dy dy 2 dy d Y 2 dx dx Substituting these into the given differential equation d 2 Y dy 2Y = 2x + 3 ( ) ( 1 + 2b)= 3

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1 Solutions 14() 1 Coplete solutions to Eerise 14() 1. (a) Sine f()= 18 is a onstant so Y = C, differentiating this zero. Substituting for Y into Y = 18 C = 18, hene C = 9 The partiular integral Y = 9. (This partiular integral an be seen learly without going through all the ehanis above. Only the last ter on the left hand side y plays a part. Sine you need 18, Y = 9). (b) Sine f()= + 3 is a linear funtion so Y = a + b d Y = a, = Substituting these into the given differential equation d Y Y = + 3 a a+ b = + 3 a a + b = + 3 Equating oeffiients of : a = hene a = 1 onstants: ( a + b)= 3 ( 1 + b)= 3 b = hene b = 1 Substituting a = 1 and b = 1 into Y = a + b Y = 1 () Sine f()= 13sin( 3) our trial funtion is (see (14.16)) Y = aos( 3) + bsin ( 3) = 3asin( 3) + 3bos( 3) by (6.13) by(6.1) = 9aos 3 9bsin 3 by (6.1) by (6.13) Substituting into Y = 13sin ( 3) 9aos 3 9bsin 3 3asin 3 + 3bos 3 aos 3 + bsin 3 = 13sin 3 (6.1) sin ( ) = os( ) (6.13) [ os( ) ] = sin( )

2 ( 9a 3b a) os( 3) + ( 3a 9b b) sin ( 3) = 13sin ( 3) ( 11a 3b) os( 3) + ( 3a 11b) sin ( 3) = 13sin ( 3) Solutions 14() Equating oeffiients of os( 3): 11a 3b = sin( 3): 3a 11b = 13 Solving these siultaneous equations a = 3 and b = 11. Putting a = 3 and b = 11 into our trial funtion Y = a os( 3)+ bsin( 3) the partiular integral Y = 3os( 3) 11sin( 3) (d) Sine f()= os() our trial funtion is given by Y = aos + bsin ( ) = asin + b os sin = aos b Substituting into Y = os ( ) aos( ) bsin ( ) asin ( ) + bos( ) aos( ) + bsin ( ) = os( a b a os + b+ a b sin = os Equating oeffiients of os : 3a b= 1 ( 3a b) os( ) + ( a 3b) sin( ) = os( ) sin : a 3b=, a= 3b Solving these siultaneous equations a = 3, b = 1. Substituting these into () Y = os( ) sin ( ) = 3os( ) + sin ( ) (e) Straightforward, y = e 3 4. The first two, (a) and (b), are unopliated. (a) y = Ae + Be + 3 (b) y = Ae 3 + Be + 7 () First we find the opleentary funtion y, (right hand side is equal to zero). The harateristi equation is + + 1= ( + 1) =, = 1 ( equal roots) By (14.5) y = ( A+ B)e. Sine f( )= 36e 5 so by (14.17), Y = Ce 5 Differentiating ) (14.5) Equal roots,, y = ( A + B)e (14.7) If f( )= Ae then y = Ce

3 Solutions 14() 3 Substituting into = 5Ce Y e 5 5 = 5Ce = Ce + 5Ce + Ce = 36e 5 36Ce 5 = 36e 5 whih C = 1 Substituting C = 1 into Y = Ce 5 the partiular integral Y = e 5 Adding the opleentary funtion, y, and the partiular integral the general solution y = A+ B e + e 5 (d) Sae proedure as part (a). The harateristi equation is + 3 4= ( 4)( 1) + = = 4, = 1 1 Sine we have distint roots so 4 y = Ae + Be Partiular integral Y: Sine f()= 34sin () we try the funtion Y = aos + bsin = asin + b os sin = aos b Substituting into + 3 4Y = 34sin ( ) aos( ) bsin ( ) + 3 asin ( ) + bos( ) 4 aos( ) + bsin ( ) = 34sin ( a+ 3b 4a os + b 3a 4b sin = 34sin ( 3b 5a) os( ) + ( 5b 3a) sin( ) = 34sin( ) Equating oeffiients of os : 3b 5a= sin : 5b 3a= 34 Solving these siultaneous equations a = 3, b = 5. Substituting for a and b into our trial funtion, Y = a os( )+ bsin( ), Y = 3os + 5sin and the general solution, y = y + Y, is 5sin 4 y = Ae + Be + 3os + )

4 Solutions 14() 4 3. By adding to both sides, the given differential equation beoes + = g Dividing through by + = g (*) Copleentary funtion : + = The harateristi equation r + = r + = By (14.8) = Aos( ωt) + Bsin ( ωt) where ω = Partiular integral X: Sine + = g and g is a onstant so X=C, X = X =. Substituting these into (*) C = g whih C = g Hene X = g. The general solution is = + X so we have 4. We have g = Aos( ωt) + Bsin ( ωt) + where ω = d 1 =.5+ 4 dt 3 3 ( ) d 3 = (.5 + 4) dividing by dt d 4 =.5 (*) dt Charateristi equation is r 4 =, solving r1 = and r =. By (14.4) = Ae t + Be t Sine f()=.5 our trial funtion is a onstant, X = C. Substituting into (*) 4C =.5 C =.15 The general solution is found by adding together the opleentary funtion, = Ae t + Be t, and the partiular integral, X =.15: t t = Ae + Be.15 r 1 r (14.4) If r1 and r then y = Ae + Be r + = y = Aos + Bsin ) (14.8) (

