Differential equation with bounded solutions What are the possible values of $c,d\in\mathbb{R}$ such that any $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f''(x)+f'(x)+cf(x)-dx=0$ is bounded?
My approach was consider $c=0$ which give that for any value of $d$ there exists an $f$ satisfying the equation but is unbounded and then I considered $c\neq 0$ and tried to solve the equation but it looks a lot messy. Is there an easy way to approach this exercise?
 A: First look at the homogeneous equation $f''(x) + f'(x) + c f(x) = 0$.  If any solution of that is unbounded, $(c,d)$ is not possible.   
A: This diff. eqn being linear it can only have solutions of the form $f(x) = Ae^{rx}$ or $f(x) = Axe^{rx}$. However, due to it being nonhomogeneous $f(x) = ex + g$ is also a solution. Eventually all solutions will be a linear combination of these three. 
If $f(x)$ is bounded on $\mathbb{R}$, then $r=0$. Furthermore, $e = 0$, which implies $d= 0$. 
If $r = 0$ is the only solution to the characteristic polynomial, then you get a term like $Ax$, which implies unboundedness. If $r_1 = 0$ and $r_2 \neq 0$, then it would be unbounded due to $r_2$. Therefore there is no bounded solution on $\mathbb{R}$.
A: I assume $c$ and $d$ are real. If $c\neq 0,\;\;c\neq  1/4$, the general solution of your equation is:
$$
f=c_1 e^{r_1x}+c_2e^{r_2x}+dx/c-d/c^2,
$$
where $c_i$ are arbitrary constants and $r_i$ are the two distinct roots of $r^2+r+c=0$. Clearly, here the solution is not bounded if $d\neq 0$. However, if $d=0$, you need the real parts of the $r_i$'s to be less or equal to zero if you want all solutions to be bounded on $(0,\infty)$. The roots of $r^2+r+c=0$ both have negative real parts if and only if $c>0$. However, none of the solutions will be bounded on $\mathbb{R}$.
If $c=0$ then the general solution is
$$
f=c_1 +c_2e^{-x}+dx^2/2-dx.
$$
The general solution is thus bounded on $(0,\infty)$ only if $d=0$. However, not all the solutions will be bounded on $\mathbb{R}$. In the particular case where $c_2=0$, the solution (the constant one) will be bounded on $\mathbb{R}$.
In the case $c=1/4$ the general solution is
$$
f=c_1 e^{-x/2}+c_2xe^{-x/2}+dx/c-d/c^2.
$$
This is bounded on $(0,\infty)$ only in the case $d=0$. However, none of the solutions will be bounded on $\mathbb{R}$.
Conclusion: only in the case $d=0$ and $c\geq 0$ do you have that all solutions are bounded solutions on $(0,\infty)$. However, you will never have that all solutions are bounded on all $\mathbb{R}$. 
Also from the analysis above: the only case where there will be a nonzero solution (not all solutions) that is bounded on $\mathbb{R}$ will be when $d=c=0$ and the bounded solution will be the constant one.
A: Find a solution of this form: $f = Ax^2 + Bx + C$. Plug in and solve for $f$:
\begin{align*}
2A + 2Ax + B + c(Ax^2 + Bx + C) & = dx \\
2A + B + cC & = 0 \\
2A + cB & = d \\
cA & = 0.
\end{align*}
We see that if $c = 0$, $A = \frac{d}{2}$, $B = -d$ and $C$ can be arbitrary. The solution is $f = \frac{d}{2}x^2 - dx + C$. Otherwise, $A = 0$, $B = \frac{d}{c}$ and $C = -\frac{d}{c^2}$. The solution is $f = \frac{d}{c}x - \frac{d}{c^2}$.
In both cases, $f$ is unbounded, so the problem has no solution.
