Can someone show me a proof of the general solution for 2nd order homogenous linear differential equations?

Question

I thought I had it figured out but there's a sort of 'leap of faith' at a pivotal point that annoys me. Can someone show me how to derive the general solution of an equation such as:

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$

I want to avoid just saying, "let's assume the solution is of the form $y=e^{mx}$". I want to see that form jumping out of an algebraic explanation.

I know I'll get

$y=Ae^{\alpha x}+Be^{\beta x}$

when I have 2 distinct roots, but explaining the general form for when I have two equal roots requires me to use this 'leap of faith' part.

Answer 1

Let $u$ and $v$ be the roots of $x^2 + bx + c =0$. Then $$\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0 = \frac{d^2y}{dx^2}-(u+v)\frac{dy}{dx}+uvy =\frac{d}{dx}\left(\frac{dy}{dx} - uy\right) - v\left(\frac{dy}{dx} - uy\right).$$

Let $\frac{dy}{dx} - uy = z (x)$. Then $$\frac{dz}{dx} - vz=0 \\ \implies\frac{dz}{dx} = vz \\ \implies z(x) = C\cdot e^{vt}.$$

Now, you have $$\frac{dy}{dx} - uy = C\cdot e^{vt}.$$ To solve this, use the integrating factor trick:

$$e^{-ux} \frac{dy}{dx} - u\cdot e^{-ux}y = C\cdot e^{(v-u)x} = \frac{d}{dx}\left(y\cdot e^{-ux}\right).$$

Integrate both sides: $$y\cdot e^{-ux} = \int C\cdot e^{(v-u)x} dx = C_1\cdot e^{(v-u)t}$$ (why?)

To finally get $$y(x) = C_1 \cdot e^{vx}.$$ But remember that we could have changed $u$ by $v$, and this would be a linear combination of the solutions (still a solution.) Then

$$ y(x) = C_1 \cdot e^{vx} + C_2 \cdot e^{ux}.$$

Edit. As Pragabhava suggested, here's the case when $u = v$:

Suppose that we have already gotten $$C\cdot e^{(v-u)x} = \frac{d}{dx}\left(y\cdot e^{-ux}\right).$$ Then $$C\cdot e^{0} = \frac{d}{dx}\left(y\cdot e^{-ux}\right) = C.$$ Integrate both sides to get $$y\cdot e^{-ux} = C\cdot x + C_1 \\ \implies y(x) = C \cdot e^{ux} + C_1 \cdot x\cdot e^{ux},$$ as desired.

Answer 2

One way to attack this is with the Exponential Shift Theorem.

For any polynomial $P$, constant $k$ and smooth function $u(x)$, $P(D) (\exp(k x) u) = \exp(k x) P(D+k) u$

Here $D$ is the differentiation operator $\dfrac{d}{dx}$. Thus e.g. for the polynomial $P(t) = t^2 + 2 t + 3$, $P(D) u = \dfrac{d^2 u}{dx^2} + 2 \dfrac{du}{dx} + 3 u$.

Now finding one nontrivial solution of the differential equation $P(D) y = 0$ would be easy if the constant term of the polynomial was $0$: $y=1$ would be a solution. But $P(t+k)$ has constant term $0$ (as a polynomial in $t$) if and only if $P(k) = 0$. Thus you look for roots of the polynomial. If $P(k) = 0$, then $P(D+k) 1 = 0$, and by Exponential Shift $P(D) \exp(k x) = 0$. Each distinct root of the polynomial $P$ gives you a solution; it is easy to check that these are linearly independent, and if the polynomial has all distinct roots you have a fundamental set of solutions.

If the roots are not all distinct, Exponential Shift can be applied again. Thus if $k$ is a root of $P$ with multiplicity $m$, the terms of $P(t+k)$ in $t^j$ for $0 \le j \le m-1$ are all $0$. Then $P(D+k) u$ involves only the $m$'th and higher derivatives of $u$, which means that it is $0$ if $u$ is a polynomial of degree less than $m$. So we get solutions $1, x, \ldots, x^{m-1}$ of $P(D+k) u = 0$ and by Exponential Shift solutions $\exp(kx), x \exp(kx), \ldots, x^{m-1} \exp(kx)$ of $P(D) y = 0$.