Exercises 4.5a particular solution has the formyp,1(x) =Acosθ+Bsinθ.Substitution into theequation yields(−Acosθ−Bsinθ)−(Acosθ+Bsinθ) =−2Acosθ−2Bsinθ= sinθ⇒−2A= 0,−2B= 1⇒A= 0,B=−1/2.So,yp,1(θ) =−(1/2) sinθ.(ii)For the equationy−y=e2θ,a particular solution has the formyp,2(θ) =Ae2θ. Substitution yields(Ae2θ)−(Ae2θ)= 3Ae2θ=e2θ⇒A= 1/3,andyp,2(θ) = (1/3)e2θ.By the superposition principle, a particular solution to the original nonhomogeneous equationis given byyp(θ) =yp,1(θ)−yp,2(θ) =−(1/2) sinθ−(1/3)e2θ,and a general solution isy(θ) =yp(θ) +y
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