Linear homogenous ODEs with constant coefficients

1. Nov 12, 2012

d.arbitman

Given the ODE of the form:
y''(x) + A*y'(x) + B*y(x) = 0

If we choose a solution such that y(x) = e$^{mx}$
and plug it into the original ODE, the ODE becomes:
(m$^{2}$ + A*m + B)e$^{mx}$ = 0

If we solve for the roots of the characteristic equation such that
m = r$_{1}$, r$_{2}$ (root 1 and root 2, respectively)

The solution to the ODE would have the form:
y(x) = c*e$^{r_{1}*x}$ + d*e$^{r_{2}*x}$, where c and d are constants

My question is, why are the constants where they are in the solution? In other words, why are they multiplying y$_{1}$ & y$_{2}$, where
y$_{1}$ = e$^{r_{1}*x}$ and y$_{2}$ = e$^{r_{2}*x}$ ?

Why are the constants not just added to y(x), such that the solution to the ODE would be as follows, where k is a constant.
y(x) = e$^{r_{1}*x}$ + e$^{r_{2}*x}$ + k

This question mainly is for second order and higher differential equations. I understand how it works for first order linear homogenous DEs because the constant is simply the constant of integration, but I am having trouble understanding how it applies to higher orders.

Last edited: Nov 12, 2012
2. Nov 12, 2012

lurflurf

That is because the differential equation is linear. That is if L[y]=0 is a linear differential equation and u and v are any two solutions so that L=L[v]=0 then L[a u+a v]=0.

3. Nov 13, 2012

HallsofIvy

Staff Emeritus
The fundamental theorem for such equations is:
"The set of all solutions to a linear homogeneous differential equation of order n form a vector space of dimension n"

That means that if we can find a set of n independent solutions, a basis for that vector space of solutions, any solution can be written as a linear combination of those solutions. And a "linear combination" means a sum of the functions multiplied by constants.

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