Chem Differential Eq HW Solutions Fall 2011 168

Chem Differential Eq HW Solutions Fall 2011 168 - A168...

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Unformatted text preview: A168 Appendix A Ordinary Differential Equations: Review of Concepts and Methods 13. An integrating factor is e e x2 2 y + xe x2 2 x2 2 y = xe , so x2 2 x2 x2 d e2 y =e2 x dx ⇒ ⇒ ye x2 2 = xe ⇒ y = 1 + Ce− x2 2 x2 2 dx = e x2 2 +C . We now use the initial condition: y(0) = 0 ⇒ 0 = 1 + C ⇒ C = −1 ⇒ y = 1 − e− x2 2 . 17. An integrating factor is sec x (see Exercise 9), so sec xy + y tan x sec x = tan x sec x ⇒ d [y sec x] = sec x tan x dx ⇒ y sec x = tan x sec x dx = sec x + C ⇒ y = 1 + C cos x. We now use the initial condition: y(0) = 1 ⇒ 1 = 1 + C ⇒ C=0 ⇒ y = 1. 21. (a) Clear. (b) ex as a linear combination of the functions cosh x, sinh x: ex = cosh x +sinh x. (c) Let a, b, c d be any real numbers such that ad − bc = 0. Let y1 = aex + be−x and y2 = cex + de−x . Then y1 and y2 are solutions, since they are linear combinations of two solutions. We now check that y1 and y2 are linearly independent: y1 y2 y1 W (y1 , y2 ) = y2 aex + be−x = cex + de−x aex − be−x = cex − de−x −ad + bc − (ad − bc) = −2(ad − bc) = 0. Hence y1 and y2 are linearly independent by Theorem 7. 25. The general solution is y = c1 ex + c2e2x + 2x + 3. Let’s use the initial conditions: y(0) = 0 ⇒ c1 + c2 + 3 = 0 (∗) y (0) = 0 ⇒ c1 + 2c2 + 2 = 0 (∗∗) Subtract (*) from (**) ⇒ c2 − 1 = 0; c2 = 1 Substitute into (*) ⇒ c1 + 4 = 0; c1 = −4. ...
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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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