p1spring09_page 7

p1spring09_page 7 - -10 = 0 12 solve for t 2 = 13 5a The...

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some answers: 1b) This is 1st order linear and the integrating factor is e t/ 5 . So you have ( e t/ 5 y ) ± = e - 4 t/ 5 . Integrating you get e t/ 5 y = - 5 4 e - 4 t/ 5 + c . The initial value gives c = 5 + 5 4 so y ( t ) = - 5 4 e - t + 25 4 e - t/ 5 . 1c) this is d ( 2 3 x 3 y ) = 0 2) x ± = kx (10000 - x ) and 25 = k · 250 · (9750) gives k = 1 / 97500. Critical points are at x = 0 and x = 10000. Since k is positive, x ± is positive when 0 < x < 10000, so 0 is not stable. Make a direction ﬁeld, or try values above 10000, or recognize the logistic equation, to say that 10000 is stable. 3) You may either use the substitution method or simply multiply by y ± as discussed in the lecture. If the latter: y ± y ±± = - y 3 y ± integrates to 1 2 ( y ± ) 2 = c - 1 4 y 4 . The initial conditions give c = 9 / 2, and require the minus sign in y ± = - q 9 - 1 2 y 4 . 4) The DE gives slope y + t 1000 = 0 . 11 + 0 . 01 at the point where the line is tangent. But the graph shows the tangent line going through point ( t 2 , 0 . 47). These only match if 0 . 47 - 0 . 11 t 2
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Unformatted text preview: -10 = 0 . 12 solve for t 2 = 13. 5a) The Chain Rule gives w x = af ± , w xx = a 2 f ±± etc. So you only need f to be twice diﬀerentiable and a 2 + b 2 = 1. It means that many wave shapes are possible. 5b) A particular solution is y p = 2 = c 3 . Then we know that the general solution is of the form y = c 1 cos(2 t ) + c 2 sin(2 t ) + 2. To get the initial conditions you choose y = 1 . 5 cos(2 t ) + 2 . 05 sin(2 t ) + 2 6) Euler’s method replaces dy dx = x 2 + y 2 by y n +1-y n h = x 2 n + y 2 n . With h = . 1 you have x = 0, x 1 = . 1, x 2 = . 2, and y =-1. Then y 1-y h = x 2 + y 2 , y 1 + 1 . 1 = 1 , so y 1 =-. 9. Then y 2-y 1 h = x 2 1 + y 2 1 , y 2 + . 9 . 1 = . 01 + . 81 so y 2 =-. 9 + . 082 =-. 818. 7...
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