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Unformatted text preview: Math 20D, Sample Final Exam Solutions December 2, 2006 Name: Section: This exam consists of 12 pages including this front page. The last page is the table of Laplace transforms. 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 two 4by6 index cards, both sides. 4. You have two hours for this exam. Score 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total 100 1 1. (a) Find the general solution of y 00 5 y = 0 . r 2 5 r = 0 gives r ( r 5) = 0, i.e. r = 0 , 5. Thus the general solution is y = C 1 + C 2 e 5 t . (b) Find the general solution of dy dt = y 2 + 1 . This is separable. So dy y 2 + 1 = dt Z dy y 2 + 1 = Z dt tan 1 y = t + C y = tan( t + C ) . 2 2. Solve the following initial value problem. y 3 y = e 3 t sin t, y (0) = 2 This is first order linear. So the integrating factor is μ = e R 3 dt = e 3 t Thus dt ( e 3 t y ) dt = sin t e 3 t y = cos t + C y = e 3 t cos t + Ce 3 t . This is the general solution. Now by the initial condition, y (0) = 1 + C = 2 . So C = 3. Hence the solution is y = e 3 t cos t + 3 e 3 t . 3 3. Determine the values of b so that the following differential equation is exact, and then solve it using that value of b . ( xy 2 + bx 2 y ) dx + ( x + y ) x 2 dy = 0 Let M = xy 2 + bx 2 y and N = ( x + y ) x 2 = x 3 + x 2 y . Then if it is exact, we must have M y = N x 2 xy + bx 2 = 3 x 2...
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 Spring '06
 Mohanty
 Math, general solution

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