The DE y''+2y'+y=1/x² has initial conditions y(1)=0,y'(1)=3 and it has complementary solutions yc=c₁e⁻ˣ+c₂xe⁻ˣ. Note: Since x≠0, assume the interval of solution is [1,[infinity]). Using y=u₁e⁻ˣ+u₂xe⁻ˣ, we obtain u₁'=W₁/W;u₂'=W₂/W where
W=[e⁻ˣ xe⁻ˣ ]=e⁻²ˣ.
[-e⁻ˣ -xe⁻ˣ+e⁻ˣ]
Find W₁ and W₂