173_pdfsam_math 54 differential equation solutions odd

173_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 4.2 9. We differentiate y = A cos 2t + B sin 2t twice to get y = −2A sin 2t + 2B cos 2t and y = −4A cos 2t − 4B sin 2t, substitute y , y , and y into the given equation, and compare coefficients. This yields y + 2y + 4y = (−4A cos 2t − 4B sin 2t) + 2(−2A sin 2t + 2B cos 2t) + 4(A cos 2t + B sin 2t) = 4B cos 2t − 4A sin 2t = 3 cos 2t + 4 sin 2t ⇒ 4B = 3, −4A = 4 ⇒ A = −1, B = 3/4 ⇒ y = − cos 2t + (3/4) sin 2t. EXERCISES 4.2: Homogeneous Linear Equations; The General Solution, page 167 1. The auxiliary equation for this problem is r 2 + 5r + 6 = (r + 2)(r + 3) = 0, which has the roots r = −2 and r = −3. Thus {e−2t , e−3t } is a set of two linearly independent solutions for this differential equation. Therefore, a general solution is given by y (t) = c1 e−2t + c2 e−3t , where c1 and c2 are arbitrary constants. 3. The auxiliary equation, r 2 + 8r + 16 = (r + 4)2 = 0, has a double root r = −4. Therefore, e−4t and te−4t are two linearly independent solutions for this differential equation, and a general solution is given by y (t) = c1 e−4t + c2 te−4t , where c1 and c2 are arbitrary constants. 5. The auxiliary equation for this problem is r 2 + r − 1 = 0. By the quadratic formula, we have √ √ −1 ± 5 −1 ± 1 + 4 = . r= 2 2 Therefore, a general solution is z (t) = c1 e(−1− √ 5)t/2 + c2 e(−1+ √ 5)t/2 . 169 ...
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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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