KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow

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1 KEY Mah 334 Miderm III Winer 008 secion 00 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering paper. Wrie your name, course, and secion number on he blue book, or on your own pile of papers. Again, do no wrie his or any oher ype of informaion on his exam. Good Pracice: A differenial equaion has he propery ha one can check wheher a given funcion saisfies i. So check your soluions! If one doesn work, ry again, or a leas noe ha your proposed soluion doesn work.

2 a. Compue he Laplace ransform L [ f ]( s) of f ( ) = e direcly from he definiion. For which values of s does your formula hold? Assume here ha a is a real number. poins Soluion By definiion of he Laplace ransform, and by using inegraion by pars, we ge L = = = = s a 0 + ( s a) + ( s a) + ( s a) = { e e d 0 0 } = e d s a s a 0 + ( s a) + ( s a) = de = e = 0 s a s a 0 s a s a ( s a + s s a ) ( ) [ f ]() s e f () d e e d e d de = ( s a). ( ) ( ) ( 0 ) s a (.) Here we assumed ha s > a, which can also be seen o be necessary for he inegral o converge.. Solve he following IVP. Using he Laplace ransform and knowing how o deal wih piecewise defined funcions in his ransform should make hings easier. Wha is he value of y (00)? Do no express his value in erms of an absrac formula, bu raher as a concree number. To compue his value concreely, you will need o know ha he funcion sin( ) is π periodic. 0, < 0, 0 < π y + y = f( ) =, y(0) = y (0) = 0. 4 π, π < 4π 0, 4π (.)

3 5 poins Soluion Taking he Laplace ransform of (.) we ge ( + ) L [ ] = L [ + ] = L [ ] s y ( s) y y ( s) f ( s) L, (.3) y () s f (). s s + [ ] = L [ ] To compue L [ f ]( s) we rewrie f in (.) as: ( 0 π ) ( π )( π 4π ) u0() ( 4π ) uπ() ( 4 π) u4π() ( ) ( π) ( π) f() = u () u () + 4 u () u () = + + = 0 u () u () + 4 u (). 0 π 4π (.4) So, according o he relevan heorem, and given L [ g] () s = when g () = u 0() (as s per he able provided you, or as per problem ), hen πs 4πs [ ] = [ ] [ ] + [ ] L f () s L g () s e L g () s e L g () s πs 4πs ( ) [ ] = e + e L g ( s) (.5) 4 ( πs πs = e + e ). s Thus, explicily, (.3) is πs 4πs L [ y] () s = L [ f ]() s = ( e + e ) (.6) s + s + s and we already know hen ha he soluion is of he form y() = h() u () h( π ) u () + h( 4 π ) u () (.7) 0 π 4π where we need hen only find h (), whose Laplace ransform is

4 3 L s + s s + 0+ s = +. s + s [ h] () s = = + (.8) Thus, according o he able, or memorized formulae, we have ha he soluion is given by (.7) and by Noe ha for any 4π, including = 00, we have h () = sin(). (.9) y () = hu () () h ( π ) u () + h ( 4 π ) u () 0 π 4π = h () h ( π) + h ( 4 π) ( ) ( π ) = sin( ) π sin( π) + 4π sin( 4 π) = sin( ) sin( ) + 4π sin( ) = 0. (.0) 3. Solve he following iniial value problem in erms of he convoluion inegral: 9 poins Soluion y + y = g( ), y(0) = y (0) = 0. (.) Here I expec you o wrie he soluion y = y ( ) of (.) as ( ) y() = yδ g () = yδ( τ ) g( τ) dτ, (.) 0 so ha he problem is effecively o deermine he funcion yδ ( ) in (.). (Hin: Use he Laplace ransform and he convoluion heorem. Noe ha he formula you obain for (.) could be used o solve he previous problem, bu is no as helpful as he mehod described in ha problem, paricularly for compuing a concree value of he soluion y a a large value of.) Noe he poin value his is sraighforward if you recall he relevan heory. By he Laplace Transform we have from (.) ha

5 4 ( + ) L [ ] = L [ + ] = L [ ] s y ( s) y y ( s) g ( s) L [ y] () s = L [ g] () s = L [ sin() ] s L [ g] () s s + = L [ sin g] (.3) so ha evidenly in (.) we have yδ () = sin(). (.4) 4. Suppose ha boh x () = and () = x (.5) solve he sysem and ha solves he sysem x = + x (.6) + x p () = (.7) Then wha is he soluion of he IVP x = + x + +. (.8) x =, (0) =? + x + + x 3 (.9) This problem ess your general undersanding of he srucure of soluions of linear sysems wihou requiring you o regurgiae a heorem (wih all of he inricacies of is ariculaion). Noe he poin value his problem is sraighforward if you recall he heory. poins

