complementary function is Ae x + Be 2 x , where A and B are arbitrary constants. 1

A trial solution of y = αe 2 x gives d x = 2 αe 2 x and d 2 y d y d x 2 = 4 αe 2 x . Substituting these into the differential equation, 4 αe 2 x − 6 αe 2 x + 2 αe 2 x ≡ e 2 x . That is, 0 ≡ e 2 x , which is impossible Since y = αe 2 x is unsatisfactory, we investigate, instead, y = F ( x ) e 2 x , where F ( x ) is a function of x instead of a constant. We have d y d x = 2 F ( x ) e 2 x + F ′ ( x ) e 2 x . Hence, d 2 y d x 2 = 4 F ( x ) e 2 x + 2 F ′ ( x ) e 2 x + F ′′ ( x ) e 2 x + 2 F ′ ( x ) e 2 x . 2

Substituting these into the differential equation, (4 F ( x ) + 2 F ′ ( x ) + F ′′ ( x ) + 2 F ′ ( x )) e 2 x + ( − 6 F ( x ) − 3 F ′ ( x ) + 2 F ( x )) e 2 x ≡ e 2 x . That is F ′′ ( x ) + F ′ ( x ) = 1 . This is satisfied by the function F ( x ) ≡ x. Thus a suitable particular integral is y = xe 2 x . Note: It may be shown in other cases that, if the standard trial solution is already contained in the complemen- tary function, then it is is necessary to multiply it by x in order to obtain a suitable particular integral. 2. Determine the complementary function and a partic- ular integral for the differential equation d 2 y d x 2 + y = sin x. 3

Solution The auxilary equation is m 2 + 1 = 0, with solutions m = ± j . Hence, the complementary func- tion is A sin x + B cos x , where A and B are arbitrary constants. A trial solution of y = α sin x + β cos x gives d 2 y d x 2 = − α sin x − β cos x. Substituting into the differential equation, 0 ≡ sin x , which is impossible. Here, we may try y = x ( α sin x + β cos x ) , giving d y d x = α sin x + β cos x + x ( α cos x − β sin x ) = ( α − βx ) sin x + ( β + αx ) cos x. Therefore, d 2 y d x 2 = ( α − βx ) cos x − β sin x − ( β + αx ) sin x + α cos x 4

= (2 α − βx ) cos x − (2 β + αx ) sin x. Substituting into the differential equation, sin x ≡ (2 α − βx ) cos x − (2 β + αx ) sin x + x ( α sin x + β cos x ) . That is, 2 α cos x − 2 β sin x ≡ sin x. Hence, 2 α = 0 and − 2 β = 1. An appropriate particular integral is now y = − 1 2 x cos x. 3. Determine the complementary function and a partic- ular integral for the differential equation 9d 2 y d x 2 + 6d y d x + y == 50 e − 1 3 x . The auxiliary equation is 9 m 2 + 6 m + 1 = 0, or (3 m + 1) 2 = 0, which has coincident solutions, m = − 1 3 . Hence, the complementary function is ( Ax + B ) e − 1 3 x . 5

In this example, both 3 x and xe − 1 e − 1 3 x are contained in the complementary function. Thus, in the trial solution, it is necessary to multiply by a further x , giving y = αx 2 e − 1 3 x . We have d y 3 x − 1 d x = 2 αxe − 1 1 3 x 2 e 3 x and d 2 y 3 x − 2 3 x − 2 3 x + 1 d x 2 = 2 αe − 1 3 αxe − 1 3 αxe − 1 9 αx 2 e − 1 3 x . Substituting these into the differential equation, 50 e − 1 3 x ≡ 18 α − 12 αx + αx 2 + 12 αx − 2 αx 2 + αx 2 e − 1 3 x . � � Hence 18 α = 50 or α = 25 9 . An appropriate particular integral is y = 25 9 x 2 e − 1 3 x . 4. Determine the complementary function and a partic- ular integral for the differential equation d 2 y d x 2 − 5d y d x + 6 y = sinh 2 x. 6

Solution The auxiliary equation is m 2 − 5 m + 6 = 0 or ( m − 2)( m − 3) = 0, which has solutions m = 2 and m = 3. Hence, the complementary function is Ae 2 x + Be 3 x . However, sinh 2 x ≡ 1 2( e 2 x − e − 2 x ) . Thus, part of sinh 2 x is contained in the complemen- tary function and we must find a particular integral for each part separately. (a) For 1 2 e 2 x , we may try y = xαe 2 x , giving d y d x = αe 2 x + 2 xαe 2 x and d 2 y d x 2 = 2 αe 2 x + 2 αe 2 x + 4 xαe 2 x . 7

Substituting these into the differential equation, (4 α + 4 xα − 5 α − 10 xα + 6 xα ) e 2 x ≡ 1 2 e 2 x . This gives α = − 1 2 . (b) For − 1 2 e − 2 x , we may try y = βe − 2 x , giving d y d x = − 2 βe − 2 x and d 2 y d x 2 = 4 βe − 2 x . Substituting these into the differential equation, (4 β + 10 β + 6 β ) e − 2 x ≡ − 1 2 e − 2 x , which gives β = − 1 40 . The overall particular integral is thus, y = − 1 2 xe 2 x − 1 40 e − 2 x . 8