Exam 1

# Exam 1 - Name: Exam 1 Instructions. Answer each of the...

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Unformatted text preview: Name: Exam 1 Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. 1. [18 Points] Consider the differential equation xy + 2 y = 8 on the interval 0 &lt; x &lt; . (a) Find the general solution ( x ). I Solution. Use the technique from Section 1.7, Page 43. First write the func- tion in standard form by dividing by x , so that ( x ) satisfies: ( * ) ( x ) + 2 x ( x ) = 8 x . Thus the coefficient of y is a ( x ) = 2 /x and we compute an antiderivative of a ( x ), namely A ( x ) = R 2 x dx = 2ln x = ln x 2 to form a multiplier function u ( x ) = e A ( x ) = e ln x 2 = x 2 which makes the left hand side of equation ( * ) into a perfect derivative: x 2 ( ( x ) + 2 x ( x )) = ( x 2 ( x )) = x 2 8 x = 8 x. Integrating with respect to x gives x 2 ( x ) = 4 x 2 + C where C is an arbitrary integration constant. Solving for ( x ) gives ( x ) = 4 + C x 2 . J (b) Find the solution that satisfies (2) = 0. I Solution. Substituting (2) = 0 in the formula found in part (a) gives 0 = (2) = 4 + C 2 2 , and solving for C gives C =- 16. Hence, the solution of the initial value problem is ( x ) = 4- 16 x 2 . J (c) What is lim x ( x ) for the general solution ( x )? Math 4027 February 24, 2005 1 Name: Exam 1 I Solution. From the formula for the general solution found in part(a), we find that lim x ( x ) = lim x 4 + C x 2 = 4 + lim x C x 2 = 4 . J 2. [12 Points] Find all real valued solutions of the differential equation y (4) + 6 y 000 + 16 y 00 + 24 y + 16 y = 0 . You may assume the following factorization: r 4 + 6 r 3 + 16 r 2 + 24 r + 16 = ( r 2 + 4 r + 4)( r 2 + 2 r + 4) . I Solution. From the given factorization, we conclude that the characteristic poly- nomial p ( r ) can be factored as p ( r ) = ( r + 2) 2 (( r + 1) 2 + 3) so that the roots of the equation p ( r ) = 0 are- 2 which occurs with multiplicity 2, and the pair of complex conjugate roots- 1 3 i . Then, according to the discussion in Section 2.9, summarized in Theorem 19, the real valued solutions of the givenin Section 2....
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## Exam 1 - Name: Exam 1 Instructions. Answer each of the...

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