Find the general solution of the differential equation y^(5) −2y^(4) + y^(3) = 0.

[tex]y^{(5)}-2y^{(4)}+y^{(3)}=0[/tex]We can reduce the order of the ODE by substituting [tex]v(x)=y^{(3)}(x)[/tex], so that [tex]v'(x)=y^{(4)}(x)[/tex] and [tex]v''(x)=y^{(5)}(x)[/tex]. Then[tex]v''-2v'+v=0[/tex]has characteristic equation[tex]r^2-2r+1=(r-1)^2=0[/tex]with root [tex]r=1[/tex], which has multiplicity 2, so that the characteristic solution is[tex]v_c=C_1e^x+C_2xe^x[/tex]Integrate both sides to solve for [tex]y''(x)[/tex]:[tex]y''=C_1e^x+C_2e^x(x-1)+C_3[/tex][tex]y''=C_1e^x+C_2xe^x+C_3[/tex]Integrate again to solve for [tex]y'(x)[/tex]:[tex]y'=C_1e^x+C_2e^x(x-1)+C_3x+C_4[/tex][tex]y'=C_1e^x+C_2xe^x+C_3x+C_4[/tex]And one last time to solve for [tex]y(x)[/tex]:[tex]y=C_1e^x+C_2e^x(x-1)+\dfrac{C_3}2x^2+C_4x+C_5[/tex][tex]\boxed{y(x)=C_1e^x+C_2e^x+C_3x^2+C_4x+C_5}[/tex]