Initial value problem for 2nd order ODE $y''+ 4y = 8x$ How can I go about solving this equation
$y''+ 4y = 8x$?
Progress
I found  the general solution for its homogeneous form. What I don't know is how to find its particular solution.
 A: Follow the steps:
1) find the solutions of the homogeneous ode $y_1,y_2$

$$y''+4y=0$$

2) find a particular solution $y_p$ by assuming 

$$y_p=Ax+B$$

then plug in back in the ode to find the constants $A$ and $B$. See table
3) construct the general solution

$$ y = c_1y_1 + c_2y_2 + y_p $$

A: You solve the homogeneous problem:
$$y''+4y=0$$
You will find: $y_h(x)=c_1 \cos(2x)+ c_2 \sin(2x)$
So,for the non-homogeneous problem:
$$y_n(x)=Ax+B$$
Replace at $y''+4y=8x$ and find the constants $A$ and $B$.
The general solution is $y(x)=y_h(x)+y_n(x)$
A: It may be worth noting that your DEQ is invariant to a translation group:
$$
G(x,y)=(x+\lambda, y+2\lambda)\lambda_o=0
$$In other words, if you substitute $x'=x+\lambda$ and $y'=y+2\lambda$ into your equation (and please note that the primes do not signify differentiation, and to avoid confusion here I'll use $\frac{dy}{dx}=\dot{y}$), then you get the same DEQ.  The stabilizers for this invariant group transformation are $\mu=y-2x$, $\nu=\dot{y}$, and $\eta=\ddot{y}$.  Rewriting your DEQ in terms of the invariant group stabilizers (Sophus Lie referred to them as differential invariants), you get
$$
\eta=-4\mu
$$Now we note that
$$
\frac{d\mu}{dx}=\nu-2
$$and
$$
\frac{d\nu}{dx}=\eta=-4\mu
$$which means
$$
\frac{d\nu}{d\mu}=\frac{\frac{d\nu}{dx}}{\frac{d\mu}{dx}}=\frac{-4\mu}{\nu-2}
$$a separable equation.  Before proceeding further it is worth noting that when the slope of the direction field is undefined a singularity (or saddle points and their interconnecting separatrix) is present. These are points that do not change under group transformation.  The only nontrivial singularities occur when  $\nu=2$ and $\mu=0$, which lead to the conclusion that $y=2x$ is a special solution for your DEQ.  If we separate the equation, it becomes
$$
(\nu-2)d\nu=(-4\mu)d\mu
$$which integrates to a quadratic for $\nu$.
$$
\nu^2-4\nu+(4\mu^2+C_1)=0
$$Playing a little fast and loose with the constant,
$$
\nu=2\pm2\sqrt{C_1-\mu^2}
$$Noting that
$$
\frac{d\mu}{dx}=\nu-2=\pm2\sqrt{C_1-\mu^2}
$$you can once again separate variables and, after a little work, arrive at a solution.
$$
y=C_1sin(\pm2x+C_2)+2x
$$Note that if the constants are set to zero you get the singularity in the $\mu\nu$-direction field of the stabilizers which is the special solution, $y=2x$.
