Spring 2006 - Zabrocki's Class - Practice Final Exam

# Spring 2006 - Zabrocki's Class - Practice Final Exam - Math...

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Math 20D - Spring 2006 - Practice for ﬁnal -SOLUTIONS Last update: Sunday June 11, 2006 - 8am 1. t is a solution to the diﬀerential equation (1 - t ) y 00 + ty 0 - y = 0 . Find a second linearly independent solution. SOLUTION: Assume that y 2 = , then y 0 2 = μ + 0 and y 00 2 = 2 μ 0 + 00 . (1 - t )(2 μ 0 + 00 ) + t ( 0 + μ ) - μt = 0 2 μ 0 + 00 - 2 μt - t 2 μ 00 + t 2 μ 0 + - = 0 Check : do all μ ’s cancel? Yup. Good. Hopefully I’m on the right track. μ 0 ( t 2 - 2 t + 2) + ( t - t 2 ) μ 00 = 0 substitute μ 0 = v , μ 00 = dv dt . t 2 - 2 t + 2 t 2 - t = 1 v dv dt 1 v dv dt = t - 2 + 2 /t t - 1 = t - 1 - 1 + 2 /t t - 1 = 1 - 1 1 - t +2 1 t ( t - 1) = 1 - 1 1 - t +2 ± 1 t - 1 - 1 t ² = 1 - 2 t + 1 t - 1 Now integrate lnv = t - 2 ln( t ) + ln( t - 1) Solve for v μ 0 = v = e t t - 1 t 2 = e t t - e t t 2 Integrate by parts (only the R e t t dt piece). μ = Z e t t dt - Z e t t 2 dt = e t t + Z e t t 2 dt - Z e t t 2 dt = e t t Therefore y 2 ( t ) = t e t t = e t (check it. .. it is also a solution to the diﬀerential equation). 2. Use the results of the previous problem and the method of variation of parameters to ﬁnd all solutions to the equation (1 - t ) y 00 + ty 0 - y = 2( t - 1) 2 e - t . SOLUTION: Assume that there is a solution of the form y ( t ) = tu 1 ( t ) + e t u 2 ( t ) and assume that tu 0 1 ( t ) + e t u 0 2 ( t ) = 0 (because we can and because it is going to help us), then we compute that y 0 ( t ) = u 1 ( t ) + tu 0 1 ( t ) + e t u 2 ( t ) + e t u 0 2 ( t ) = u 1 ( t ) + e t u 2 ( t ), y 00 ( t ) = u 0 1 ( t ) + e t u 2 ( t ) + e t u 0 2 ( t ). (1 - t )( u 0 1 ( t ) + e t u 2 ( t ) + e t u 0 2 ( t )) + t ( u 1 ( t ) + e t u 2 ( t )) - ( tu 1 ( t ) + e t u 2 ( t )) = 2( t - 1) 2 e - t (1 - t )( u 0 1 ( t ) + e t u 0 2 ( t )) = 2( t - 1) 2 e - t u 0 1 ( t ) + e t u 0 2 ( t ) = u 0 1 ( t ) - tu 0 1 ( t ) = 2( t - 1) e - t

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u 0 1 ( t ) = 2 e - t u 1 ( t ) = - 2 e - t + c 1 u 0 2 ( t ) = - 2 te - 2 t u 2 ( t ) = te - 2 t + 1 2 e - 2 t + c 2 Hence the solutions to the diﬀerential equation are all of the form - te - t + 1 2 e - t + c 1 t + c 2 e t . 3. Find the solutions to the following ﬁrst order diﬀerential equations (a) ( x 2 + 3 xy + y 2 ) dx - x 2 dy = 0 SOLUTION: This one is homogeneous dy dx = (1 + 3 y/x + y 2 /x 2 ) substitute v = y/x and x dv dx + v = dy dx . Hence, x dv dx + v = (1 + 3 v + v 2 ) 1 (1 + v ) 2 dv dx = 1 x - 1 1 + v = ln( x ) + c 1 + y/x = 1 + v = - 1 ln( x ) + c y = - x ln( x ) + c - 1 (b) y 0 + (2 /t ) y = cos t t 2 SOLUTION: This one can be recognized as exact after one ﬁnds an integrating factor e R 2 /tdt =
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## This note was uploaded on 06/12/2008 for the course MATH 20D taught by Professor Mohanty during the Spring '06 term at UCSD.

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Spring 2006 - Zabrocki's Class - Practice Final Exam - Math...

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