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AMath350.W13.A4.Sol

Substitute these into the de 2a 2de x dxe 8ax 4b

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Unformatted text preview: + Dxe +8Ax + 4B + 4De x x 4Dxe +3Ax2 + 3Bx + 3C + 3Dxe x x = x2 + e x . Matching coeﬃcients, we have 3A = 1 8A + 3B = 0 2A + 4 B + 3C = 0 2D = 1 and solving these yields A = 1/3, B = 8/9, C = 26/27, and D = 1/2. Putting everything together, the solution to the given ODE is y = C1 e 3x + C2 e x 1 + x2 3 8 26 1 x+ + xe x . 9 27 2 AMATH 350 3. Page 3 Assignment #4 Solutions- Winter 2013 a) Show that the price satisﬁes the second-order DE d2 p dp + 8 + 25p = 100. 2 dt dt dq dS dp = = D S with D = 10 8p and = 25p 4. (a) We have dt dt dt ﬁrst equation and then using the second gives 100. Di erentiating the dp d2 p dp d2dp dDdq dS p = = S8 with D5p + 100 8p and dS = 25p 8 100+Di pe= 100. 2 = 10 or + 25 rentiating the 4. (a) We have 2 = dt = D 2 dtdt dt dt . dt dt dt dt ﬁrst equation and then using the second gives 00 (b) If the initial value of the price ispp0+and + 25p = 0. derivative of the price 8p0 the initial b) Associated homogeneous equation: 2 8+ 6 2 d2 dp How does d (t . What is8 + 25 0 ) behaviour 2 4 100p price? 3 Auxili...
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