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AMath350.W14.A3.Sol

# Integrating t12 2 1e 12 dt et2 1 ce v linear

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Unformatted text preview: 1/2 ) v 0 = 1 p 3/2 p0 . Using the Let v = = 2 2 18 1 + Ce t/2 DE: 1 = 1 + C1 e t/2 = vv + = 4 1 8 4/2 Linear DE 1 0 v =2 p1 + 16 2 8 Standard form: v 0 p 1 v v 8 factor: General solution:+ 2= = 12. =Integrating t12 2 1e 1/2 dt = et/2 . /) (1 + Ce v + Linear DE = 1 t/2 16 1 t/2 8 4 2e t/2 0 ) 1. Apply IC: p(0) = p0 e) v 0+ 2 e v = 2 8 C = p= p0 1/2 dt 1 1 (1 + C ) 0 Standard form: v + 2 v = d . t/2 Integrating 2factor: e = et/2 . 1 t/ 8 [e v ] 16 = e Solution of IVP: p(t) = dt 8 2. e (1 + ( 14 t/ 1)1 t/1)2 p v eC et/2 v 0 + t/2e0 2= =et/2 +t/2 e2 v 1 48 (b) Note that limt p(t) = 16. d t/2 1 t/2 t/2 1 If the initial price is greater than \$16 the pricet/will 1 + Ce with limiting value \$16. [ev v ] = + Cee 2 = decrease = 1 dt 48 If the initial price is equal to \$16, it will remain at \$16. 4 1 t/2 16 v price ewill+ C1 If the initial price is less than \$16 2the = increase with limiting value \$16. et/ General solution: p = v 2 = 4 t/2 )2 (1 + Ce 14 1 + Ce t/2 16 v C = + C11. t/2 = e )= Apply IC: p(0) = p0 ) p0 = (1 + C )2 4 4 4 p0 16 16 Solution of IVP: p(t) = . General solution: p = (1 + = 4 0 1)e t/t2/)2 2 v 2( p (1 + Ce 2 ) (b) Note that limt p(t) = 16. 4 16 b) (Discuss the(0) =is 0 ) p0 of your ﬁndings for the application. That is, \$16. implications = )C= 1 Applyinitial price pgreater than \$16 the price will decrease .with limiting value If the IC: p 2 p0 describe the price is equal to \$16, it will remainand how this depends on p0 . behaviour of the(1 + Cin time at \$16. price ) If the initial You need of IVP: is(less than \$16 ) 16 If the initial price p t) = the increase with limiting value \$16. Solution only consider p0 > 0. 4 price will 2 . (1 + ( p0 1)e t/2 ) Solution: (b) Note that limt p(t) = 16. 4 If the initial price is greater than \$16 the price will decrease with limiting value \$16. If the initial price is equal to \$16, it will remain at \$16. If the initial price is less than \$16 the price will increase with limiting value \$16. 4...
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