Recitation 7, February 25, 2010 Solutions to second order ODEs Solution suggestions 1. Check that both = cos(wt) and z = sin(wt) satisfy the second order linear differential equation +w² = 0 This equation is called the harmonic oscillator. Ifzcos(wt), then wsin(wt) and w² cos(wt) war. If z = sin(at), then i wcos(wt) and w² sin(wt) = w 2. In fact, check that any sinusoidal function with circular frequency wi Acos(wt), satisfies the equation += 0. If Acos(wt), then Aw sin(wt), and Aw² cos(wt-) = 3. Among the functions x(t) = A cos(at), which have (0)=0? Doesn't this contradict the uniqueness theorem for differential equations? r(0) Acos. When A = 0, then (1) 0 for every t; when A≠ 0, (0)=0 implies cos 0, and hence can be any odd multiple of /2, Le.,=/2, 3/2, 5/2, This does not contradict the uniqueness theorem, because the uniqueness theorem as stated only applies to first order equations. 4. Given numbers to and io, can you find a solution to +wr=0 for which (0) and (0)ż? How many such solutions are there? Suppose z (1) a cos (wt)+bsin (wt), x(0) a, so001. (0) -aw sin 0+ bw cos 0 hwto. Then b2/w. The solution is then zo cos(wt) + to sin(wt)/w. There is only one such solution. 5. Suppose that is a (perhaps complex) constant such that et is a solution to+kz=0. What does have to be? Lete. Then re,ret, and + kr (r²+k)e. We want this to be zero, but since et is never zero, it must berk. Suppose k≥ 0. then must have the form ivk. 6. Find a solution z, to a=0 note the sign! such that (0)=1 and (0) 0. Find another solution z₂ such that 22(0)=0 and 12(0)=1. We use a similar approach as we studied both (t) and x(t) est are solutions to 0. First, notice that ar0. Then for any constants c₁ and ca, z(t) = ce + ce are also solutions to az = 0. Moreover, z (0) c+c, and (0)=a(c). Assuming a ≠ 0, to get (t), we need c +1 and a(c)=0, which implies c₁== 1/2. Sor (t) cosh (at). For ry (t), set 0 and a (102) = 1, 80 and 22 (t)=sinh (at) /a.

18.03 Differential Equations, Recitation 7 (Solutions to Problems)

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