Math 266, Practice Midterm Exam 2

Transcription

1 Mh 66, Prcic Midrm Exm Nm: Ground Rul. Clculor i NOT llowd.. Show your work for vry problm unl ohrwi d (pril crdi r vilbl). 3. You my u on 4-by-6 indx crd, boh id. 4. Th bl of Lplc rnform i vilbl h l pg.

2 Pr I: Mulipl Choic (5 poin) ch. For ch of h following quion circl h lr of h corrc nwr from mong h choic givn. (No pril crdi.). To u h mhod of undrmind cofficin o find priculr oluion of h diffrnil quion y (3) 3y + 3y y = 4, which of h following form of Y () hould w ry? () Y () = A (b) Y () = A (c) Y () = A (d) Y () = A 3 () Y () = (A A 4 ). A m wighing 4 lb rch pring f. h m i chd o vicou dmpr wih dmping conn lb-c/f. Th m i pulld down n ddiionl 3 in, nd hn rld. L u = u() dno h diplcmn of h m from h quilibrium. (Th grviy conn i g = 3f/c. ) Thn u ifi h diffrnil quion nd iniil condiion: () u + 6u + 3u = 0, u(0) = 0, u (0) = 0. (b) u + 6u + 3u = 0, u(0) = 0.5, u (0) = 0. (c) 4u + u + 3u = 0, u(0) = 0.5, u (0) = 0. (d) 4u + u + u = 0, u(0) = 0, u (0) = 0. () 4u + u + u = 0, u(0) = 0.5, u (0) = Which of h following form fundmnl of oluion o h homognou diffrnil quion y (4) + y + y = 0. () {co, in,, } (b) {,,, } (c) {co, in, co, in } (d) {, } () { co, in, co, in } 4. Suppo h following h iniil vlu problm dcrib h moion of crin pring-m ym: 4u () + u () + u() = 3 co(), u(0) =, u (0) = 0. Thn wh w cn conclud for lim u()?

3 () (b) (c) (d) lim u() do no xi. lim = +. lim u() =. lim u() = 0. () lim u() = 0 <, 5. Conidr h funcion f() = in,. Thn f() = () + u ()(in ) + u ()( in ) (b) + u () in + u () (c) + u () in( ) + u () (d) + u ()(in ) + u ()( in ) () + u ()(in( ) ) + u ()( in( )) Anwr Ky: (d), (b), (c), (), (). Pr II: Wrin nwr quion 6. Vrify h y (x) = x i oluion o x y + xy y = 0, x > 0. Thn find h gnrl oluion of h bov quion. (5 poin) Soluion: Sinc y (x) = x =, w h x y + xy y = x x = 0. So y (x) = x i oluion of givn quion. To find h cond oluion, w u h mhod of rducion of ordr. Aum h y(x) = y (x)v(x) i oluion of h quion. Thn w found h v(x) ifi h quion whr p(x) = x/x = /x hr. y v + (y + p(x)y )v = 0 Sinc y (x) = x, w hv h diffrnil quion xv + ( + x x)v = 0, h i xv + 3v = 0. Hnc w olv v = Cx 3. So w find h v = Cx + C. 3

4 In priculr, w cn choo v(x) = x. So y(x) = y (x)v(x) = x i oluion of h quion. I i y o chck h h Wronkin W (x, x ) = x x x = x 0 Hnc x nd x form fundmnl of oluion. Thu h gnrl oluion i givn by y(x) = C x + C x. 7. Conidr h diffrnil quion y y = 3, > 0. () () Show h y () = nd y () = form fundmnl of oluion for h homognou quion y y = 0. (0 poin) Soluion: I i ndrd o chck y () nd y () r oluion of () (Wll, you hv o how m h dil of compuion in h xm.) To chck h y () nd y () form fundmnl of oluion, i uffic o chck h Wronkin i nonzro. W hv W (y, y ) = = = 3 0 (b) Find priculr oluion of quion ().(0 poin) Soluion: L u find h priculr oluion by vriion of prmr: y() = y () y g W (y, y ) + y () y g W (y, y ). whr g = 3 = 3. From h bov, w hv known h W (y, y ) = 3. Now w g y() = 3 ( (3 ) + (3 )) = + ln 3. 4

