how to solve this inhomogeneous second order differential equation I want to solve this inhomogenenous differential equation of second order: 
$$ x''+2x'+x= \sqrt{t+1}e^{-t} $$
with initial conditions $ x(0)= \pi $ and $ x'(0)=  \sqrt{2} - \pi $
The solution is $ y= y_p+y_h $, so the particular solution added with the homogenous solution.
Because $ D:= 2^2-4 =0 $
the solution to the homogeneous part is $$ y(x)=(C_1+C_2x)e^{-x} $$
How do I solve the particular solution?
 A: The fact that $e^{-t}$ is a solution of the homogeneous equation might suggest the substitution $x(t) = e^{-t} u(t)$, which (after a bit of simplification) turns the differential equation into
$$  u''(t) = \sqrt{t+1} \ .$$
Now integrate twice.
A: Well, you can determine the particular solution by variation of parameters. List the basis solutions in $x_\text{c}(t)$:
$$x_{b_1}(t)=e^{-t}\space\wedge\space x_{b_2}(t)=te^{-t}\tag1$$
Now, compute the Wronskian of $x_{b_1}(t)$ and $x_{b_2}(t)$:
$$\mathcal{W}(t)=\left|\begin{matrix}
  e^{-t} & te^{-t} \\
  \frac{\text{d}}{\text{d}t}\left(e^{-t}\right) & \frac{\text{d}}{\text{d}t}\left(te^{-t}\right)
 \end{matrix}\right|=\left|\begin{matrix}
  e^{-t} & te^{-t} \\
  -e^{-t} & e^{-t}-te^{-t}
 \end{matrix}\right|=e^{-2t}\tag2$$
Let $f(t)=e^{-t}\sqrt{1+t}$. Let $v_1(t)=-\int\frac{f(t)x_{b_2}(t)}{\mathcal{W}(t)}\space\text{d}t$ and $v_2(t)=\int\frac{f(t)x_{b_1}(t)}{\mathcal{W}(t)}\space\text{d}t$. The particular solution will be given by:
$$x_p(t)=v_1(t)x_{b_1}(t)+v_2(t)x_{b_2}(t)\tag3$$
A: $$x''+2x'+x= \sqrt{t+1}e^{-t}$$
With Operator Method:
$$(D+1)^2x= \sqrt{t+1}e^{-t}$$
The particular solution is:
$$x_p=\dfrac 1{(D+1)^2} \sqrt{t+1}e^{-t}$$
$$x_p=e^{-t}\dfrac 1{D^2} \sqrt{t+1}$$
Integrating twice $\sqrt{t+1}$.
$$x_p=\dfrac 23e^{-t}\dfrac 1{D} (t+1)^{3/2}$$
Finally:
$$x_p=\dfrac 4{15}e^{-t}(t+1)^{5/2}$$
