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186_pdfsam_math 54 differential equation solutions odd

# 186_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 and substitute y , y , and y into the initial conditions. This yields c 1 + c 2 = 1 , 2 c 1 + c 2 + 2 c 3 = 0 , 4 c 1 c 2 + 2 2 c 3 = 0 c 1 = 1 , c 2 = 0 , c 3 = 2 . With these values of the constants c 1 , c 2 , and c 3 , the solution becomes y ( t ) = e 2 t 2 e t sin 2 t . 29. (a) As it was stated in Section 4.2, third order linear homogeneous differential equations with constant coeﬃcients can be handled in the same way as second order equations. Therefore, we look for the roots of the auxiliary equation r 3 r 2 + r + 3 = 0. By the rational root theorem, the only possible rational roots are r = ± 1 and ± 3. By checking these values, we find that one of the roots of the auxiliary equation is r = 1. Factorization yields r 3 r 2 + r + 3 = ( r + 1)( r 2 2 r + 3)
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Unformatted text preview: . Using the quadratic formula, we ±nd that the other two roots are r = 2 ± √ 4 − 12 2 = 1 ± √ 2 i. A general solution is, therefore, y ( t ) = c 1 e − t + c 2 e t cos √ 2 t + c 3 e t sin √ 2 t . (b) By inspection, r = 2 is a root of the auxiliary equation, r 3 + 2 r 2 + 5 r − 26 = 0. Since r 3 + 2 r 2 + 5 r − 26 = ( r − 2) ( r 2 + 4 r + 13 ) , the other two roots are the roots of r 2 + 4 r + 13 = 0, that is, r = − 2 ± 3 i . Therefore, a general solution to the given equation is y ( t ) = c 1 e 2 t + c 2 e − 2 t cos 3 t + c 3 e − 2 t sin 3 t . 182...
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