Chapter 4yp(θ)=2 (Acosθ+Bsinθ)e−θ+θ(Acosθ+Bsinθ)e−θ=2 [(B−A) cosθ−(B+A) sinθ]e−θ+θ(Acosθ+Bsinθ)e−θ.(Note that we did not evaluate the terms containing the factorθbecause they give zero resultwhen substituted into the original equation.) Therefore,yp+ 2yp+ 2yp=2 [(B−A) cosθ−(B+A) sinθ]e−θ+ 2(Acosθ+Bsinθ)e−θ=2 (Bcosθ−Asinθ)e−θ=e−θcosθ .HenceA= 0,B= 1/2,yp(θ) = (1/2)θe−θsinθ, and a general solution is given byy(θ) =12θe−θsinθ+ (c1cosθ+c2sinθ)e−θ.23.The corresponding homogeneous equation,y−y= 0, is separable. Solving yieldsdydt=y⇒dyy=dt⇒ln|y|=t+c⇒y=±ecet=Cet,whereC= 0 is an arbitrary constant.By inspection,y≡0 is also a solution.Therefore,yh(t) =Cet, whereCis an arbitrary constant, is a general solution to the homogeneousequation. (Alternatively, one can apply the method of solving first order linear equations in
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