Solving $ y'' - y' = -3$ via undetermined coefficients. I stumbled on a problem when solving this ODE. The method of undetermined coefficients say that I can suppose the solution in the form $y_p = k$.
Then:
$(k)''-(k)'=-3 \Rightarrow 0=3$, which is absurd.
When I suppose the solution in the form $kx$ it works, and I find the particular solution $3x$, but why I need to suppose a particular solution in a degree 1 if the RHS of the ODE is of degree 0?
 A: You've stumbled on one of the key difficulties in solving non-homogeneous equations using the method of undetermined coefficients.
The homogeneous equation $y''-y' = 0$ has the solutions $y_1 = A$ and $y_2 = B\mathrm{e}^t$, where $A$ and $B$ are constants. 
This means using a guess for a particular solution of a constant cannot work, as we would be guessing a solution that is already a part of the homogeneous solution!
To remedy this, we need to multiply our guess by the simplest function that will make our guesss linearly independent with our two homogeneous solutions, namely $x$. 
So your guess should be $y_p=Cx$. In general, whenever your initial choice of the form of $y_p$ has any term in
common with the homogeneous solution, then you must alter it by
multiplying your initial choice of $y_p$ by $x$, as many times as necessary but no more than necessary. 
As a general note, you always want to solve the homogeneous equation first to check if this kind of thing is going to be an issue. 
A: You can use general method 
or you can just integrate this way:
$$y''-y'=-3$$
$$y'-y=-3x+K$$
$$e^{-x}y'-e^{-x}y=(-3x+K)e^{-x}$$
$$e^{-x}y=\int(-3x+K)e^{-x}dx$$
$$y(x)=K_1+K_2e^{x}-3e^x\int xe^{-x}dx$$
$$y(x)=K_1+K_2e^{x}+3x$$
