AMath350.W13.A3.Sol

# Longterm behaviour lim pt 5 equilibrium solutions pt

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Unformatted text preview: (1 + cos(2⇡ t))et+ 2 dt 1 1 et+ 2 sin(2 t) p = 5et+ 2 sin(2 t) + C General Solution: p(t) = 5 Ce t 1 2 sin(2 t) sin(2 t) sin(2 t) IC: p(0) = p0 ) p0 = 5 C ) C = 5 p0 . 1 Solution of IVP: p(t) = 5 (5 p0 )e t 2 sin(2 t) . Longterm behaviour: lim p(t) = 5. Equilibrium solutions: p(t) = 5. dp + Standard form: dp (1t+ cos(2⇡ t))p = 5(1 + cos(2⇡ t)). (1 + t) 1 Standard form: dt + +cos(2 cos(2⇡ tt))p = 5(1 + cos(2⇡ t)). 1 1 +1 Integrating dp = dt 1 supply and 2 sin(2 t) . ( (a) The a commodity eD 1+cos(2 t = e demand t)20+4p)) by 6.6.(Consider DE is Factor: (withS ) = dt)(20+8p+ 1p3/2 curves given= (4p p3/2 ), with IC p(0) = p0 . 4 4 dt = et 2 sin(2 . 4 Integratingdt Factor: e t+ 21 sin(2 t) dp t 2p Equilibrium solutions:(1 p) cos(2⇡20 ++41 sp(4 t) p p1/2 ) 5(1 + = 0, 16. et+ 21 sin(2 t) 4+ p ) p cos(2⇡ t)) S ( p =3/2 t= e0 ) in(2 e + )) = 1 1 0 3 0 dt Bernoulli+equationdp D(pn +=3/20⇡Let tv 2=3in(2 3t/2 == 1/2 ) cos(2⇡ t))pt+ /12 pin(2Using the 2. t 8e 1 p1 et 2 sin(2 t) with(1 =cos(2 +))p + p s/2 ) p p 5(1 + v = 1 2 e 2 s . t) +) d t+ 1 sin(2 t) dt DE: e2 p = 5(1 + cos(2⇡ t))et+ 2 sin(2 t) dt d rate of sin(2 t) 1 The rate of increase of price is 1/4 the t+e1t1sin(2 decrease = 1quantity and t))et+ 21 sin(2 t) +2 1 of 5(1 +) cos(2⇡ the t) p t+ sin(2 t + C v 0 = e 2 p 1/2p+ = 5e 2 dt initial price is p0 . ) 1 2 in(2 General Solution:21psin(2 = p 5 = C 5et +221 ssin(2 t)t) + C t8 et et+ (1) t) 1 a) (Set up an initial value problem = that models thisLinear DE and ﬁnd all situation 1 v+ IC: p(0) = p ) p0 = 5 C )General Solution: p(t) = 5 C e t 2 sin(2 t) C = 5 p0 . equilibrium 0 solutions of the diﬀerential2in(2 t) 8 equation. Then ﬁnd the solution 1 Solution of IVP: p(t) = 5 (5 p )e t 2 s . 1 of the(0) =form: v 00+ lim = t1 .= 50C = 5 p0 . solutions:2 dt(t) =t/2 . IC: p initial 0value problem. ) Integrating factor: e 1/ p = e 5. Standard behaviour: =25 p(C ) . Equilibrium v8 Longterm p ) p ) t Solution: IVP: p(t) = 5 (5 p0 )e t 21 sin(2 t) . Solution of 1 t...
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