9.6 Enginring Diffrntial Equation Practic Final Exam Solution Fall 0 Problm. (0 pt.) Solv th following initial valu problm: x y = xy, y() = 4. Thi i a linar d.. bcau y and y appar only to th firt powr. pt. Firt writ th d.. in tandard form: y y = x x. pt. Th intgrating factor i ρ(x) = /x dx = ln(x) = x. pt. Multiplying both id of th tandard form of th d.. by th intgrating factor, w hav ( x y y = x x) ( x ) pt. x y x y = x d x y = x dx x y = x dx = x + c. pt. y() = 4 (4) = + c c =. pt. Thrfor, x y = x + y = x + x. Problm. (0 pt.) Solv th following initial valu problm: xy x + x {{ M + ln x + y = 0 {{ N xy x + x + ln x + y = 0 with y(5) = 0. Thi d.. i both linar (bcau y and y appar only to th firt powr) and xact (bcau M/ y = N/ x. pt. ( ( x/ x To trat th d.. a linar, firt writ th d.. in tandard form: y + ) ) x + ln(x y = + ) ln(x + ). pt. ( x/ ( x + ) Th intgrating factor i ρ(x) = xp ln(x + ) dx {{ u=ln(x +) ) ) = lnln(x +) = ln (x +. pt. Multiplying both id of th tandard form of th d.. by th intgrating factor, w hav x ln x + y + x y = x pt. d { ln x + y = x + dx ln x + y = x dx = x + c. pt. y(5) = 0 ln 5 + (0) = 5 + c c = 5. pt. Thrfor, ln x + y = x 5.
If you rgard th d.. a an xact quation, th olution of th d.. i f(x, y) = c, whr th function f atifi th condition f x = M = xy f x x and + y = N = ln x +. f x = xy xy x + x f = x + x dx = y ln x + x + g(y) pt. f y = ln x + + g (y). But f y = N = ln x +, o ln x + + g (y) = ln x + g (y) = g(y) = y pt.. Thrfor, f = y ln ( x + ) x y, o th olution of th d.. i y ln ( x + ) x y = c. pt. Th initial condition y(5) = 0 0 ln 5 + 5 (0) = c c = 5 pt.. Thrfor, th olution of th givn IVP i y ln ( x + ) x y = 5 Problm. (5 point) A 000 gallon holding tank that catch runoff from om chmical proc initially ha 600 gallon of watr with Q(0) = ounc of pollutant diolvd in it. Pollutd watr flow into th tank at a rat of gal/hr and contain 5 ounc/gal of pollutant in it. Th wll mixd olution lav th tank at gal/hr a wll. Aftr how many hour will th amount of pollutant in th tank rach 500 ounc? dq = rat in - rat out = (flow rat in)(concntration in) - (flow rat out)(concntration out), dt pt. o dq ( dt = gal ) ( 5 oz ) ( gal ) Q oz. 5 pt. hr gal hr 600 gal (Th volum of liquid in th tank i contant bcau th flow rat in qual th flow rat out.) Thrfor, th d.. modling thi mixing problm i dq dt = 5 Q Thi d.. i both parabl and linar d.. pt. Trating th d.. a a parabl quation, w hav dq dt = 5 Q = (Q 000) dq Q 000 = dt dq dt Q 000 = t ln Q 000 = + c 5 pt. Th initial condition Q(0) = ln 000 = 0 + c c = ln(998) pt. ln Q 000 = t + ln(998) To dtrmin whn th amount of pollutant in th tank rach 500 ounc, t Q = 500 and olv for t: ln 500 000 = t + ln(998) t = (ln(500) ln(998)) 6 hour pt. To trat th d.. a a linar quation, firt put it into tandard form: dq dt + Q = 5. pt. Th intgrating factor i ρ(t) = / dt = t/. pt. Multiplying both id of th tandard form of th d.. by th intgrating factor, w hav dq t/ dt + Q = 5 t/ t/dq dt + t/ Q = 5 t/ d t/ Q = 5 t/ dt t/ Q = 5 t/ dt = 000 t/ + c. pt. Q(0) = 0 () = 000 0 + c = c = 998. pt. Thrfor, t/ Q = 000 t/ 998 Q = 000 998 t/.
