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Unformatted text preview: Math 20D, Sample Midterm 1 Solutions October 14, 2006 Name: Section: This exam consists of 7 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use one 4by6 index card, both sides. Score 1 10 2 10 3 10 4 10 5 10 6 10 Total 60 1 1. (a) Solve the following initial value problem. y 2 y = e 2 t , y (0) = 2 The integrating factor is = e 2 t . So the given equation becomes ( y ) = 1 . By integrating, we get y = t + C. Thus y = te 2 t + Ce 2 t . Since y (0) = 0 e + Ce = C , we have C = 2. So the solution is y = te 2 t + 2 e 2 t . (b) Solve: dy dx = x e x y + e y dy dx = x e x y + e y ( y + e y ) dy = ( x e x ) dx Z ( y + e y ) dy = Z ( x e x ) dx 1 2 y 2 + e y = 1 2 x 2 + e x + C 2 2. Show that the following differential equation is exact and solve it y x + 6 x dx + (ln  x   2) dy = 0 Let M = y x + 6 x and N = ln  x   2. Then M...
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 Spring '06
 Mohanty
 Math

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