5 Solutions 14() 5 5. By adding to both sides, the differential equation beoes + = F αt sin Copleentary funtion : + =, + = Thus the harateristi equation is By (14.8) A ( t) B ( t) Partiular integral X: By (14.16) r + = = os ω + sin ω where ω = ( α ) sin ( α ) X = aos t + b t Differentiating X = α asin ( αt) + bos( αt) X = α aos( αt) bsin ( αt) = α aos( αt) + bsin ( αt) Substituting into F X + X = sin ( αt) F α aos( αt) + bsin ( αt) + aos( t) bsin ( t) sin α + α = αt F α aos( αt) + α bsin ( αt) = sin ( αt) Equating oeffiients of os( αt): α a =, a = beause α sin( αt): α b = F F F b = α = α / / F Substituting a = and b = into X = aos( αt)+ bsin αt α F X = sin ( αt ) α The general solution, = + X, is F = Aos( ωt) + Bsin ( ωt) + sin ( α t) where ω = α If f ( ) Asin then Y aos bsin ) (14.16) = = + (

6 Solutions 14() 6 6. Very siilar to solution Rearranging the equation to + + = g + L (*) Two parts, first find the opleentary funtion; + + = Dividing by + + = The harateristi equation is r + r + = Using the quadrati equation forula, (1.16), ± r = = ± 1 = ± 4 1 = ± 4 1 = ± ( 4 ) 1 1 r = ± j 4 = ± j 4 beause < 4 Using (14.6) with α = and β = 1 4 t ( β ) ( β ) = e Aos t + Bsin t where represents the opleentary funtion. We need to find the partiular integral; Sine the right hand side of (*) is a onstant, g + L, so by (14. 11) X = C, X = and X = Substituting into X Ý + X Ý + X = g + L C = g + L g C = + L (1.16) b± b 4a r = a (14.6) α If = α ± jβ then = e t Aos( βt) + Bsin ( βt) (14.11) If f ( ) = onstant then Y =C

7 Solutions 14() 7 g Hene the partiular integral is X = + L. The general solution, = + X, is t g = e Aos( βt) + Bsin ( βt) + + L 1 where β = Dividing the given differential equation by CL we have di 1 di i I + + = (*) dt RC dt LC LC Putting R = 1 3, C = 9 and L = = = RC = = 3 9 LC 5 Substituting these and I = 5 3 into (*) di 5 di ( 1 ) + ( ) i = ( )( 5 ) dt dt Copleentary funtion i : The harateristi equation is ( 1 ) + ( ) = Putting a = 1, b = 1 5 and = 9 into (1.16) By (14.4) ( 1 ) ( 1 ) ( 4 ) ± = 4 4 = ( 5 ) ± (.36 ) =.764 and = ( ) t ( ) t i = Ae + Be Partiular integral I : Sine the right hand side of (**) is 9 trial funtion. 5 3 Substituting into ( (**), a onstant, so our I = K ( K is onstant) di 5 di I = 5 ) dt dt K = 5 K = 5 I = (1.16) (14.4) b b 4a ± = a t If and then y = Ae + Be 1 1 t

8 Solutions 14() 8 We now i = i + I 4 4 (. ) t (. ) t i = Ae + Be By looing at EXAMPLE 7 of this hapter, we now the opleentary funtion y = Ae + Be. (Sae harateristi equation). The trial funtion is Y = Ce beause Ce is already part of the opleentary funtion. Y = Ce ( ) Differentiating by using the produt rule (6.31) = Ce Ce = Ce ( Ce Ce ) = Ce + Ce Substituting into Y = 5e Ce + Ce Ce + Ce Ce = 5e C + C C + C C e = 5e 3C = 5 C = 5 / Hene Y = e putting C = into ( ) 3 3 The general solution is given by y= y + Y, so we have y = Ae + Be e 3.(a) Charateristi equation is = 5 ( )( ) = = 3, = 1 1 Sine we have two distint roots so y = Ae 3 + Be Hene the trial funtion is Y = Ce 3 (b) Charateristi equation is + 9= + 3 = By (14.8), y = Aos( 3)+ Bsin( 3). The trial funtion ould be aos( 3) + bsin ( 3 ) Sine a os( 3) is already part of the opleentary funtion, so we try Y = aos( 3) + bsin ( 3) (6.31) ( uv) = u v + v u (14.8) + = then y = Aos( )+ Bsin( )

9 Solutions 14() The differential equation an be rearranged to d y P P π + y = sin (*) EI EI L Copleentary funtion y ; As before P y = Aos( )+ Bsin( ) where = EI Partiular integral Y ; Sine the right hand side of (*) is P π sin EI L (14.16) our trial funtion is π π Y = Cos + Dsin (**) L L π π π = Csin Dos L + L L π π π = Cos Dsin L + L L Substituting into d Y + P EI Y = P π sin EI L π π π P π π Cos Dsin Cos Dsin L + L L + EI + L L P π = sin EI L Equating oeffiients of os π L ; P π C = EI L C = Equating oeffiients of sin π L ; P π P D EI L = EI PL π EI P D = ( EI ) L EI so by Multiply through by EI D ( PL π EI ) Substituting C = and D = L PL = P D = EIπ PL PL EIπ into (**) PL (14.16) If f()= Asin( ) then Y = a os( ) + bsin( )

10 Solutions 14() PL sin π Y = EIπ PL L The general solution, y, is given by PL π y = Aos( ) + Bsin ( ) + sin EIπ PL L