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Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 2 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 Problem 1 : Solve the following initial value problems (this includes specifying the maximal (i.e. biggest) interval on which the solution exists). If in the process you can’t find an antiderivative, leave the integral. (16 points) (a) ty + 2 y = t 2 t + 1; y (1) = 1 Solution: Rewriting ty + 2 y = t 2 t + 1 as a linear first order equation gives y + 2 t y = t 1 + 1 t , the general solution for which is given by y ( t ) = e I R Qe I dt + Ce I , where I = R Pdt , P ( t ) = 2 t , and Q ( t ) = t 1 + 1 t . We note I = Z P dt = Z 2 t dt = 2log  t  , so that e I = e 2 log  t  = t 2 , and hence Z Qe I dt = Z t 3 t 2 + tdt = 1 4 t 4 1 3 t 3 + 1 2 t 2 . Thus y ( t ) = e I Z Qe I dt + Ce I = t 2 1 4 t 4 1 3 t 3 + 1 2 t 2 + Ct 2 , which simplifies as y ( t ) = 1 4 t 2 1 3 t + 1 2 + C t 2 . Since y (1) = 1, we conclude 1 = 1 4 1 3 + 1 2 + C, so that C = 7 12 . Finally we note the general solution is valid for t 6 = 0, i.e on the intervals (0 , + ∞ ) and (∞ , 0). But since our initial condition is at t = 1, we must take (0 , + ∞ ) as our interval of definition. (b) (1 e t ) y + e t y = 1; y (2) = 1 Solution: Rewriting (1 e t ) y + e t y = 1 as a linear first order equation gives y + e t 1 e t y = 1 1 e t , the general solution for which is given by y ( t ) = e I R Qe I dt + Ce I , where I = R Pdt , P ( t ) = e t 1 e t and Q ( t ) = 1 1 e t . We note I = Z P dt = Z e t 1 e t dt = log  1 e t  , 1 so that e I = (1 e t ), and hence Z Qe I dt = Z 1 1 e t (1 e t ) dt = t. Thus y ( t ) = e I Z Qe I dt + Ce I = t 1 e t + C 1 e t = t + C 1 e t . Since y (2) = 1, we conclude 1 = 2 + C 1 e 2 , so that C = 1 e 2 . Finally we note the general solution is valid for t 6 = 0, i.e on the intervals (0 , + ∞ ) and (∞ , 0), so with our initial condition we must take (0 , + ∞ ) as our interval of definition. (c) y = 2 t + t 2 1 y ; y (0) = 0 Solution: We can separate the equation y = 2 t + t 2 1 y to give (1 y ) dy = (2 t + t 2 ) dt. Integrating both sides of this equation yields y 1 2 y 2 = t 2 + 1 3 t 3 + C. Since y (0) = 0, this implies 0 1 2 2 = 0 2 + 1 3 3 + C , so that C = 0. Hence we have y 2 2 y + 2 t 2 + 2 3 t 3 = 0 , which from the quadratic formula yields y ( t ) = 2 ± q 4 4 ( 2 t 2 + 2 3 t 3 ) 2 = 1 ± r 1 2 t 2 2 3 t 3 , and since y (0) = 0 we see that we must choose the negative sign to conclude y ( t ) = 1 r 1 2 t 2 2 3 t 3 . The solution is valid when 1 2 t 2 2 3 t 3 > 0, which by computation occurs roughly in the interval ( . 83 ,. 64)....
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This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Math, Differential Equations, Statistics, Equations, Probability

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