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# 341h2 - M341 H2(S Zhang 1 EP Separable equations and...

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M341 H2 (S. Zhang) . 1. EP – Separable equations and applications. § 1.4: 3, 12, 17, 23, 30, 33, 34, 49, 52 ans: 2. Find the general solution of dy dx = y sin x ans: § 1.4:3. Z dy y = Z sin xdx ln y = - cos x + c y = Ce - cos x 3. Find the general solution of y dy dx = x ( y 2 + 1) ans: § 1.4:12. Z ydy y 2 + 1 = Z 2 xdx 1 2 ln( y 2 + 1) = 1 2 x 2 + C y 2 = ce x 2 - 1 4. Find the general solution of dy dx = 1 + x + y + xy ans: § 1.4:17. Z dy 1 + y = Z (1 + x ) dx ln | 1 + y | = x + 1 2 x 2 + C 5. Solve the IVP dy dx + 1 = 2 y ; y (1) = 1 ans: § 1.4:23. Z dy 2 y - 1 = Z dx 1 2 ln | 2 y - 1 | = x + c y = 1 2 ( 1 + Ce 2 x ) By the initial condition, C = e - 2 , y = 1 2 ( 1 + e 2 x - 2 ) 6. Solve the differential equation dy dx 2 = 4 y to find the general solution and singular solution. Deter- mine the points ( a, b ) for which the initial problem dy dx 2 = 4 y ; y ( a ) = b has (a) no solution, (b) infinitely many solutions that are defined for all x , (c) on some neighborshood of the point x = a , only finitely many solutions. ans: § 1.4:30. Note that due to nonlinearity with y 0 , we get two slope (direction) fields! See the two graphs. 1 0 -2 -3 y(x) 0 3 -1 2 3 1 -1 2 x -2 In order to divide the equation by y , we need to check if y 6 = 0 and if y = 0 is a solution. Yes, y 0 is a singular solution! ± Z dy 2 y = Z dx ± y = x - C ; y = ( x - C ) 2 Note that y ( x ) is always nonnegative. Note also that the whole curve of y = ( x - C ) 2 is a solution curve, not only the right half! (a) No solution when b < 0. (b) If b 0, we can combine y = 0 and y = ( x - C ) 2 ( x > C ) for any C > a to get infinitely many solutions, also we can combine y = ( x - C ) 2 and y = 0 (for x > C ) to get infinitely many solutions. (c) If b > 0, locally, we have only two integral curves, one increasing curve y = ( x - C ) 2 passing through point ( a, b ) and one decreasing curve y = ( x - C 1 ) 2 passing through point ( a, b ). 1

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y(x) 0 -2 2 -3 2 -2 3 -1 3 1 1 -1 x 0 7. A certain city had a population of 25000 in 1960 and a pop- ulation of 30000 in 1970. Assume that its population will continue to grow exponentially at a contant rate. What population can its city planners expect in the year 2000? ans: § 1.4:33. P 0 = kP, P = P 0 e kt Let t = 0 be the year 1960. k = 10 ln 30000 ln 25000 = 0 . 01823 So at year 2000, P (40) = 25000 e . 01823 × 40 = 51840 .
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