# hw3s - 2.4 53(d ote that ylx =(1 cos 8(8 sin 8 Setting x =...

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2.4 53 (d) ote that ylx = (1 - cos 8)/(8 - sin 8). Setting x = 1, Y = 2, the solution of the equation (1 - cos 8)/(8 - sin 8) = 2 is 8 ~ 1.401. Substitution into either of the expressions yields k ~ 2.193. 2.4 2. Rewrite the differential equation as I 1 = a y + t(t - 4) y . It is evident that the coefficient 1/t(t - 4) is continuous everywhere except at t = 0,4. Since the initial condition is specified at t = 2, Theorem 2.4.1 assures the existence of a unique solution on the interval a < t < 4. 3. The function tan t is discontinuous at odd multiples of ~. Since ~ < It < 3; , the initial value problem has a unique solution on the interval (~ , 3;). 5. p(t) = 2tl(4 - t 2 )and g(t) = 3t 2 /(4 - t 2 ). These functions are discontinuous at x = ±2. The initial value problem has a unique solution on the interval (-2,2). 6. The function In t is defined and continuous on the interval (0,00). At t = 1, In t = 0, so the normal form of the differential equation has a singularity there. Also, cot

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hw3s - 2.4 53(d ote that ylx =(1 cos 8(8 sin 8 Setting x =...

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