358_pdfsam_math 54 differential equation solutions odd

358_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 yields the roots r =1, 2, and 2. Thus a general solution has the form y ( x )= c 1 e x + c 2 e 2 x + c 3 e 2 x . Next, we Fnd constants c 1 , c 2 ,and c 3 such that the solution satisFes the initial conditions. Di±erentiating y ( x ) and substituting the initial conditions, we obtain the system y (0) = ( c 1 e x + c 2 e 2 x + c 3 e 2 x ± x =0 = c 1 + c 2 + c 3 = 4 , y 0 (0) = ( c 1 e x 2 c 2 e 2 x +2 c 3 e 2 x ± x =0 = c 1 2 c 2 c 3 = 1 , y 0 (0) = ( c 1 e x +4 c 2 e 2 x c 3 e 2 x ± x =0 = c 1 c 2 c 3 = 19 . Solving yields c 1 =1 ,c 2 = 2 3 = 3 . With these coefficients, the solution to the given initial problem is y ( x e x 2 e 2 x 3 e 2 x . 21. By inspection, r = 2 is a root of the characteristic equation, r 3 4 r 2 +7 r 6 = 0. ²actoring yields r 3 4 r 2 r 6=( r 2)( r 2 2 r +3)=0 . Therefore, the other two roots are the roots of r 2 2 r +3=0,which are r ± 2 i so a general solution to the given di±erential equation is given by
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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