Given the ODE of the form:(adsbygoogle = window.adsbygoogle || []).push({});

y''(x) + A*y'(x) + B*y(x) = 0

If we choose a solution such that y(x) = e[itex]^{mx}[/itex]

and plug it into the original ODE, the ODE becomes:

(m[itex]^{2}[/itex] + A*m + B)e[itex]^{mx}[/itex] = 0

If we solve for the roots of the characteristic equation such that

m = r[itex]_{1}[/itex], r[itex]_{2}[/itex] (root 1 and root 2, respectively)

The solution to the ODE would have the form:

y(x) = c*e[itex]^{r_{1}*x}[/itex] + d*e[itex]^{r_{2}*x}[/itex], 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[itex]_{1}[/itex] & y[itex]_{2}[/itex], where

y[itex]_{1}[/itex] = e[itex]^{r_{1}*x}[/itex] and y[itex]_{2}[/itex] = e[itex]^{r_{2}*x}[/itex] ?

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[itex]^{r_{1}*x}[/itex] + e[itex]^{r_{2}*x}[/itex] + 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Linear homogenous ODEs with constant coefficients

**Physics Forums | Science Articles, Homework Help, Discussion**