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Unformatted text preview: Chapter 4 Hence, a second linearly independent solution is y 2 ( t ) = t 2 Z t 6 dt ( t 2 ) 2 = t 2 Z t 2 dt = t 3 . One can also take y 2 ( t ) = t 3 , because the given equation is linear and homogeneous. 48. Putting the equation in standard form yields p ( t ) = t 1 2. Hence, exp Z p ( t ) dt = exp Z ( 2 t 1 ) dt = e 2 t ln t = t 1 e 2 t . Therefore, by Theorem 8, a second linearly independent solution is y 2 ( t ) = e t Z t 1 e 2 t ( e t ) 2 dt = e t Z t 1 dt = e t ln t. 50. Separation variables in (16) yields w w = 2 y 1 + py 1 y 1 = 2 y 1 y 1 + p . Integrating, we obtain Z w w dt = Z 2 y 1 y 1 + p dt = 2 Z y 1 y 1 dt Z pdt ⇒ ln  w  = 2 ln  y 1   Z pdt ⇒  w  = y 2 1 exp Z pdt . Obviously, w ( t ) does not change its sign on I (the righthand side does not vanish on I ). Without loss of generality, we can assume that w > 0 on I and so v = w = y 2 1 exp Z pdt ⇒ v = Z exp R pdt y 2 1 dt ⇒ y 2 = y 1 v = y 1 Z exp R pdt y 2 1 dt. 52. For y ( t ) = v ( t ) f ( t ) = tv ( t ), we find y = tv + v , y 00 = tv 00 + 2 v , y 000 = tv 000 + 3 v 00 . Substituting y and its derivatives into the given equation, we get t ( tv 000 + 3 v 00 ) + (1 t ) ( tv 00 + 2 v ) + t ( tv + v ) tv 156 Exercises 4.8 = t 2 v 000 ( t 2 4 t ) v 00 + ( t 2 2 t + 2) v = 0 . Hence, denoting v = w (so that v 00 = w and v 000 = w 00 ) yields t 2 w 00 ( t 2 4 t ) w + ( t 2 2 t + 2) w = 0 , which is a second order linear homogeneous equation in w ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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