By Jerrold Marsden, Alan Weinstein

The second one of a three-volume paintings, this is often the results of the authors'experience instructing calculus at Berkeley. The ebook covers innovations and functions of integration, limitless sequence, and differential equations, the full time motivating the examine of calculus utilizing its purposes. The authors comprise a number of solved difficulties, in addition to broad workouts on the finish of every part. additionally, a separate pupil consultant has been ready.

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**Example text**

Com 62 Calculus Analyse 1c-5 Linear differential equations of higher order and of constants coefficients When we collect these things we see that the characteristic polynomial is (R2 + 4)R2 = R4 + 4R2 , corresponding to the linear diﬀerential equation of fourth order and of constant coeﬃcients d4 x d2 x + 4 = q(t). dt4 dt2 Since x = t2 is a solution of the inhomogeneous equation, we must have d2 x d4 x + 4 2 = 8, 4 dt dt and the equation becomes q(t) = d4 x d2 x + 4 = 8. 8 Find the complete solution of the diﬀerential equation d3 x d2 x dx + x = et cos 2t, + 3 +3 t ∈ R.

We shall only ﬁnd a particular solution. 8, so we also refer to this example. D. Guess a complex exponential x = c · e(1+2i)t . Although it is not required, we shall nevertheless also solve the homogeneous equation. I. If we put x = c · e(1+2i)t into the left hand side of the equation, we get d3 x d2 x dx +x + 3 +3 3 2 dt dt dt = c (1 + 2i)3 + 3(1 + 2i)2 + 3(1 + 2i) + 1 e(1+2i)t = c{1 + (1 + 2i)}3 e(1+2i)t = 8c(1 + i)3 e1+2i)t = 8c · 2i(1 + i)e(1+2i)t = 16(−1 + i)c e(1+2i)t . This is equal to e(1+2i)t for c = equation is Re −1 − i (1+2i)t e 32 =− 1 −1 − i 1 · = , so a solution of the corresponding real 16 −1 + i 32 1 t 1 t 1 t e Re (1 + i)e2it = e sin 2t − e cos 2t.

A) If R = 1 is a root in the characteristic equation, then (9) does not have a solution of the form x = c · et , t ∈ R. b) If R = 1 is a simple root in the characteristic equation, then (9) has a solution of the form x = c · t et , t ∈ R. c) If R = 1 is a double root in the characteristic equation, then (9) has a solution of the form x = c · t2 et , t ∈ R. A. Linear diﬀerential equation of second order and of constant coeﬃcients with a guideline. D. Insert the given functions into (1) and (2). Then, we generalize in (3).