practice-prelim-1 - Solutions to the practice prelim...

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Solutions to the practice prelim, Math-293 1. A) a) No . (1 + x ) d (ln( C + x )) dx = (1+ x ) ( C + x ) = ln( C + x ) . b) Yes . (1 + x ) d (0) dx = 0 = 0 . c) Yes . (1 + x ) d ( C (1+ x )) dx = C (1 + x ) = C (1 + x ) B) i) Yes . x 2 d dx d ( x 2 ) dx = 2 x 2 = 2 x 2 ii) No . x 2 d dx d ( e x - 1 ) dx = x 2 e x - 1 = 2 e x - 1 2. A) dy dt = e t + y = e t e y . This is a separable equation, dy e y = e t dt + C. (1) We get from (1), - e y = e t + C . Using initial condition, we get C = - 2. Hence, our solution is, e y = - e t + 2 . (2) Taking logarithms, y ( t ) = ln(2 - e t ) . B) Clearly y ( t ) blows up if 2 - e t = 0, or t = ln 2 which is finite. 3. Given equation, dy dt + y (1 + t ) = b. (3) Clearly, (3), is a linear differential equation and can be solved using integrating factor. ρ = e 1 1+ t dt = e ln(1+ t ) = (1 + t ). Multiplying both sides of (3) by ρ , we get, d dt y (1 + t ) = b (1 + t ) . (4) Integrating (4), both sides with respect to t , we get, y (1 + t ) = bt + bt 2 / 2 + C. (5) This can be rearranged by dividing throughout by (1 + t ), y ( t ) = bt 1 + t + bt 2 1 + t + C 1 + t (6) 1
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Using y (0) = 0 on (6), we get C = 0. Using y (1) = 2 on the resulting particular solution, we get b = 8 / 3.
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