235_pdfsam_math 54 differential equation solutions odd

# 235_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 4.7 (g) No, because the “stiﬀness”, −1, is negative. 17. For the radius, r (t), we have the initial value problem r (t) = −GMr −2 , r (0) = a, r (0) = 0. Thus, in the energy integral lemma, f (r ) = −GMr −2 . Since f (r )dr = −GMr −2 dr = GMR−1 + C, we can take F (r ) = GMR−1 , and the energy integral lemma yields 1 GM 2 [r (t)] − = C1 = const. 2 r (t) To ﬁnd the constant C1 , we use the initial conditions. C1 = Therefore, r (t) satisﬁes GM GM 1 2 [r (t)] − =− 2 r (t) a ⇒ 1 GM GM 2 (r ) = − 2 r a ⇒ r =− 2GM a a−r . r 1 GM 1 GM GM 2 [r (0)] − = · 02 − =− . 2 r (0) 2 a a (Remember, r (t) is decreasing, and so r (t) < 0.) Separating variables and integrating, we get r dr = a−r − 2GM a dt ⇒ a arctan r − a−r r (a − r ) a =− 2GM t + C2 . a We apply the initial condition, r (0) = a, once again to ﬁnd the constant C2 . But this time we have to be careful because the argument of “arctan” function becomes inﬁnite at r = a. So, we take the limit of both sides rather than making simple substitution. lim a arctan r (t) − a − r (t) t→+0 t→+0 r (t)[a − r (t)] a r (t) − lim a − r (t) t→+0 r (t)[a − r (t)] a =a π π −0 =a , 2 2 231 =a lim arctan ...
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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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