To dtrmin whn th amount of pollutant in th tank rach 500 ounc, t Q = 500 and olv for t: 500 = 000 998 t/ t = ln(500/998) 6 hour pt. Problm 4. (0 point) Find th gnral olution to ach of th following diffrntial quation: a. (4 point) y + 6y + 9y = 0 Th charactritic quation i r + 6r + 9 = 0 (r + ) = 0 r = (doubl root). pt. Thrfor, y = c x + c x x. pt. b. (6 point) y () 6y + y = 0 Th charactritic quation i r 6r + r = 0 r r 6r + = 0 r = ( 6) ± ( 6) 4()() () r 6r + = 0. = 6 ± 6 Thu, th root of th charactritic quation ar 0 and ± i. pt. = 6 ± 4i = ± i Thrfor, y = c 0x + c x co(x) + c x in(x) or y = c + c x co(x) + c x in(x). 4 pt. Problm 5. (5 point) Solv th following initial valu problm: y + 4y = 6 x 8x with y(0) = and y (0) = 0. Stp. Find y c by olving th homognou d.. y + 4y = 0. Charactritic quation: r + 4 = 0 r = 4 r = ± 4 = 0 ± i. Thrfor, y c = c 0x co(x) + c 0x in(x) = c co(x) + c in(x). pt. Stp. Find y p uing ithr th Mthod of Undtrmind Cofficint or th Mthod of Variation of Paramtr. Mthod. Undtrmind Cofficint. Sinc th nonhomognou trm in th d.. (6 x 8x ) i th um of an xponntial function and a polynomial of dgr, w gu that y p i th um of an xponntial function and a polynomial of dgr : y p = A x +Bx +Cx+D. No trm in thi gu duplicat a trm in y c, o thr i no nd to modify th gu. pt. y = A x +Bx +Cx+D y = A x +Bx+C y = 4A x +B. Thrfor, th lft id of th d.. i y + 4y = 4A x + B + 4 A x + Bx + Cx + D = 8A x + 4Bx + 4Cx + B + 4D. W want thi to qual th nonhomognou trm 6 x 8x, o 8A = 6, 4B = 8, 4C = 0, and B + 4D = 0 A =, B =, C = 0, and D =. Thrfor, y p = x x +. 6 pt. Mthod. Variation of Paramtr. From y c w obtain two indpndnt olution of th homognou d.: y = co(x) and y y = in(x). Th Wronkian i givn by W(x) = y y y = co(x) in(x) in(x) co(x)
= co(x) ( co(x)) ( in(x)) in(x) = co (x) + in (x) =. pt. ( y f(x) in(x) 6 x 8x ) u = dx = dx = 4x in(x) dx 8 x in(x) dx W(x) For th firt intgral, u ubtitution with u = x and u ntri 40 (with n = ) and 9 from th Tabl of Intgral: 4x in(x) dx = u in(u) du = u co(u) + u co(u) du = u co(u) + (co(u) + u in(u)) = x co(x) + co(x) + x in(x) For th cond intgral in th formula for u, u ntry 49 from th Tabl of Intgral with a = and b = : 8 x x in(x) dx = 8 + in(x) co(x) = x in(x) + x co(x) Thrfor, u = x co(x) + co(x) + x in(x) 4 x in(x) + 4 x co(x) pt. ( y f(x) co(x) 6 x u = W(x) dx = 8x ) dx = 8 x co(x) dx 4x co(x) dx For th firt intgral, u ntry 50 from th Tabl of Intgral with a = and b = : 8 x x co(x) dx = 8 + co(x) + in(x) = x co(x) + x in(x) For th cond intgral in th formula for u, u ubtitution with u = x and u ntri 4 (with n = ) and 8 from th Tabl of Intgral: 4x co(x) dx = u co(u) du = u in(u) u in(u) du = u in(u) (in(u) u co(u)) = x in(x) + in(x) x co(x) Thrfor, u = x co(x) + x in(x) x in(x) + in(x) x co(x) pt. and y p = u y + u y = x co(x) + co(x) + x in(x) x in(x) + x co(x) co(x) + x co(x) + x in(x) x in(x) + in(x) x co(x) in(x) = x co (x) + co (x) + x in(x) co(x) x in(x) co(x) + x co (x) + x co(x) inx) + x in (x) x in (x) + in (x) x co(x) in(x) = x co (x) + in (x) + co (x) + in (x) + x co (x) + in (x) = x + + x pt. Stp. y = y c + y p, o y = c co(x) + c in(x) + x x +. pt. Stp 4. U th initial condition to find c and c. y = c co(x) + c in(x) + x x + y = c in(x) + c co(x) + 4 x 4x y(0) = = c co(0) + c in(0) + 0 (0) + = c + c = 0 pt. y (0) = 0 0 = c in(0) + c co(0) + 4 0 4(0) = c + 4 c = pt. Thrfor, y = in(x) + x x +.
