20dhw5 - Solutions to Assignment #5, Math 20d February 17,...

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Unformatted text preview: Solutions to Assignment #5, Math 20d February 17, 2006 Section 3.5: 9) The characteristic equation is 0 = 25 r 2- 20 r + 4 = (5 r- 2) 2 , which gives a repeated root, r = 2 5 . Thus the general solution is y = c 1 e 2 5 t + c 2 te 2 5 t . 12) The characteristic equation is 0 = r 2- 6 r + 9 = ( r- 3) 2 , and thus the general solution is y = c 1 e 3 t + c 2 te 3 t . To solve the initial value problem, we find that y = 3 c 1 e 3 t + 3 c 2 te 3 t + c 2 e 3 t . Inserting the initial conditions tells us that c 1 = 0, and c 2 = 2. Thus our final answer is y = 2 te 3 t . Thus as t , the t term is insignificant, and y . 14) The characteristic equation is 0 = r 2 + 4 r + 4 = ( r + 2) 2 . Thus the general solution is y = c 1 e- 2 t + c 2 te- 2 t . Thus y =- 2 c 1 e- 2 t- 2 c 2 te- 2 t + c 2 e- 2 t . Substituting in the initial conditions gives us 2 = e 2 ( c 1- c 2 ) 1 = e 2 (- 2 c 1 + 3 c 2 ) Multiplying the second equation by 2 and subtracting, we find that c 1 = 7 5 c 2 . Similarly, multiplying the first equation by 2 and adding yields c 2 = 5 e- 2 , and thus c 1 = 7 e- 2 . So our final solution is y = 7 e- 2- 2 t + 5 te- 2- 2 t . Clearly, then, as t , y 0. 1 18) Given that y 1 = t is a solution to our equation, we assume that tv ( t ) t is also a solution. We divide by t 2 (this is ok since we assume t > 0.)to write the original equation in general linear form: y 00 + 2 t y- 2 t 2 y = 0. From equation (30) in the book, taking derivatives and substituting into the original equation yields: tv 00 + 4 tv = 0 . This is actually a linear equation, but in the variable v . To make this explicit, we let u = v , and our equation becomes tu +4 u = 0. This equation is separable, so we write u u =- 4 t ln | u | = ln | t- 4 | + c u = Ct- 4...
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20dhw5 - Solutions to Assignment #5, Math 20d February 17,...

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