• Support PF! Buy your school textbooks, materials and every day products Here!

Solving general linear ODE's

  • Thread starter foxjwill
  • Start date
  • #1
354
0

Homework Statement



Is there a standard method for finding the general solution to any given linear homogeneous ODE?

I tried working it out myself. Here's what I tried. I'm almost positive it's wrong because it doesn't take into account most of the coefficients. But maybe I'm on the right track?

The Attempt at a Solution


[tex]a_0(x)y + a_1(x)y' + \cdots + a_{n-1}(x)y^{(n-1)} + y^{(n)} = b(x)\qquad\qquad\qquad(1)[/tex]​

Now, recall that the product rule for derivatives of order [tex]n[/tex] is

[tex]\frac{d^n}{dx^n}[f(x)g(x)] = {\sum^n_{k=0} \binom{n}{k} f^{(k)}(x)g^{(n-k)}}(x).[/tex]​

We want to find some function [tex]u(x)[/tex] that, when multiplied to both sides of [tex](1)[/tex], will put the left side in the form of the [tex]n[/tex]th degree product rule. In other words,

[tex]u(x)a_0(x)y + u(x)a_1(x)y' + \cdots + u(x)y^{(n)} = \frac{d^n}{dx^n}[u(x)y]\qquad\qquad\qquad(2)[/tex]​

After expanding [tex](2)[/tex], we can set each term of the left side equal to the corresponding term on the right (e.g. [tex]u(x)a_2(x) y' = u^{(n-1)}(x)y'[/tex] ) and, after canceling out the all the [tex]y[/tex]'s, form a system. Also note that since [tex]u(x)y^{(n)}[/tex] appears on both sides of [tex](2)[/tex] it cancels out. From here until I say otherwise, all functions will be referenced, for convenience, without the "[tex](x)[/tex]" appended to it.

[tex]
\left \{
\begin{array}{lcl}
u^{(n)} &=& u a_0\\
u^{(n-1)} &=& \dbinom{n}{1} u a_1\\
& \vdots\\
u'' &=& \dbinom{n}{n-2} u a_{n-2}\\
u' &=& \dbinom{n}{n-1} u a_{n-1}\\
\end{array}
\right .
\qquad\qquad\qquad(3)[/tex]​


Rearranging [tex](3)[/tex], we have

[tex]
\left \{
\begin{array}{lcl}
\dfrac{u^{n}}{u^{(n-1)}} &=& \dfrac{a_1}{a_0} \\
\\
\dfrac{u^{(n-1)}}{u^{(n-2)}} &=& \dbinom{n}{1} \dfrac{a_2}{a_1}\\
& \vdots\\
\dfrac{u'''}{u''} &=& \dbinom{n}{n-2} \dfrac{a_{n-2}}{a_{n-3}}\\
\\
\dfrac{u''}{u'} &=& \dbinom{n}{n-1} \dfrac{a_{n-1}}{a_{n-2}}\\
\end{array}
\right .
\qquad\qquad\qquad(4)[/tex]​


Solving each equation in [tex](4)[/tex], we get

[tex]u^{(k)} = e^{\binom{n}{n-k} {\int \frac{a_{n-k}}{a_{n-(k+1)}}dx }}[/tex].​


Setting [tex]k=0[/tex],

[tex]u(x) = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}\qquad\qquad\qquad(5)[/tex]​


Now, going back to [tex](1)[/tex] and [tex](2)[/tex], since, by definition,

[tex]\frac{d^n}{dx^n}[u(x)y] = u(x)b(x),[/tex]​

we can plug in [tex](5)[/tex],
[tex]\frac{d^n}{dx^n}\left[ e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx} y \right] = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}b(x)\qquad\qquad\qquad(6)[/tex]​

and then antidifferentiate, which, at the moment, I am feeling too lazy to actually do.
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?
 
  • #3
354
0
Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?
Yes, but I was wondering about methods to solve general degree ODE's with variable coefficients.
 
  • #4
You have introduced one function [tex]u(x)[/tex] and you also have n equations in equation (3).

In general your solution for [tex]k=0[/tex], i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no [tex]u(x)[/tex] in general.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Series solutions (including Frobenius' method) are the most general methods of solving linear equations.
 
  • #6
354
0
You have introduced one function [tex]u(x)[/tex] and you also have n equations in equation (3).

In general your solution for [tex]k=0[/tex], i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no [tex]u(x)[/tex] in general.
Yeah. I had a feeling it was something like that.
 
  • #7
354
0
Series solutions (including Frobenius' method) are the most general methods of solving linear equations.
Hm. Darn. I don't like using series solutions. They're so cumbersome. Ah, well.

Also, I'm assuming the laplace and fourier transforms, in general, can't be used to solve linear ODE's with variable coefficients? I know that there are a few specific cases where they work, but, yeah.
 

Related Threads on Solving general linear ODE's

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
746
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
5
Views
1K
Replies
7
Views
2K
Replies
3
Views
1K
Top