Solve one of the following non-homogeneous Cauchy-Euler equations using whatever technique you prefer. Put an "X" through whichever equations you would not put an "X" through either equation, I will grade whichever one I prefer. a) x^2 y" + 10xy' + 8y = x^2 b) x^2 y" - 3xy' + 13y = 4 + 3x

Answer:a.[tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]b.[tex]y(x)=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]Step-by-step explanation:1.[tex]x^2y''+10xy'+8y =x^2[/tex]It is Cauchy-Euler equation where [tex]x=e^t[/tex]Then auxillary equation[tex]D'(D'-1)+10D'+8=0[/tex][tex]D'^2+9D'+8=0[/tex][tex](D'+1)(D'+8)=0[/tex]D'=-1 and D'=-8Hence, C.F=[tex]C_1e^{-t}+C_2e^{-8t}[/tex]C.F=[tex]C_1\frac{1}{x}+C_2\frac{1}{x^8}[/tex]P.I=[tex]\frac{e^{2t}}{D'^2+9D'+8}=\frac{e^{2t}}{4+18+8}[/tex] Where D'=2[tex]P.I=\frac{1}{30}e^{2t}=\frac{1}{30}x^2[/tex][tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]b.[tex]x^2y''-3xy'+13y=4+3x[/tex]Same method apply Auxillary equation[tex] D'^2-D'-3D'+13=0[/tex][tex]D'^2-4D'+13=0[/tex][tex]D'=2\pm3i[/tex]C.F=[tex]e^{2t}(C_1cos 3t+C_2sin 3t)[/tex]C.F=[tex]x^2(C_1cos (3lnx)+C_2sin(3lnx))[/tex][tex]e^t=x[/tex]P.I=[tex]\frac{4e^{0t}}{D'^2-4D'+13}+3\frac{e^t}{D'^2-4D'+13}[/tex]Substitute D'=0 where[tex] e^{0t}[/tex] and D'=1 where [tex]e^t[/tex]P.I=[tex]\frac{4}{13}+\frac{3}{10}e^t[/tex]P.I=[tex]\frac{4}{13}+\frac{3}{10}x[/tex][tex]y(x)=C.F+P.I=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]