257_pdfsam_math 54 differential equation solutions odd

# 257_pdfsam_math 54 differential equation solutions odd -...

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Review Problems The back substitution e t = x (or t =ln x )resu lts y ( x )= c 1 x + c 2 x 2 2 x 2 ln x + x ln x. 29. The roots of the auxiliary equation in this problem are r = 4 ± p 4 2 4(1)(7) 2 = 2 ± 3 i. Therefore, a general solution is given by y ( t )= ± c 1 cos 3 t + c 2 sin 3 t ² e 2 t . Substituting the initial conditions, we obtain y (0) = ( c 1 cos 3 t + c 2 sin 3 t ) e 2 t ³ ³ t =0 = c 1 =1 , y 0 (0) = ´ ( 2 c 1 + 3 c 2 )cos 3 t ( 3 c 1 +2 c 2 )sin 3 t µ e 2 t ³ ³ t =0 = 2 c 1 + 3 c 2 = 2 . Solving this system yields c 1 =1 , c 2 = 0. The solution to the given initial value problem is y ( t )= e 2 t cos 3 t. 31. We solve the corresponding homogeneous equation. Its auxiliary equation, r 2 2 r +10=0, has the roots r =1 ± 3 i .Thu s y h ( t )=( c 1 cos 3 t + c 2 sin 3 t )
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Unformatted text preview: Now, we apply the method of undetermined coeﬃcients and look for a particular solution to the original nonhomogeneous equation of the form y p ( t ) = A cos 3 t + B sin 3 t . DiFerentiating y p ( t ) twice, we obtain y p ( t ) = − 3 A sin 3 t +3 B cos 3 t , y p = − 9 A cos 3 t − 9 B sin 3 t and substitute these expressions into the original equation. Thus we get ( − 9 A cos 3 t − 9 B sin 3 t ) − 2( − 3 A sin 3 t + 3 B cos 3 t ) + 10( A cos 3 t + B sin 3 t ) = 6 cos 3 t − sin 3 t 253...
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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 Berkeley.

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