MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we can wrie i as dx an x = d. 2 Inegrae i o obain ln ( sin x ) = 2 /2 + C which gives afer exponeniaion and relabeling he consan sin x = Ce 2 /2. We could solve i explicily wih he inverse sine funcion so x() = sin (Ce 2 /2 ). (b) y () = y 2 + + 2 Soluion: This is he linear equaion y () y 2 = + 2 µ() = e /(2)d =. which has he inegraing facor We muliply he equaion by he inegraing facor and obain ( y() ) = + 2. Afer inegraion we obain ha y() = ( ) () + C = + C /2. (c) x = x + Soluion: This is he same as he linear equaion x x = wih he inegraing facor µ() = e /d =. Afer muliplying he equaion by he inegraing facor we can wrie i as so x() = ln + C for > 0. (d) (6xy y 3 )dx + (4y + 3x 2 3xy 2 )dy = 0 Soluion: This is an exac equaion as ( x ) = y (6xy y3 ) = 6x 3y 2 = x (4y + 3x2 3xy 2 ).
We will hen o ry o find F (x, y) such ha F x = 6xy y3 and F y = 4y +3x2 3xy 2. Inegraing boh equaions we obain ha F (x, y) = 3x 2 y y 3 x + C (y) = 2y 2 + 3x 2 y xy 3 + C 2 (x). This implies ha he soluions are given (implicily) by where C is deermined by he iniial condiions. 2y 2 + 3x 2 y xy 3 = C, 2. An acciden a a nuclear power plan has lef he surrounding area pollued wih radioacive maerial ha decays naurally. The iniial amoun of radioacive maerial presen is 5 su (safe unis), and 5 monhs laer is sill 0 su. (a) Wrie a formula giving he amoun A() of radioacive maerial (in su) remaining afer monhs. (b) Wha amoun of radioacive maerial will remain afer 8 monhs? (c) How long - oal number of monhs or fracion hereof - will i be unil A = su, so i is safe for people o reurn o he area? Soluion: This is a model of naural decay, so A() saisfies A () = ka(), A(0) = 5 which has he soluion A() = 5e k. To obain he consan k we use he fac ha A(5) = 0 = 5e 5k. Hence k = 5 ln 2 3. We compue hen ha A() = 5 ( ) /5 2. 3 Afer 8 monhs A() = 5 ( 2 3) 8/5 = 7.8405. The ime 0 when A( 0 ) = is given by 5 ( 2 3 ) 0/5 = or, 0 = 5 ln(/5) ln(2/3) = 33.394367. 3. Solve y 3 y = e, wih y(0) = a, and discuss how he behavior of y as depends on he iniial value a. Soluion: This is a linear equaion wih inegraing facor µ(x) = e /3 so he soluion is given by ( ) 3 y() = e /3 4 e 4/3 + C = Ce /3 3 4 e. Imposing he iniial condiion gives C 3 4 = a, hence C = a + 3 4. The soluion is hen y() = (a + 3 4 )e/3 3 4 e. We disinguish he following cases: (a) if a = 3 4 hen lim y() = 0 (b) if a > 3 4 hen lim y() = (c) if a < 3 4 hen lim y() =.
4. Solve y (x) = + 3x2 3y 2 6y, y(0) = and idenify he inerval where he soluion is valid. Soluion: This is a separable equaion, so afer separaing we obain which afer inegraing becomes (3y 2 6y)dy = ( + 3x 2 )dx y 3 3y 2 = x + x 3 + C The iniial condiion yields C = 2. The soluion is hen defined implicily by he equaion y 3 3y 2 = x + x 3 2 The soluion can no exis when he derivaive does no exis, hence we impose ha 3y 2 6y which yields y 0 or y 2. From he equaliy y 3 3y 2 = x+x 3 2 we obain ha y = 0 implies x+x 3 2 = 0 which means (x )(x 2 +x+2) = 0 wih he only soluion x =. When y = 2 we obain x+x 3 +2 = 0 equivalen o (x + )(x 2 x + 2) = 0 wih he only real soluion x =. In order o avoid he poins x = ±, bu o have an inerval ha includes x = 0 (where he IC is given) we mus have ha he inerval of exisence is (, ). 5. An invesmen is modeled by he ODE y = y(6 y), where y is he amoun (in housands) of dollars a ime (in monhs). (a) Wha kind of growh does he invesmen follow? (b) Assume ha he invesmen sars losing $5,000 per monh. For he new equaion, discuss he sabiliy of he criical poins wih a phase line analysis. Find he long erm oucome of an iniial invesmen of $500, $3,000, $5,000, and respecively $6,000. Soluion: The iniial invesmen follows logisic growh wih growh consan k = and carrying capaciy M = 6 (his is $6,000 for he model). Once money sars being aken ou, he new model is given by y = y(6 y) 5 = (y )(y 5) which has criical poins y = and y = 5. By sudying he sign of y = (y )(y 5) we obain ha y is decreasing for y (, ) (5, ) and increasing for y (, 5). Thus, y = is a source (hence, unsable) and y = 5 is a sink (hence, sable). The long ime behavior for an iniial invesmen of y(0) = 0.5 (we conver in unis, which are housands) goes o (assuming ha no invesor would ake an infinie deb, we will sop he invesmen a y = 0) y(0) = 3 will grow o y = 5 (i.e. $5,000) y(0) = 5 will say consan. y(0) = 6 will decrease o y = 5. 