This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
 Spring '10
 Edgars

Click to edit the document details