The particular integral :

Transcription

1 Second order linear equation with constant coefficients The particular integral : d f df Lf = a + a + af = h( x)

2 Solutions with combinations of driving functions d f df Lf = a + a + af = h( x) + h( x) d f df Lf = a + a + af = h( x) d f df Lf = a + a + af = h( x) Since the equation is linear a solution to the original equation is given by f = f+ f

3 Sinusoidal h d f df Lf = a a af h( x) + + =. h= Hcos x so Lf a f + a f+ a f = Hcosx Lg x ag ag ag H i ( ) '' + ' + = e x e( Lg) = L[ e( g)] = e( He ) = H e(e ) = Hcosx f =e( g) is the solution to the real equation Solution: i g Pe x H = + + P=. a + ia + a = Lg ( a i a a) Pe f He =e( ) ( a a ) + ia ( a a )cosx+ asinx = H. ( a a) + a

4 Ex 5 f + 3f+ f = cosx. i g'' + 3 g' + g = e x. g = Pe where P=. + 3i+ PI e f = e( ) = (cosx+ 3sin x). + 3i CF x x f Ae = + Be General solution x f = Ae + Be + (cosx+ 3sin x) x

5 f + 3f+ f = cosx. x f = Ae + Be + (cosx+ 3sin x) x x() = 4, x'() = 7 3 A= B= 5

6 Ex 5 f + 3f+ f = cosx. What if cos(x) sin(x)? i g'' + 3 g' + g = e x. f =m( g) g = Pe where P=. + 3i+ PI e f = m( ) = (sinx3cos x). + 3i CF mx mx f Ae = + Be General solution mx mx f Ae = + Be + x x (sin 3cos )

7 Ex 6 f + 3f+ f = 3cosx+ 4sinx. = + = i( x+ ) 5cos( x ) 5 e(e ) where = arctan( 4/ 3) Proof: cos( x+ ) = cosxcossinxsin A B A x B x A B x x A + B A + B cos + sin = + ( cos + sin ) A B cos( x ) = + +,. cos = A A + B, sin = B/ A + B and tan = BA /

8 Ex 6 f + 3f+ f = 3cosx+ 4sinx. = + = i( x+ ) 5cos( x ) 5 e(e ) where = arctan( 4/ 3) g'' + 3 g' + g = 5e x i( +) Trial solution : g= Pe x i( +) 5 5 P= =, + 3i+ + 3i i( x+ ) e f = 5 e( ) = [cos( x+ ) + 3sin( x+ )]. + 3i

9 Ex 7 f+ f = x g + g = i cos '' e x d d C.F. ( + i)( i) g = e g=ce, f=acos(x)+bsin(x) d d P.I. ( i)( i) e + g =. i Try g= Pxe x Then d d d e = ( + i)( i) Pxe = ( + i) Pe = ipe xe P= f= e( ) = xsinx i i

10 Ex 8 x f+ f = e (3cosx+ 4sin x) = -x i( x+) 5 e(e e ) Trial function g = Pe (i ) x+ i where = arctan( 4/ 3) P= 5 = 5. (i ) + i (i ) x+ i e x PI f = 5 e( ) = e [cos( x+ ) sin( x+ )]. i

11 Recap nd-order linear ODEs with constant coefficients: a f + a f + a f = h(x) General solution = PI + CF CF = c u + c u, u and u linearly independent solutions of the homogeneous equation Complementary function CF by solving auxiliary equation Particular integral PI by trial function with functional form of the inhomogeneous term Next: physical application to forced, damped oscillator

12 Oscillators = +. spring friction forcing mx m x mx mfcost The associated complex equation is Transients : z+ z + z = F. CF - Auxiliary equation : i e t z = e t. 4 > + + = = ± i 4 = ± i where /. 4 Complementary function Constant phase shift x= A t + B t = N t+ t/ t/ e [ cos( ) sin( )] e cos( ) Since >, the CF as t... CF describes "transients"

13 Steady state solutions z+ z + z = F. i e t No damping exponential Particular integral x it = Fe e( ) + i. i.e. the Particular integral describes the steady state solution after the transients have died away. Since the denominator = ( ) + e where arctan( ) the i particular integral can be written as i( t) Fe(e ) Fcos( t) x= =. ( ) + ( ) + For >, x achieves the same phase as F at t greater by - is called the " phase lag" of the response.

14 i( t) Fe(e ) Fcos( t) x= = ( ) + ( ) + The amplitude of the response is A F =, ( ) + This has a maximum when da = 4( ) + =. d R / is called the resonant frequency The frictional coefficient causes the resonant frequency to be less than the normal frequency arctan( )

15 Oscillators = +. spring friction forcing mx m x mx mfcost F= mfcos t, x i( ) Fe(e ) Fcos( ) = = ( ) + ( ) + Power Input (steady state) P= W =Fx W = t P= Fx = mfcost Fsin( t) ( ) + mf = [ cos( t)sin( t)] ( ) + mf = [sin( t ) + sin( )]. ( ) + x(t) x(t ) F' Average over a period P mf sin =. ( ) +

16 Energy dissipated = +. spring friction forcing mx m x mx mfcost m F D = mxx = ( ). + x= Fcos( t) ( ) + sin ( t ) = D= P= mf sin ( ) + since sin = / ( ) + ( arctan( ) )

17 Quality Factor = +. spring friction forcing mx m x m x mfcos t Energy content of transient motion that the CF describes E = mx + m x ( ) x= A t+ t/ e cos( ) = co ] t+ ) t mae [ 4 cos cos sin sin s ( E m A ( ) e t small ( / ) 4 Quality factor Q E() t = csc h( / ) / / E( t / ) E( t+ / ) e e (for small / ). Q is the inverse of the fraction of the oscillator s energy that is dissipated in one period - approximately the number of oscillations before the energy decays by factor e

18 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS more than unknown function: y (x), y (x),...,y n (x) set of ODEs that couple y,...,y n physical applications: systems with more than degree of freedom. dynamics couples differential equations for different variables. Example. System of first-order differential equations: y = F (x, y, y,...,y n ) y = F (x, y, y,...,y n ) y n = F n (x, y, y,...,y n )

19 An nth-order differential equation y (n) = G(x,y,y,y,...,y (n ) ) can be thought of as a system of n first-order equations. Set new variables y = y; y = y ;... ; y n = y (n ) Then the system of first-order equations y = y y n = y n y n = G(x, y, y,...,y n ) is equivalent to the starting nth-order equation.

20 Systems of linear ODEs with constant coefficients can be solved by a generalization of the method seen for single ODE: General solution = PI + CF Complementary function CF by solving system of auxiliary equations Particular integral PI from a set of trial functions with functional form as the inhomogeneous terms

21 Warm-up exercise The variables ψ(z) and φ(z) obey the simultaneous differential equations 3 dφ + 5ψ = z dz 3 dψ + 5φ =. dz Find the general solution for ψ. Next time we will consider explicit examples of solution of systems of ODE s with constant coefficients