differential equation : non-homogeneous solution, finding YP hi i have a problem for this Differential Equations :
$$
\frac{d^{3}y}{dx^3} - 9\frac{dy}{dx} = 10 - 4x
$$
i know first we must solve the homogeneous equation:
and my result is : $C_1 + C_2e^{3x} + C_3e^{-3x}$
i'm really confused to find yp from $10 - 4x$, 
my friend said it must : $(ax+b)x$
my question is : 
1. why it's not $c - (ax + b)$, 
2. if it same with $c1$ from yh(homogeneous solution ) why we not multiply $c$ with x? 
   so $yp = cx - (ax + b) $
3. and why $(ax+b)$ must be multiplied with $x$?? 

can someone explain me?
thanks for anyone helped me.
 A: It is straightforward to see that if a polynomial solves the inhomogeneous ODE
$$\tag{ODE}
y'''-9y'=10-4x,
$$
it must be of degree $2$. Hence, we have
$$
p'''(x)-9p'(x)=10-4x \iff -9p'(x)=10-4x \iff p'(x)=-\frac{10}{9}+\frac49x,
$$
i.e.
$$
p(x)=-\frac{10}{9}x+\frac29x^2
$$
Setting
$$\tag{2}
z=y-p,\ u=z'
$$
we see that $z$ solves the homogeneous equation
$$\tag{ODE'}
z'''-9z'=0,
$$
and therefore $u$ solves the 2nd order ODE
$$
u''-9u=0
$$
whose solution is given by
$$
u(x)=-3c_1e^{-3x}+3c_2e^{3x}, \ c_1,c_2 \in \mathbb{R}.
$$
It follows that
$$
z(x)=c_3+c_1e^{-3x}+c_2e^{3x},\ c_3 \in \mathbb{R}
$$
and 
$$
y(x)=c_3-\frac{10}{9}x+\frac29x^2+c_1e^{-3x}+c_2e^{3x}.
$$
A: Related problems: (I). We use the annihilator method. Let $D=\frac{d}{dx}$. Now, apply $D$ twice to both sides of the differential equation to transform it to a homogeneous differential equation, so we have
$$ D^2(D^3-9D)y=D^2(10-4x)=0\implies D^3(D^2-9)y=0.  $$
The last equation gives you the auxiliary equation of the new homogeneous differential equation which allow us to write the general solution of the new homogeneous differential equation
$$ m^3(m^2-9)=0 \implies m=0,0,0,3,-3,$$
which gives the general solution 
$$ y(x) = c_1+c_2x+c_3x^3+c_4e^{3x}+c_5e^{-3x}=(c_1+c_4e^{3x}+c_5e^{-3x})+c_2x+c_3x^2 \rightarrow(1).$$
You can recognize that the expression between the brackets in equation $(1)$ corresponds to the solution of the original homogeneous differential equation
$$ \frac{d^{3}y}{dx^3} - 9\frac{dy}{dx} = 0.  $$
So, the other terms in $(1)$ tell you about the form of $y_p$ of the original differential equation which you assume as
$$ y_p=Ax +Bx^2. $$
A: Inhomogeneous solution here will be of the form
$$y_p(x) = a (10-4 x)^2 + b$$
such that $y_p(0) = 0$ (subsequent values of  $y_p'(0)$ and $y_p''(0)$ will have to be taken into account into overall initial conditions on solution $y$).  The reason it takes a quadratic form is that you have a $y'$ but no $y$ in your equation.
Putting this form into the equation and solving for $a$ and $b$ (from the initial condition), I get
$$y_p(x) = \frac{1}{72} (10-4 x)^2 - \frac{25}{18}$$
