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Unformatted text preview: EA4 Workshop questions with answers  set 1  week of 9/29/08 1. Consider the equation dy dt = y n , y (0) = 1 , (1) where n is a positive integer. (a) Find the solution for all values of n ≥ 1 . Explain why you have to use a different procedure for n = 1 than for n > 1 . Answer the equation is separable so we can write dy y n = dt. Consider first n > 1 . Integrating both sides we have 1 1 n y 1 n = t + c, where c is an integration constant. Now use the initial condition to obtain c = 1 n 1 . Next, multiply both sides of the equation by 1 n to get y 1 n = 1 ( n 1) t. Inverting the fraction we get y = 1 (1 ( n 1) t ) 1 n 1 . (2) When n = 1 the integrals become dy y = dt Integrating both sides now gives natural logs rather than a polynomial term in y . This is why the situation for n = 1 is different from the n > 1 case. We get ln(  y  ) = t + c. Evaluating c from the initial condition, we get c = 0 . Raising both sides to the power e gives y = e t . (3) (b) Show that for n ≥ 2 the solution blows up at a finite time. Answer the denominator in equation (2) becomes zero when t = 1 / ( n 1) . This is the blowup time. For n = 1 , it is clear from (3) that the solution is defined for all t although it approaches infinity for large t . 1 (c) Show that the blowup time gets smaller as n increases. Explain why this is so. Answer From the differential equation (1) for any fixed value of y , dy/dt gets bigger and bigger as n increases. Thus, you expect the solution to grow more rapidly as n increases. This means that the blowup occurs earlier. This is verified by the solution which shows that the blowup time is 1 /...
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 Fall '08
 TAFLOVE
 Derivative, Constant of integration

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