Solve the following differential equation : $\frac{d^3y}{dx^3}+y=3+e^{-x}+5e^{2x}.$
My solution goes like this:
Given, $\frac{d^3y}{dx^3}+y=3+e^{-x}+5e^{2x}$. We write this, as $(D^3+1)y=3+e^{-x}+5e^{2x}.$ We first, calculate the value of the complementary function, i.e general solution of the differential equation, $(D^3+1)y=0.$ The roots of $f(x)=x^3+1$ are :$-1,a_1=\frac12+\frac{\sqrt 3i}{2},a_2=\frac12-\frac{\sqrt 3i}{2}.$ Thus, the complementary function $CF=c_1e^{-x}+c_2e^{a_1x}+c_3e^{a_2x},$ where $c_1,c_2,c_3$ are arbitary constants. Now, we evaluate the particular integral $y$ of the original equation $$\frac{d^3y}{dx^3}+y=3+e^{-x}+5e^{2x}$$ i.e \begin{align} &y=\frac{1}{D^3+1}(3+e^{-x}+5e^{2x}) \\ &\implies y=\frac{1}{D^3+1}(3)+\frac{1}{D^3+1}e^{-x}+5\frac{1}{D^3+1}(e^{2x}) \\ &\implies y = 3+\frac{1}{D+1}(\frac{1}{D^2-D+1}e^{-x})+5\frac{e^{2x}}{9} \\ &\implies y = 3+\frac{1}{D+1}\frac{e^{-x}}{3}+5\frac{e^{2x}}{9} \\ &\implies y= 3+\frac{xe^{-x}}{3}+5\frac{e^{2x}}{9}. \end{align} Thus, the particular integral is, $$y= 3+\frac{xe^{-x}}{3}+5\frac{e^{2x}}{9}.$$ Now, the complete solution of the given equation $\frac{d^3y}{dx^3}+y=3+e^{-x}+5e^{2x}$ is : $$y_1=CF+y=c_1e^{-x}+c_2e^{a_1x}+c_3e^{a_2x}+3+\frac{xe^{-x}}{3}+5\frac{e^{2x}}{9}.$$
Is the above solution correct? If not, where is it going wrong ? Is the method, valid?
Your CF or null space solution is wrong.Do you see why? ( Hint:think about complex roots)
$y_{\text{complete}}=y_{\text{null}}+y_{\text{particular}}$
Auxiliary equation : $m^3+1=0$
Auxiliary roots:
$m=-1,\frac{1}{2}\pm\frac{\sqrt 3i}{2}$
$y_{\text{null}}=c_1e^{-x}+e^{\frac{x}{2}}(c_2 \sin(\frac{\sqrt{3}x}{2})+c_3\cos(\frac{\sqrt{3}x}{2}))$
You have calculated $y_{\text{particular}}$ correctly by the method of $D$ operator.
Let's compute it using the method of undermined coefficients.
Let $y_{p}=A+Bxe^{-x}+Ce^{2x}$
$y_{p}'=Be^{-x}-Bxe^{-x}+2Ce^{2x}$
$y_{p}''=-2Be^{-x}+Bxe^{-x}+4Ce^{2x}$
$y_{p}'''=3Be^{-x}-Bxe^{-x}+8Ce^{2x}$
$y_p'''+y_p=3+e^{-x}+5e^{2x}$
$A+3Be^{-x}+9Ce^{2x}=3+e^{-x}+5e^{2x}$
Equating coefficients: $A=3, B=\frac{1}{3}, C=\frac{5}{9}$
Hence $y_{\text{particular}}=\frac{1}{3}+\frac{1}{3}xe^{-x}+\frac{5}{9}e^{2x}$
Note:After the edit your solution looks perfect to me.I also knew that you have made a typo.But my intention to write this answer is to provide you an easy alternative.