I am facing some difficulties understanding the difference between a general solution and the most general solution of a 2nd order ODE
for example for homogeneous constant coefficients case with 2 real roots, is $y=C_1 e^{r_1x}+C_2 e^{r_2x}$ a general solution or the most general solution?
and how can I prove that a general solution is the most general one?
$\endgroup$1 Answer
$\begingroup$There are many kinds of 2nd Order ODEs. Some are (In)homogeneous, some are (Non-)linear. So this question is kind of vague on what you want as a general solution.
Note: I mostly use $\mathbb{R}$ for constants as i'm not always sure whether i can use $\mathbb{C}$ but generally it should work too. All functions here are of the form $y(x)$. Every equation here for which i provide solutions is second order here. Inhomogeneous means the same as non-homogeneous.
The 'most general' ODE
$$ y''(x)= F(x,y,y') $$
has no general solutions as non-linear equations may never be solved generally. Some can be solved but there is no general rule for it.
The Linear Inhomogeneous ODE((independent) variable coefficient)
$$ (ax^{k}+b)y''+ (cx^{l}+d)y' + (ex^m+g)y = F(x) $$
where $a,b,c,d,e,g \in \mathbb{R} $ and $ k,l,m \in \mathbb{R} $may or may not have a solution, involves some sort trick to be solved usually and even if there is a solution it's not given like you classically learnt.
When $k=2, l=1, m=2$ and $a=1, b=0, c=1, d=0, e=1, g=-\alpha^2 \hspace{4mm} \alpha \in \mathbb{C}$ and $F(x)=0$you get the Bessel-equation
$$ x^2y''+xy'+(x^2-\alpha)y=0 $$
whose solution cannot be given with elementary functions and you need a special series with the gamma function to calculate specific values of it.
The Bessel functions of the first kind being $J_\alpha$ are defined as such:
$$ J_{\alpha}(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!\Gamma(n+\alpha+1}\left(\dfrac{x}{2}\right)^{2m+\alpha},$$
where $\Gamma(x)$ is the Gamma-function.
A more 'regular' ODE would be the (Cauchy-)Euler equation whose general form is:
$$ a_nx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + a_{n-2}x^{n-2}y^{(n-2)} + ... +a_0y=0$$
Here the (n) in the superscript means the n-th order derivative.
The second order version of it let's say is in the form:
$$ ax^2y''+ bxy'+cy=0$$
where $a,b,c \in \mathbb{R} $
The general solution for this kind of equation is reached like this:
First substitute $y=x^m,\;\; y'=mx^{m-1},\;\; y''=m(m-1)x^{m-2} $
Then the transformed equation will look like this:
$$ax^2\cdot m(m-1)x^{m-2} + bx\cdot mx^{m-1} + c x^m =0$$
Notice how the x powers fall out and we can factor out $x^m$.
Thus we get: $$x^m[am(m-1)+bm+c]=0 $$
Divide by $x^m$ which can never be $0$ and we arrive to this simple problem:
$$am^2-(a+b)m+c=0$$
Solve for $m$ and plug your results into this: $$y = c_1x^m_1 + c_2x^m_2$$
If there is a resonance($m_1=m_2$) we multiply our results by $\ln(x)$ so they will be linearly independent(take note how this is similar when this happens with a constant coefficient ODE)
$$ y= c_1x^m\ln(x) + c_2x^m $$
If our solutions are of the complex form $ \alpha \pm \beta i$
$$ y=c_1x^\alpha \cos(\beta\ln(x)) + c_2x^\alpha \sin(\beta\ln(x)) $$
If F(x) is not zero then you need to find the particular solution first, you can find more of this at the bottom of this answer.
Constant Coefficient Linear Homogeneous ODE
$$ ay''+by'+cy=0 $$
where $a, b, c \in \mathbb{C} $
We transform this into an algebraic problem:
$$ ar^2+br+c=0 $$
Solve it for r, get your results in $\alpha$ and $\beta$
the solution of this is as you wrote:
$$ y= c_1e^{\alpha x}+c_2 e^{~beta x}$$
If there is a resonance($\alpha=\beta$)
$$ y= (c_1+c_2x)e^{\alpha x / 2} $$
If the two roots are of the complex form $\alpha \pm \beta i$
$$ y= e^{\alpha x}(c_1\cos(\beta x)+c_2\sin(\beta x)) $$
Constant Coefficient Linear Inhomogeneous ODE
$$ ay''+by'+cy=F(x) $$
When $F(x) \neq 0$, you solve the inhomogeneous term $F(x)$ with Variation of Parameters or Method of undetermined coefficients (some schools also teach the Annihilator Method), then you add this to the homogeneous general solution thus getting the whole inhomogeneous general solution for that kind of problem.
$$ y_{hom.gen.}+y_{inhom.part.}=y_{inhom.gen.} $$
These are the ones that i had the top off my head. I am not really sure what you could prove about this. Maybe you could (re)formulate your question a bit better.
$\endgroup$
