How can I solve this ODE $ \frac {d^2x}{dt^2} + 8\frac {dx}{dt} + 25x = 10u(t)$? The problem its that I don't know how to treat the $ 10u(t) $ to obtain the particular solution since I don't know what the $ u(t) $ function represents.
I've already have the complementary solution of the homogeneous equation associated which is: 
$$ X_c(t)=C_1e^{-4t}\cos {3t} + C_2e^{-4t}\sin {3t} . $$
Here again, I leave the equation to solve. 
$$ \frac {d^2x}{dt^2} + 8\frac {dx}{dt} + 25x = 10u(t)$$
Thank you.
EDIT:
Thanks SplitInfinity, Yes, the $ u(t) $ was the Step Function and something I forgot to mention that the problem says: consider initial conditions in $ 0 $. So the step function states: 
$$ u(t)=
\begin{cases} 
0 & t<0 \\ 
1 & t \ge 0 
\end{cases} $$
in this case $ t=1 $, therefore the equation ends up like this
$$ \begin{align}
x''+8x'+25x&=10(1) \\
x''+8x'+25x&=10
\end{align}$$ 
because the non-homogeneous part its a linear polynomial the $X_p$ (proposal solution) need to be in this way:
$$ X_p=A \\
X'_p=0 \\
X''_p=0 $$
So the equation ends up like:
$$ \begin{align}
x_p''+8x_p'+25x_p & =10 \\
0+8(0)+25(A) & = 10 \\
A & = \frac {10}{25} \\
A & = \frac 25
\end{align}$$
Now the General Solution is given by $ X=X_c+X_p $ which is:
$$ X=C_1e^{-4t}\cos {3t} + C_2e^{-4t}\sin {3t}+\frac 25 $$
To find out the $C_1$ and $C_2$ values we need to apply the initial values, which are:
$$ x(0)=0,~~
x'(0)=0 $$
in order to do so, we differentiate $X$
$$ X'=C_1e^{-4t}(-3\sin{3t}-4\cos{3t})+C_2e^{-4t}(3\cos{3t}-4\sin{3t}) $$
Now, applying initial values to X:
$\require{cancel}$
\begin{align} 
X(0) & = C_1\cancelto{1}{e^{-4(0)}}\cancelto{1}{\cos{3(0)}}+\cancelto{0}{C_2e^{-4(0)}\sin{3(0)}}+\frac25 \\
0 &= C_1 + \frac 25 \\
C_1 &= -\frac25
\end{align} 
Now to $ X' $:
\begin{align}
X'(0) &= C_1\cancelto{1}{e^{-4(0)}}(\cancelto{0}{-3\sin{3(0)}})+C_2\cancelto{1}{e^{-4(0)}}(3\cancelto{1}{\cos{3(0)}}-\cancelto{0}{4\sin{3(0)}}) \\
0 &= C_1(-4)+C_2(3) \\
0 &= -4C_1 + 3C_2 \\
3C_2 &= 4C_1 \\
\text{Replacing the $C_1$ value} \\
C_2 &= \frac{4(-\frac25)}{3} \\
C_2 &= -\frac8{15}
\end{align}
Replacing the constants $ C_1 $ and $ C_2 $ in $X$ we found the solution:
\begin{align}
X &= -\frac25 e^{-4t}\cos{3t}-\frac8{15}e^{-4t}\sin{3t}+\frac25 \\
X &= \frac1{15} \left[ e^{-4t}(-6\cos{3t}-8\sin{3t})+6 \right]
\end{align}
 A: Solving this with Laplace Transform, with $t>0$:
$$x''(t)+8x'(t)+25x=10u(t)\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[x''(t)+8x'(t)+25x(t)\right]_{(s)}=\mathcal{L}_{t}\left[10u(t)\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[x''(t)\right]_{(s)}+\mathcal{L}_{t}\left[8x'(t)\right]_{(s)}+\mathcal{L}_{t}\left[25x(t)\right]_{(s)}=\mathcal{L}_{t}\left[10u(t)\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[x''(t)\right]_{(s)}+8\mathcal{L}_{t}\left[x'(t)\right]_{(s)}+25\mathcal{L}_{t}\left[x(t)\right]_{(s)}=10\mathcal{L}_{t}\left[u(t)\right]_{(s)}\Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8\left(sx(s)-x(0)\right)+25x(s)=10u(s)\Longleftrightarrow$$
$$s^2x(s)-sx(0)-x'(0)+8sx(s)-8x(0)+25x(s)=10u(s)\Longleftrightarrow$$
$$x(s)\left[s^2+8s+25\right]-sx(0)-x'(0)-8x(0)=10u(s)\Longleftrightarrow$$
$$x(s)\left[s^2+8s+25\right]=10u(s)+sx(0)+x'(0)+8x(0)\Longleftrightarrow$$
$$x(s)=\frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}\Longleftrightarrow$$
$$\mathcal{L}_{s}^{-1}\left[x(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}\right]_{(t)}\Longleftrightarrow$$
$$x(t)=\mathcal{L}_{s}^{-1}\left[\frac{10u(s)+sx(0)+x'(0)+8x(0)}{s^2+8s+25}\right]_{(t)}$$
Notice, when $u(t)=\theta(t)$ with $\theta(t)$ is the 'Unit step function' (Heaviside step function) then we get that the Laplace Transform of that function:
$$u(s)=\frac{1}{s}$$
So:
$$x(t)=\mathcal{L}_{s}^{-1}\left[\frac{\frac{10}{s}+sx(0)+x'(0)+8x(0)}{s^2+8s+25}\right]_{(t)}\Longleftrightarrow$$
$$x(t)=\frac{12+2e^{-4t}\left(3\left(5x(0)-2\right)\cos(3t)+\left(20x(0)+5x'(0)-8\right)\sin(3t)\right)}{30}$$
A: The homogeneous solution is 
$$y_-(t)=e^{-4t}(A_-\cos(3t)+B_-\sin(3t).$$
A particular solution for the inhomogeneous right side $10$ is a constant function that then has to satisfy $25C=10$ or $C=\frac25$, so that the full solution family is
$$
y_+(t)=\frac25+e^{-4t}(A_+\cos(3t)+B_+\sin(3t).
$$
If, as SplitInfinity comments, $u(t)$ is the Heaviside or unit step function, then one has to arrange that both branches of the solution meet at $t=0$ in value and first derivative, that is
$$
y_-(0)=A_-\;=\;y_+(0)=\frac25+A_+\\
y_-'(0)=-4A_-+3B_-\;=\;y_+'(0)=-4A_++3B_+\\
\implies B_-=\frac8{15}+B_+
$$
which gives the full solution as
$$
y(t)=y_-(t)+u(t)(y_+(t)-y_-(t))
=e^{-4t}(A_-\cos(3t)+B_-\sin(3t)+\frac2{15}u(t)\left(3-e^{-4t}(3\cos(3t)+4\sin(3t))\right)
$$