5 8. A m wighing 4 lb rch pring.5 in. Th m i diplcd in. in h poiiv dircion from i quilibrium poiion nd rld wih no iniil vlociy. Auming h hr i no dmping nd h h m i cd on by n xrnl forc of co 3 lb. () Formul h iniil vlu problm dcribing h moion of h m.(0 poin) Soluion: W hv mg = 4 nd L =.5/ = /8. Hnc k = mg/l = 3 nd m = 4/3 = /8. No h u(0) hould b in = /6 f. Now w g h iniil vlu problm: 8 u + 3u = co 3, u(0) = 6, u (0) = 0. (b) Find h oluion of h iniil vlu problm.(0 poin) Soluion: Th bov quion i quivln o u + 6 u = 6 co 3. I i y o h chrcriic polynomil of h homognou quion i r + 6. Thn h gnrl oluion for h homognou quion i C co 6 + C in 6. To find priculr oluion, w um h Y () = A co 3 + B in 3 wih A, B undrmind cofficin. Thn w hv B = 0 nd A = 6. So h gnrl oluion i 6 9 u() = C co 6 + C in co Now by h iniil vlu condiion, w g C = 6, nd C = 0. Thn w find h oluion of hi iniil vlu problm. u() = ( ) co 6 + co (c) If h givn xrnl forc i rplcd by forc 4 in ω of frquncy ω, find h vlu of ω for which ronnc occur.(0 poin) Soluion: By h formul, w know ronnc occur whn ω = ω mx ifi h following quion ω mx = ω 0( γ mk ). Bu w know γ = 0 in hi c. So ω mx = ω 0 = 6. 5

6 9. Conidr h iniil vlu problm y y + y = ; y(0) =, y (0) = 0. () Find h Lplc rnform Y () = L(y). Soluion: From h bov quion, w hv L(y ) L(y ) + L(y) = L( ) =. So Y () y(0) y (0) (Y () y(0)) + Y () =. Plug in y(0) =, y (0) = 0. W g Y () = (b) Find y = L (Y ()). + + ( )( + ). Soluion: Fir no h So Aum h + = ( ) + ( ) +. L ( + ) = co in. ( )( + ) = + b + c +. W olv h =, nd b = c =. So ( )( + ) = ( ) +. Thn h invr Lplc rnform of h bov i co. So y = in +. 6

7 Tbl of Lplc Trnform - f f - ( ) = L F( ) F( ) = L ( ) ( ) = L F( ) F( ) = L f ( ) f { } { }.. 3. n, n=,,3, K n! n + 4. p, p > in ( ) + 9. in ( ) +. in( ) - co( ) 3. co( ) - in ( ) p ( ) 3 ( + ) ( - ) ( + ) in 5. in ( + b) ( b) + co( b) + inh - b in b - + b 7. ( ) 9. ( ). inh ( b) 3. n, n=,,3, K 5. u ( c ) = u ( - c ) Hviid Funcion ( ) b ( ) - -b n! n ( - ) + -c -c F 6. n- 8. ( ) { }, n=,,3, K - G p + ( ) p + { } ( n ) 35 L - n n + co + - co + 0. ( ). in( ) + co( ) 4. co( ) + in ( ) ( ) ( + ) ( + 3 ) ( + ) co 6. co( + b) ( b) - in ( b) + coh - - co b - + b 8. ( ) 0. ( ). coh ( b) 4. f ( c ) d 6. ( - c) Dirc Dl Funcion ( ) - ( ) - -b Ê ˆ F Á c Ëc -c 7. uc ( ) f ( - c) ( ) 8. uc ( ) g( ) L g( + c) 9. c n ( f ( ) F( - c) 30. f ( ), n=,,3, K ( ) n n - F ) ( ) F Ú F( u) du 3. Ú f ( v) dv ( ) 3. f () 33. Ú f ( -) g( ) d F( ) G( ) 34. f ( T) f ( ) 0 0 -c { } + = T - Ú f () d 0 -T - F -f 0 - f f ( ) F( ) - f ( 0) 36. f ( ) ( ) ( ) ( ) ( n 37. ) n n- n- ( n f ( ) ( ) ( ) ( ) ) ( n F - f 0 - f 0 L -f - ( 0) - f - ) ( 0) p