6. A ank conains 50 Kg of sal dissolved in 00 l of waer. The ank capaciy is 400 l. From = 0, /4 kg of sal/l is enering a a rae of 4 l/min, and he well-mixed mixure is drained a 2 l/min. Find: (a) he ime when he ank overflows; (b) amoun of sal in he ank before overflow; (c) he concenraion of sal in he ank a overflow. Soluion: Firs, noe ha here are 2l of liquid gained per minue, and since he ank had 400 00 = 200l available in volume, i will ake 50 minues o reach he poin of overflow. Se up he model and denoe Q() = amoun (kg) of sal in he ank a ime (min)
so Q(0) = 50. Nex we compue We formulae he IVP: In-rae: = 4 l/min /4 kg/l = kg /min Ou-rae: = 2 l/min Q() 00 + 2 kg /l = Q() 50 + kg/min dq = [In-rae] [Ou-rae] = d 50 + Q Q(0) = 50. Solve he linear equaion Q + 50 + Q = by compuing he inegraing facor µ() = eln(50+) = 50+. Afer inegraing and using he IC we obain ha he soluion is Q() = 50 + (50 + 22 + 2500 ) Noe ha as (assuming an infiniely large ank) we have 7. Consider he DE lim Q() =. y = ( )3 y 4 +. (a) Compue he slopes of he miniangens in he slope field a he poins (0, a) and (b,0) in he y coordinae sysem. Soluion: The slope a (0, a) is given by a 4 which is negaive and i increases as a ges large. + On he O axis he slope a (b, 0) is (b ) 3 which increases as b. (b) Draw he miniangens o he soluion curves a he poins (-,), (2,), (,), and (0,0). Soluion: The miniangen a (-,) has slope 4, a (2,) has slope, a (,) he slope is 0, 2 and a (0,0) is. (c) Draw he curve of poins (x, ) for which y = 0 and indicae on he diagram he regions where he soluions of he given DE are increasing, respecively decreasing. Soluion: The derivaive y = 0 whenever = (see he horizonal slopes below). The soluions are increasing for all and decreasing for <.
8. Solve he following IVP and find he inerval of validiy for he soluion xy = y(ln x ln y), y() = 4, x > 0. This differenial equaion becomes homogeneous afer using a quick logarihm propery y = y ( ) x x ln y Applying he subsiuion v(x) = y x wih y (x) = xv (x) + v(x) gives xv = v( ln v ) which is separable, so dv dx v( ln v ) = x. For he inegral on he LHS use he subsiuion u = ln v wih du = dv. We obain v ln(ln( v) ) = C ln x. Solve for v and noe ha well need o exponeniae boh sides a couple of imes and play fas and loose wih he consans from which we have Therefore ln( v) = e ln x+c = C x, v(x) = e C x. y(x) = xe C x where C is deermined from he iniial condiion: 4 = e C, i.e. C = ( + ln 4). The soluion is hen y(x) = xe + ln 4 x which in order o exis needs o avoid x = 0.Therefore, he inerval of exisence is (0, ).
9. Find he inerval of exisence for soluions o he following IVP: (an ) v = 2 v ln(7 ), v(4) = 2. Soluion: This equaion is linear and i can be wrien as 2 v ln(7 ) = v. an an From he exisence heorem for linear ODEs we know ha he soluion exiss as long as he coefficiens are coninuous around he iniial ime. Therefore, we need o impose 2, k π 2 for all k Z, < 7. Noe ha we had o impose ha an exiss, bu also ha i is nonzero. The larges inerval around = 4 which avoids all hese disconinuiies is (π, 3π 2 ). 0. Solve he following Bernoulli equaion: y + xy = xy n, n 0, by ransforming i ino a linear equaion hrough he following seps: (a) muliply he equaion by y n (b) use he subsiuion w(x) := y n (x). (Hin: The resuling linear equaion should be n w + xw = x.) Soluion: Using he suggesed subsiuion wih w = ( n)y n y we obain n w + xw = x. This is a linear equaion wih inegraing facor φ(x) = e ( n)x2 /2 so we can rewrie i as which afer inegraion yields Therefore ( ) e ( n)x2/2 w = ( n)xe ( n)x 2 /2 y(x) = w(x) /( n) = w(x) = Ce ( n)x2 /2 + 2. ( Ce ( n)x2 /2 + 2) /( n).