Where does the general solution for second order linear ODEs come from?

In summary, the general solution to a linear differential equation with constant coefficients can be found by solving for y in terms of e^mx, where x is a multiple root of the characteristic equation.
  • #1
sandy.bridge
798
1
If
[tex]ay+b\int^y_0ydy+cy'=0[/tex]
then
[tex]ay'+by+cy''=0[/tex]
now, let
[tex]y=e^{sx}[/tex]
thus,
[tex]s^2+a/cs+b/c=0[/tex]
and then one solves for s. It is then plugged into what sources are deeming a "general solution"
[tex]y=C_1e^{s_1x}+C_2e^{s_2x}[/tex]
however, none of these texbooks explain or derive where this comes from, and I have not taken any linear algebra or differential equations. Can someone explain where [tex]y=C_1e^{s_1x}+C_2e^{s_2x}[/tex] comes from?
Thanks!
 
Physics news on Phys.org
  • #2
sandy.bridge said:
If
[tex]ay+b\int^y_0ydy+cy'=0[/tex]
then
[tex]ay'+by+cy''=0[/tex]
now, let
[tex]y=e^{sx}[/tex]
thus,
[tex]s^2+a/cs+b/c=0[/tex]
and then one solves for s. It is then plugged into what sources are deeming a "general solution"
[tex]y=C_1e^{s_1x}+C_2e^{s_2x}[/tex]
however, none of these texbooks explain or derive where this comes from, and I have not taken any linear algebra or differential equations. Can someone explain where [tex]y=C_1e^{s_1x}+C_2e^{s_2x}[/tex] comes from?
Thanks!

Hey sandy.bridge.

Lets take it from the step if y = e^mx, y' = me^mx, y'' = m^2 e^mx then you end up getting an equation of the form:

(am^2 + bm + c) x e^(mx) = 0. Now divide by e^(mx) which gives you the quadratic term:

am^2 + bm + c = 0.

Now if you have two solutions for m, what does that imply for the final solution?

The real 'guts' of proving this general property for linear ODE's or systems of ODE's is based on the superposition principle. The idea uses linear algebra, in particular spectral theory to show the solution in terms of superpositions. If you want to prove it yourself, you need to setup the ODE in terms of a matrix system, and then do some eigen-analysis to decompose the matrix.

I didn't have to do any proving when I took my prior DE course, but if you want to understand why you get this kind of formula in a general sense, look at the superposition principle with reference to ODE's and a detailed proof of it.

Also I didn't link to Wikipedia since they are having a blackout for SOPA awareness.
 
  • #3
Note that while "trying" [itex]e^{rx}[/itex] will reduce a linear differential equation with constant coefficients to its "characteristic equation", a polynomial, that does not necessarily mean the solution will involve exponentials.

If the solution to the characteristic equation is a complex number, a+ bi, then (assuming all coefficients are real) a- bi is also a root and [itex]e^{(a+ bi)x}= e^{ax}\left(cos(bx)+ sin(bx)\right)[/itex]. Also, if r is a multiple root (If the differential equation is y'''+ 3y''+ 3y'+ y= 0, the characteristic equation is [itex]r^3+ 3r^2+ 3r+ 1= (r+1)^3= 0[/itex] which has r= -1 as a triple root. The general solution to the differential equation is [tex]y= Ae^{-x}+ Bxe^{-x}+ Dx^2e^{-x}[/tex].
 

What is a second order linear ODE?

A second order linear ODE is a type of differential equation that involves a second derivative of a dependent variable and can be expressed in the form of a linear combination of the dependent variable, its first derivative, and its second derivative.

What is the general form of a second order linear ODE?

The general form of a second order linear ODE is: y'' + p(x)y' + q(x)y = f(x), where y is the dependent variable, p(x) and q(x) are functions of the independent variable x, and f(x) is a function of x.

What are the solutions to a second order linear ODE?

The solutions to a second order linear ODE depend on the types of roots of the characteristic equation. If the roots are real and distinct, the solution is expressed as y(x) = c1e^(r1x) + c2e^(r2x). If the roots are complex, the solution is y(x) = e^(ax)(c1cos(bx) + c2sin(bx)). If the roots are equal, the solution is y(x) = c1e^(rx) + c2xe^(rx).

What is the method of undetermined coefficients?

The method of undetermined coefficients is a technique used to find a particular solution to a second order linear ODE when the non-homogeneous term, f(x), is a polynomial, exponential, sine, or cosine function. It involves guessing a particular solution based on the form of f(x) and then solving for the undetermined coefficients.

What is the method of variation of parameters?

The method of variation of parameters is a technique used to find a general solution to a second order linear ODE when the non-homogeneous term, f(x), is a general function. It involves finding two linearly independent solutions to the homogeneous equation and then using them to construct a particular solution using integrals.

Similar threads

  • Differential Equations
Replies
7
Views
1K
Replies
2
Views
2K
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
2
Views
1K
Replies
2
Views
2K
Replies
3
Views
738
  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
4
Views
2K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
3
Views
2K
Back
Top