Chem Differential Eq HW Solutions Fall 2011 169

Chem Differential Eq HW Solutions Fall 2011 169 - Section...

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Unformatted text preview: Section A.1 Linear Ordinary Differential Equations A169 Thus, y = −4ex + e2x + 2x + 3. 29. As in the previous exercise, here it is easier to start with the general solution y = c1ex−2 + c2e2(x−2) + 2x + 3. From the initial conditions, y(2) = 0 ⇒ c1 + c2 + 7 = 0 (∗) y (1) = 1 ⇒ c1 + 2c2 + 2 = 1 (∗∗) Subtract (*) from (**) ⇒ c2 − 5 = 1; c2 = 6 Substitute into (*) ⇒ c1 + 13 = 0; c1 = −13. Thus, y = −13ex−2 + 6e2(x−2) + 2x + 3. ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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