Need explanation on 2nd DE

Dec 2015
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169
Earth
Cannot find explanation for this :
Why \(\displaystyle e^{rx} \) satisfies the equation \(\displaystyle y’’+py’+qy=0\) ? p,q-constants.
How to prove it and how to derive it , where it comes from ?

My thoughts are to substitute for \(\displaystyle y=\sum_{}^{\infty} c_n x^{n}\) and see the results.
Method required !
 

v8archie

Math Team
Dec 2013
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2,682
Colombia
One way is to consider the case where $p=0$. That is $y'' + ay = 0$. This can be solved by inspection in the three cases $q < 0$, $q = 0$ and $q > 0$ and it turns out that $$y=e^{\pm\sqrt{q}}$$
are the solutions.

Now, when $p \ne 0$ we solve $$y'' + py' + qy = 0$$
\begin{equation}\text{let} \quad \left\{
\begin{aligned}
y &= ve^{ax} \\
y' &= (av + v')e^{ax} \\
y'' &= ae^{ax}(av + v') + e^{ax}(av' + v'') \\
&= (a^2v + 2av' + v'')e^{ax}
\end{aligned}
\right\} \quad \text{so that}\end{equation}
\begin{align}
(a^2v + 2av' + v'')e^{ax} + p(av + v')e^{ax} + qve^{ax} &= 0 \\
(a^2+ap+q)ve^{ax} + (2a+p)v'e^{ax} + v''e^{ax}) &= 0
\end{align}
And now, selecting $p=-2a$ we find that $v(x)$ satisfies the equation
\begin{equation}
v'' + \left(q - \frac{p^2}4\right)v = 0
\end{equation}
Which brings us back to the simpler equation with no term in $v'$. That is that $$v = e^{\pm\frac12x\sqrt{p^2-4q}}$$
This means that $$y = e^{ax}e^{\pm\frac12x\sqrt{p^2-4q}}$$

And we are thus left with $e^{rx}$ being a solution to $y'' + py' + qy = 0$ where $$r = \frac{-p \pm \sqrt{p^2 - 4q}}{2}$$
 
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v8archie

Math Team
Dec 2013
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You could also simply plug $y=e^{rx}$ into the equation and discover that, since $e^{rx} \ne 0$ you must have $r^2 + pr + q = 0$ for a solution.
 
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Dec 2015
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\(\displaystyle z'=zp\) then \(\displaystyle y'z+pyz=0\) or \(\displaystyle y'z+yz'=(yz)'=0\).
 
Jun 2019
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Any linear ODE is going to have derivatives proportional to the function (or linear combinations of the function and other derivatives), and the solution is always going to be of the form $y(x) \propto e^{\lambda x}$. So much so that I can't actually remember any of the other techniques for solving ODEs or PDEs analytically, other than assuming this form and plugging it into the differential equation.
 
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SDK

Sep 2016
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USA
At the risk of offending someone, I'm going to disagree with most of the other responses so far since they seem to amount to "just plug it into the ODE and you can see it works". This is true but is not an explanation.

I have never actually seen an intuitive explanation of this which relies on examining the scalar ODE/solution. I'll even be more controversial and say I strongly doubt such an intuitive explanation exists. Since I have strong opinions about the way that linear ODEs are introduced in most undergraduate courses (poorly imo) I'll take a crack at this in what I think is the "right" way. If anyone disagrees however I'd be very happy to hear another point of view.

First, I'm going to claim that the equation $y'' + py' + qy = 0$ is never a useful equation to study. The reason I say this is because this is really a linear equation defined on $\mathbb{R}^2$ so understanding the equation and its solution requires using linear algebra. Working with high order scalar equations completely obscures everything that is going on. So lets start by writing the "correct" equation: $x' = Ax$ where $x = (x_1, x_2) \in \mathbb{R}^2$ is a vector and
\[
A =
\begin{pmatrix}
0 & 1 \\
-q & -p
\end{pmatrix}
\]
In fact, this is completely general. Every linear $n^{\rm th}$ order ODE can be equivalently written as a vector equation of the form $x' = Ax$ where $A$ is an $n \times n$ matrix.

I won't list all the reasons why this is a better way to think about linear ODEs than high order scalar equations. But let me explain the reason this makes understanding the solutions easier. It is based on the following theorem.

Theorem: Suppose $A$ is an $n \times n$ matrix. Then the solution to the linear ODE, $x' = Ax$ is given explicitly by the formula $x(t) = e^{tA} x_0 $ where $e^{tA}$ is the matrix exponential defined by
\[
e^{tA} = \sum_{k = 0}^\infty \frac{(tA)^k}{k!}
\]
This is already amazing I think since solving nothing more than this equation is what takes up essentially an entire semester long undergraduate ODE course and the solution is just a single line.

Now, I'll finally get to the justification for the solutions being nice exponential functions whose parameters are roots of some polynomial. Suppose $A$ is diagonalizable (this assumption is not necessary. I take it here only to avoid talking about generalized eigenvectors) with eigenvalues $\{\lambda_1, \dotsc,\lambda_n\}$ and eigenvectors $\{\xi_1,\dotsc,\xi_n\}$. Lets look at what happens when you solve the ODE with an eigenvector, say $\xi_i$ for the initial condition. By the above formula, the solution is
\[ x(t) = e^{tA}\xi_i = \sum_{k = 0}^\infty \frac{(tA)^k}{k!}\xi_i = \sum_{k = 0}^\infty \frac{t^k}{k!}A^k \xi_i = \sum_{k = 0}^\infty \frac{(\lambda_i t)^k}{k!}\xi_i = e^{\lambda_i t} \xi_i.\]
What this computation shows is that eigenvectors are "straight line" solutions to the ODE since if $x(0)$ is an eigenvector, then $x(t)$ is an eigenvector for all $t \in \mathbb{R}$. To put it another way, eigenvectors are the solutions for which the flow of the ODE acts by simply scaling the initial condition by a real number. Moreover, since $\{\xi_1, \dotsc, \xi_n\}$ form a basis for $\mathbb{R}^n$ by assumption, you can easily show they form a basis for the solution space of the ODE.

You easily recover the scalar solution of the $n^{\rm th}$ order ODE by recalling that the scalar variable is given simply by $x_1(t)$. Its also not hard to show using the matrix exponential formula above that the polynomial whose roots you computed to arrive at the scalar solution is just the characteristic polynomial for $A$. So its roots are the eigenvalues of $A$ which explains why they appear in the solution. They appear because they are actually eigenvalues!

This linear algebra approach also provides insight into a lot of other familiar properties of linear ODEs.

tldr; IMHO, an intuitive explanation for the formula obtained for higher order scalar linear ODEs is based on the fact that eigenvectors are special solutions of the ODE which can be seen once it is rewritten "correctly".
 
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