The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' + 2y' + y = 0; y1 = xe−x

First confirm that [tex]y_1=xe^{-x}[/tex] is a solution to the ODE,[tex]y''+2y'+y=0[/tex]We have[tex]{y_1}'=e^{-x}-xe^{-x}=(1-x)e^{-x}[/tex][tex]{y_1}''=-e^{-x}-(1-x)e^{-x}=(-2+x)e^{-x}[/tex]Substituting into the ODE gives[tex](-2+x)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0[/tex]Suppose [tex]y_2(x)=v(x)y_1(x)[/tex] is another solution to this ODE. Then[tex]{y_2}'=v'y_1+v{y_1}'[/tex][tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]and substituting these into the ODE yields[tex](v''y_1+2v'{y_1}'+v{y_1}'')+2(v'y_1+v{y_1}')+vy_1=0[/tex][tex]xe^{-x}v''+2e^{-x}v'=0[/tex][tex]xv''+2v'=0[/tex]Let [tex]w(x)=v'(x)[/tex]. Then the remaining ODE is linear in [tex]w[/tex]:[tex]xw'+2w=0[/tex]Multiply both sides by the integrating factor, [tex]x[/tex], and condense the left hand side as a derivative of a product:[tex]x^2w'+2xw=(x^2w)'=0[/tex]Integrate both sides with respect to [tex]x[/tex] and solve for [tex]w[/tex]:[tex]x^2w=C_1\implies w=C_1x^{-2}[/tex]Back-substitute and integrate both sides with respect to [tex]x[/tex] to solve for [tex]v[/tex]:[tex]v'=C_1x^{-2}\implies v=-C_1x^{-1}+C_2[/tex]Back-substitute again to solve for [tex]y_2[/tex]:[tex]\dfrac{y_2}{y_1}=C_2-\dfrac{C_1}x[/tex][tex]\implies y_2=C_2xe^{-x}-C_1e^{-x}[/tex][tex]y_1[/tex] already captures the solution [tex]xe^{-x}[/tex], so the remaining one is[tex]\boxed{y_2=e^{-x}}[/tex]