Solve the given differential equation by undetermined coefficients.y''' − 3y'' + 3y' − y = ex − x + 21

Answer:Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18Step-by-step explanation:y''' − 3y'' + 3y' − y = ex − x + 21Homogeneous solution:First  we propose a solution:Yh = [tex]e^{r*t}[/tex]Y'h = [tex]r*e^{r*t}[/tex]Y''h = [tex]r^{2}*e^{r*t}[/tex]Y'''h = [tex]r^{3}*e^{r*t}[/tex]Now we solve the following equation:Y'''h - 3*Y''h + 3*Y'h - Yh = 0[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]To solve the equation we must propose a solution to the  polynomial :r = 1To find the other r we divide the polynomial by (r-1) as you can see  attached:solving the equation:(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0[tex]r^{2} - 2r + 1[/tex] = 0r = 1So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1 replacing in the main solutionYh =  [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]The t and [tex]t^{2}[/tex] is used because we must have 3 solution  linearly independentParticular solution:We must propose a Yp solution:Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex][tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t -  c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]equalizing coefficients of the same function: - 12c_{1} = 09c_{1} = 03c_{1} = 0c_{1} = 03c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]-c_{2} = -1c_{2} = 1c_{3} = -18Then we have:Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18