Query related to solutions of a linear differential equation

Question

Consider the differential equation

$$ y''-5y'+6y=0$$

To solve this, we assume a solution of the form $e^{rx}$, and fitting it in the equation we get

$$e^{rx}(r^2-5r+6)=0$$

which gives values of $r=2,3$, which gives possible solutions $e^{2x}$ and $e^{3x}$. My questions are:

1. Why do we let some function like $e^{rx}$? Is it from experience? What is the motivation behind taking such function?

2. We can see that functions of the form $c_1e^{2x}+c_2e^{3x}$ are also solutions of this differential equation. How are we sure that there are no other solutions? I am not getting any explanation in my textbooks. Can anyone explain?

Answer 1

This is a very general answer, and it requires knowledge from abstract algebra and linear algebra, with a bit of calculus. However, I think this is the simplest proof I know that gives all solutions of linear ordinary differential equations with constant coefficients. In the answer below, $\mathcal{C}^\infty$ can be replaced with $\mathcal{C}^r$, where $r$ is the degree of the differential equation.

From Proof that the real vector space of $C^\infty$ functions with $f''(x) + f(x) = 0$ is two-dimensional, you can see that the solutions to $$p(D)\,f=0$$ for $f\in\mathcal{C}^\infty(\Omega,\mathbb{C})$ lie within $$\ker\big(p(D)\big)=\bigoplus_{i=1}^k\,\ker\big(p_i(D)\big)\,.$$ It can be easily seen that $\ker\big(p_i(D)\big)$ consists of all functions $f:\Omega\to\mathbb{C}$ satisfying $$f(x)=g(x)\,\exp\left(z_ix\right)\text{ for }x\in\Omega\,,$$ where $g(x)\in\mathbb{C}[x]$ is a polynomial of degree less than $m_i$. This shows that $\ker\big(p(D)\big)$ is composed by $f:\Omega\to \mathbb{C}$ of the form $$f(x)=\sum_{i=1}^k\,g_i(x)\,\exp\left(z_ix\right)\text{ for }x\in \Omega\,,$$ where, for each $i=1,2,\ldots,k$, $g_i(x)\in\mathbb{C}[x]$ is a polynomial of degree less than $m_i$.

Answer 2

Let $y$ be any other solution. Fix some $x_0$. Then you can find constants $c_1,c_2$ so that $$ y(x_0)=c_1e^{2x_0}+c_2e^{3x_0}\\ y'(x_0)=2c_1e^{2x_0}+3c_2e^{3x_0} $$ Now by linearity $u(x)=y(x)-(c_1e^{2x}+c_2e^{3x})$ is also a solution, with $u(x_0)=0$, $u'(x_0)=0$. By uniqueness $u$ is the zero solution, thus $$y(x)=c_1e^{2x}+c_2e^{3x}$$ everywhere.

Answer 3

Consider $u=y'-2y$. Then $u'-3u=0$. Thus $$0=e^{-3x}(u'-3u)=\frac{d}{dx}(e^{-3x}u).$$ Thus $e^{-3x}u$ is constant: $u=Ae^{3x}$ for some constant $A$.

Likewise, if $v=y'-3y$ then $v=Be^{2x}$. Then $y=u-v=Ae^{3x}-Be^{2x}$.