6 5 Soluion According o he informaion given, he general soluion of (.9) can be expressed as x + = x () = + c, (.0) Choosing = 0 and using he iniial daa in(.9), (.0) becomes (0) 3 = x = 0 c + 0 = c + 0 c = =. 3 (.) So (.0) becomes x = x () =. + = + = 3 (.) 5. Find he fundamenal marix of soluions Φ =Φ( ) o he sysem 0 he one ha has he propery haφ (0) = I = 0. poins Soluion 0 x = x, (.3) Firs we find a represenaion of he general soluion of he sysem(.3). This can be expressed as + λ x= x() = cξ e + c ξ e, (.4) λ + provided he ξ s are independen eigenvecors associaed wih he eigenvalues λ + and λ of he marix in (.3). To deermine his we noe

7 6 0 λ = = λ ξ 0 ξ 0 unless 0 λ 0= de = λ λ+ = ( λ ) + λ λ = ± i= + i, i = : λ, λ. + (.5) So ( i) 0 0 ± i = ± = ± ± =. ( ± i ) i ± i ξ m m ξ ξ (.6) Thus, explicily, (.4) is x ( + i ) ( i ) = x () = c e + c e. + i i (.7) As per he usual heory, we can find a real-valued represenaion by finding he real and imaginary pars of eiher of he above complex-valued soluions: ( + i ) cos sin x(): = e = e ( cos+ isin ) = e + i e, + i + i cos sin cos+ sin (.8) whence a real-valued represenaion of he general soluion is x cos sin () = c e + c e. cos sin cos+ sin (.9) A fundamenal marix of soluions Ψ =Ψ (), one no necessarily having he desired propery, can be found from he above general soluion (.9) as in cos sin Ψ () = e cos sin cos+ sin. (.30) The desired fundamenal marix Φ =Φ() can be obained from Ψ =Ψ() via

8 7 cos sin 0 Φ=Φ ( ) =Ψ( ) Ψ (0) = e cos sin cos sin + cos sin 0 = e cos sin cos sin + cos sin sin = e. sin cos+ sin (.3) 6. Solve he iniial value problem given by he sysem of problem 5 and he iniial daa 6 poins Soluion x (0) =. (.3) (Hin: raher han reinvening he wheel, jus use he fundamenal marix of problem 5.) Using he fundamenal marix of problem 5 we have ( cos sin ) + ( sin ) ( ) + ( + ) cos sin sin x() =Φ () x(0) = e e sin cos sin = + sin cos sin (.33) cos = e. cos sin 7. Calculae 0 π (Hin: use he resul of problem 5). 6 poins Soluion We have, from problem 5 and he general heory, e. (.34)

9 8 0 π π π cosπ sinπ sinπ π 0 e 0 e =Φ ( π ) = e e. π sinπ cosπ sinπ = = e (.35) 8. Find a represenaion of he general soluion of he sysem poins Soluion 3 x = x. (.36) The marix in (.36) has a repeaed eigenvalue wih only one eigenvecor. Hence he general soluion is of he form λ ( ) λ x = x() = cξe + c ξ+ η e (.37) where λ is he sole eigenvalue,ξ is one of is eigenvecors, and ηis an associaed pseudo eigenvecor: Firs 3 λ = = λ ξ 0 ξ 0 unless = 3 λ = λ + = ( ) 0 de λ 4λ 4 λ λ =,. (.38) So 3 0 =, ξ = ξ ξ = and η. η= ξ = = = 0 0 η 0 η η (.39) Thus, explicily, (.37) is

10 9 η x= x () = c e + c + e. η (.40) 9. Find he fundamenal marix of soluions Φ =Φ( ) for he sysem of problem 8 0 ha saisfies Φ (0) = I =. 0 5 poins Soluion From (.40) we have a fundamenal marix of soluions whence he one desired is e + η Ψ () = + η, (.4) η η + Φ () =Ψ() Ψ (0) =Ψ () = e + η η e = η η η e = η = e e + η η η η η η + ( η) + ( + η) ( η) + ( + η) ( η ) + ( + η ) ( η ) +( + η ) + =. (.4) 0. Solve he iniial value problem obained from combining he differenial equaion of problem 8 wih he iniial daa x (0) =. (.43) (Hin: do no reinven he wheel, bu raher use he resul from problem 9.) 6 poins

11 0 Soluion From problem 9 we have + + x() =Φ () x(0) = e = e ( ) ( ) ( ) + = = e e. (.44)

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