Problm 6. (5 point) Conidr a dampd, forcd ma/pring ytm. Lt t dnot tim (in cond) and lt x(t) dnot th poition (in mtr) of th ma at tim t, with x = 0 corrponding to th quilibrium poition. Suppo th ma m = kg, th damping contant c = N /m, th pring contant k = N/m, th xtrnal forc i F(t) = 0 co(t), th initial poition x(0) = m, and th initial vlocity x (0) = m/. a. ( point) Find th poition function x(t). Th d.. modling thi ytm i mx + cx + kx = F(t), or x + x + x = 0 co(t). pt. Stp. Find x c by olving th homognou d.. x + x + x = 0. Charactritic quation: r + r + = 0 (r + )(r + ) = 0 r = or r = x c = c t + c t. pt. Stp. Find x p uing ithr th Mthod of Undtrmind Cofficint or th Mthod of Variation of Paramtr. Mthod. Undtrmind Cofficint. Sinc th nonhomognou trm in th d.. (0 co(t)) i a coin, w gu that x p i th um of a coin and in with th am frquncy: x p = A co(t) + B in(t). No trm in thi gu duplicat a trm in x c, o thr i no nd to modify th gu. pt. x = A co(t) + B in(t) x = A in(t) + B co(t) x = 4A co(t) 4B in(t). Thrfor, th lft id of th d.. i x +x +x = 4A co(t) 4B in(t)+ A in(t) + B co(t)+ A co(t) + B in(t) = A + 6B co(t) + 6A Bin(t). W want thi to qual th nonhomognou trm 0 co(t), o A + 6B = 0 and 6A B = 0 A = and B =. Thrfor, x p = co(t) + in(t). 4 pt. Mthod. Variation of Paramtr. From x c w obtain two indpndnt olution of th homognou d.: x = t and x = t x. Th Wronkian i givn by W = x x x = t t t t ( = t t) ( t) t = t. pt. x F(t) t (0 co(t)) u = dt = W t dt = 0 t co(t) dt t = 0 + co(t) + in(t) = 5 t co(t) 5 t in(t) pt. whr w hav ud ntry 50 from th Tabl of Intgral. x F(t) t u = W dt = (0 co(t)) t dt = 0 t co(t) dt t = 0 + co(t) + in(t) = 4 t co(t) + 8 t in(t) pt. whr again w hav ud ntry 50 from th Tabl of Intgral. x p = u x + u x = 5 t co(t) 5 t in(t) t + 4 t co(t) + 8 t in(t) t = co(t) + in(t) pt. Stp. x = x c + x p, o x = c t + c t co(t) + in(t). pt. Stp 4. U th initial condition to find c and c. x = c t + c t co(t) + in(t) x = c t c t + in(t) + 6 co(t) x(0) = = c 0 + c 0 co(0) + in(0) = c + c c + c = x (0) = = c 0 c 0 + in(0) + 6 co(0) = c c + 6 c c = 4 c + c =, c c = 4 c =, c = 0. pt. Thrfor, x = t co(t) + in(t).
