Solving a third order ODE, unsure on which method to use Consider the ODE $$y'''+3y''+3y'+y=e^x-x-1$$
Do you solve the complimentary solution first? If so how do I go about the particular?
 A: This ODE can be written as $$(D+1)^3=e^x-1-x~~~~(1)$$, The solution of the homogeneous part
$$(D+1)^3y=0~~~~~~(2)$$ is $$y_1=(A+Bx+Cx)e^{-x}~~~~~~(3).$$ The solution of $$(D+1)^3 y=e^x ~~~~~(4).$$ is $$y_2=\frac{e^x}{(1+1)^3}=\frac{e^x}{8}~~~~~(5).$$ The Solution of $$(D+1)^3 y=-1+x ~~~~~~(6).$$ is $$y_3=(1+D)^{-3} (-1-x) \Rightarrow y_3= (1- 3D+6 D^2+...)(-1-x) \Rightarrow y_3=-(1-3D)(1+x)=2-x.$$ So finally the total solution of (1) is
$$y=(A+Bx+Cx^2)e^{-x}+\frac{e^x}{8}+2-x.$$
A: Hint: The complimentary solution is given by $$y \left( x \right) ={\it \_C1}\,{{\rm e}^{-x}}+{\it \_C2}\,{{\rm e}^{-
x}}x+{\it \_C3}\,{{\rm e}^{-x}}{x}^{2}
$$
A: For the particular solution, handle the exponential and the polynomial separately.


*

*exponential: with the Ansatz $ae^x$, solve $ae^x+3\,ae^x+3\,ae^x+ae^x=e^x$;

*polynomial: a linear polynomial $ax+b$ will yield a linear response. You solve $(0)+3\,(0)+3\,(a)+(ax+b)=-x-1.$
For the homogeneous solution, notice that the characteristic polynomial has the triple root $-1$ and the solution will be of the form $Q(x)e^{-x}$ where $Q$ is a quadratic polynomial ($3$ coefficients).
