MATH 2202 – Exam 1 Solutions

t 2 is de ned for t 2 The component function ln(8 t) is de ned for t

Method 1: Using the formulas

1 lnt 1 t2 1 t2 lnt+ 1 t2 = 1 t3 lnt+ 1 t3 + 1 t3 lnt = 1 t3: Now, note g(t) is the right-hand side when the equation is in the form y00+ p(t)y0+ q(t)y = g(t): Our equation is t2y00(t) + 3ty0(t) + y(t) = 1 t so we need to divide both sides by t2 Doing so, we obtain the equation y00(t) + 3 t y0(t) + 1 t2 y(t) = 1 t3: Now, we see that g(t) = 1

Properties of the Beurling generalized primes

lnt 1 t2 t (s 1) dt+ Z 1 p 1 ˇ(t) t2 (t 1) dt+ Z 1 p 1 t s lnt dt+ o(1) as s1+:The third integral can be simpliﬁed using the change of variables u= (s 1)lnt;followedbyintegrationbyparts Theresultingformulais Z 1 p 1 t s lnt dt= ln 1 s 1 lnlnp 1 + Z 1 0 e ulnudu+ o(1): However,itiswellknown(seeforinstance[4])that R 1 0 e ulnudu= 0(1

1

D Y(t) = 2 5 e t E Y(t) = 2 5 te t 2 The form of a particular solution to the equation 1t2 + C 2t3 + t4 lnt 1 4 A mass weighing 2 lb stretches a spring 6 in

Solutions to Practice Problems II

2 t2 1=2 2 lnt+ 1 t5=2 lnt Simplify the right hand side: 3t =2 lnt 2t3=2 t5=2 lnt = 1 t 2 tlnt Integrate both sides: Z dw w = Z 1 t 2 tlnt dt On the right hand side you will need to use the substitution a = lnt;da = 1 t dt: lnjwj= lnjtj 2lnjlntj Now solve for w But rst we need to combine the log terms on the right hand side: lnjwj= ln 1 t + ln

1(Line integrals{Using parametrization Two types and the ux

We parametrize the curve as ~x(t) = (t;lnt);1 t 2 We can compute that ds= j~x0jdt= p 1 + 1=t2dt Hence, the integral is 2 1 t2 p 1 + 1=t2dt= 2 1 t p t2 + 1dt= 5 2 p u 1 2 du= 1 3 u3=2ju=5 u=2 (c) Let F~ = ( 2y+ 2x;2x 2y) and Cis ~x(t) = (t;t2);0 t 1 Compute the work done by F~along the curve The eld is not conservative and the curve is not

Practice Problems to Midterm Exam 1

t 2 lnt 1 4 t 2 + C +1 pt Math 0230 - Spring 2021 - Tro mov Solutions to Practice Problems to Exam 1 Page 2 of 19 2 (10 points) Evaluate the integral I= Z x p 4 x2 dx