b. ( point) What part of th olution you found in part a i th tranint part of th olution, x tr? Th tranint part of th olution i th part that approach 0 a t, o x tr = t c. ( point) What part of th olution you found in part a i th tady-tat (tady priodic) part of th olution, x p? Th tady-tat part of th olution i th part that do not approach 0 a t, o x p = co(t) + in(t) Problm 7. (0 point) Find L {F() whr a. F() = 6 U a partial fraction dcompoition: 6 = ( )( + ) = A + B Multiplying both id of th quation ( )( + ) = A + B by th dnominator ( )(+), w obtain = A( + ) + B( ) = A + A + B B = (A + B) + A B A + B =, A B = 0 A = 9/5, B = 6/5. pt. Thrfor, L { 6 { { { = L 9/5 + 6/5 = 9 5 L + 6 5 L + = 9 5 t + 6 5 t uing th Laplac Tranform tabl ntry L { a b. F() = 8 4 + ( + 4) = at pt. U a partial fraction dcompoition: 8 4 + ( = A + 4) + B + C + 4 Multiplying both id of thi quation by th dnominator + 4 w obtain 8 4 + = A + 4 + (B + C) = (A + B) + C + 4A A + B = 8, C = 4, and 4A = A =, B = 5, and C = 4. pt. Thrfor, { 8 L { 4 + ( = L + 4) + 5 4 { { { = L + 5L + 4 L + 4 = + 4 + 5 co(t) in(t) uing th Laplac Tranform tabl ntri { { { L =, L + k = co(kt), and L k + k = in(kt) pt. Problm 8. (5 point) a. Find L{h(t) whr h(t) i th picwi-dfind function givn by { if 0 t < 0 h(t) = 0 if t 0. h(t) = u(t 0) pt. o L {h(t) = L { u(t 0) = L { L {u(t 0) ( 0 = = ) 0 pt.
b. Suppo y(t) i th iz of a population of fih (maurd in ton) t day aftr th tart of th fihing aon, with a % individual growth rat. Suppo that y(0) = 00 and that w harvt h(t) ton pr day whr h(t) i th picwi-dfind function givn in part a (i.., w allow a 0 day harvt bfor impoing a ban on harvting). Th balanc law for population growth thu giv y (t) = 0.0y(t) h(t), y(0) = 00 Solv thi initial valu problm for y uing Laplac Tranform. Solution to thi IVP not uing th Laplac tranform mthod will not rciv any crdit. y (t) = 0.0y(t) h(t) y 0.0y = h(t) L { y 0.0y = L { h(t) L { y 0.0L{y = L {h(t) pt. L{y y(0) 0.0L{y = 0 pt. L{y 00 0.0L{y = + 0 ( 0.0)L{y = 00 + 0 L{y = 00 0.0 ( 0.0) + 0 ( 0.0) pt. { 00 y = L 0.0 ( 0.0) + 0 { L { 00 y = L 0.0 ( 0.0) { = 00 0.0t f(t) + u(t 0)f(t 0) whr f(t) = L (W ud th Laplac Tranform tabl ntry L { a ( 0.0) + L { 0 ( 0.0) pt. ( 0.0) = at with a = 0.0.) To find f(t) w u a partial fraction dcompoition: ( 0.0) = A + B 0.0 A = + B ( 0.0) = A ( 0.0) + B 0.0 = A 0.0A + B = (A + B) 0.0A A + B = 0, { 0.0A = A = 00, B = 00 00 f(t) = L + 00 { { = 00L + 00L = 00() + 00 0.0t 0.0 0.0 { { (W ud th Laplac Tranform tabl ntri L = at with a = 0.0 and L =.) a Thrfor, y = 00 0.0t 00 + 00 0.0t + u(t 0) y = 00 0.0t + 00u(t 0) 0.0(t 0) pt. 00 + 00 0.0(t 0)