Can $e^{\Gamma x}$ approach solve $y''(x)=0$? As far as I understood, every differential equation that can be written as $f''(x) + p\cdot f'(x) + q\cdot f(x) = 0$ can be completely solved in the way shown in this article. By "completely" I mean that this method should find all $f(x)$s that satisfies the equation.
You should:

*

*Assume that a solution of $f(x)$ is proportional to $e^{\Gamma x}$, so $f(x) = e^{\Gamma x}$

*Substitute $f(x)$ in your differential equation:

$$\frac {d^2 e^ {\Gamma x}} {dx^2} + p\cdot \frac {d e^ {\Gamma x}} {dx} + q \cdot e^ {\Gamma x} = 0$$
$$\Gamma^2 e^ {\Gamma x} + p\cdot \Gamma e^ {\Gamma x} + q\cdot e^ {\Gamma x} = 0$$
$$\require {cancel} \cancel{e^{\Gamma x}} (\Gamma^2 + p\cdot \Gamma + q) = 0 \rightarrow \text{find $\Gamma_1$ and $\Gamma_2$}$$

*

*$f(x) = c_1 e^{\Gamma_1 x} + c_2 e^{\Gamma_2 x}$
By using this method, I managed to solve any differential equation with $p\ne 0 \vee q \ne 0$, but I can't understand how to make it work if $p=0 \wedge q=0$. I expect it to work because, as far as I understood, $p, q \in \mathbb R$ without any other contraint.
Here's my steps:

*

*Assume the differential equation $f''(x) = 0$

*Assume that a solution of $f(x)$ is proportional to $e^{\Gamma x}$, so $f(x) = e^{\Gamma x}$

*Substitute $f(x)$:

$$\frac {d^2 e^ {\Gamma x}} {dx^2} = 0$$
$$\Gamma^2 e^ {\Gamma x} = 0$$
$$\require {cancel} \cancel{e^{\Gamma x}} \Gamma^2 = 0 \rightarrow \Gamma_{1,2} = 0$$

*

*$f(x) = c_1 e^{\Gamma_1 x} + c_2 e^{\Gamma_2 x} = c_1 e^{0 x} + c_2 e^{0 x} = c_1 + c_2 = c$
The solution should be $f(x) = c_1 x + c_2$ and not $f(x) = c$, which is only a partial solution.
I've surely done something wrong, but I can't find what is the mistake.
PS: I know that I could solve this differential equation just by integrating the left side and the right side. My question is to understand why my steps that seems legit to me lead into an incomplete solution
 A: In general not every solution to a homogeneous linear differential equation with constant coefficients is a linear combination of exponentials. When the characteristic polynomial has multiple roots, you get some solutions which are a polynomial times the corresponding exponential. The degree of the polynomial ends up being the multiplicity of the root minus one. There are different ways to derive this, such as guessing and reduction of order. (Reduction of order is closely related to variation of parameters for finding particular solutions.)
One nice way to get some intuition for this is to look at just the second order case and consider $(D-\lambda_1)(D-\lambda_2) y = 0$ with fixed initial conditions; a convenient case is $y(0)=0,y'(0)=1$ although any fixed initial conditions will do. The general solution is of the form $c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}$. So $c_1+c_2=0,\lambda_1 c_1 + \lambda_2 c_2 = 1$. Therefore $c_1=-\frac{1}{\lambda_2-\lambda_1},c_2=\frac{1}{\lambda_2-\lambda_1}$, so you have $\frac{e^{\lambda_2 x}-e^{\lambda_1 x}}{\lambda_2-\lambda_1}$. If you fix one of them equal to say $\lambda_0$ and send the other to it, then you have a derivative with respect to $\lambda$, yielding a solution of $x e^{\lambda_0 x}$ to $(D-\lambda_0)^2 y = 0$.
