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Unformatted text preview: Chapter 4
yp (θ) = 2 (A cos θ + B sin θ)e−θ + θ (A cos θ + B sin θ)e−θ = 2 [(B − A) cos θ − (B + A) sin θ] e−θ + θ (A cos θ + B sin θ)e−θ . (Note that we did not evaluate the terms containing the factor θ because they give zero result when substituted into the original equation.) Therefore, yp + 2yp + 2yp = 2 [(B − A) cos θ − (B + A) sin θ] e−θ + 2(A cos θ + B sin θ)e−θ = 2 (B cos θ − A sin θ) e−θ = e−θ cos θ . Hence A = 0, B = 1/2, yp (θ) = (1/2)θe−θ sin θ, and a general solution is given by y (θ ) = 1 −θ θe sin θ + (c1 cos θ + c2 sin θ) e−θ . 2 23. The corresponding homogeneous equation, y − y = 0, is separable. Solving yields dy =y dt ⇒ dy = dt y ⇒ ln |y | = t + c ⇒ y = ±ec et = Cet , where C = 0 is an arbitrary constant. By inspection, y ≡ 0 is also a solution. Therefore, yh (t) = Cet , where C is an arbitrary constant, is a general solution to the homogeneous equation. (Alternatively, one can apply the method of solving ﬁrst order linear equations in Section 2.3 or the method discussed in Problem 21, Section 4.2.) A particular solution has the form yp (t) = A. Substitution into the original equation yields (A) − A = 1 ⇒ A = −1. Thus y (t) = Cet − 1 is a general solution. To satisfy the initial condition, y (0) = 0, we ﬁnd 0 = y (0) = Ce0 − 1 = C − 1 So, the answer is y (t) = et − 1. 25. The auxiliary equation, r 2 + 1 = 0, has roots r = ±i. Therefore, a general solution to the corresponding homogeneous equation is zh (x) = c1 cos x + c2 sin x, and a particular solution 200 ⇒ C = 1. ...
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