Register to reply

2nd order Linear DE

by Apteronotus
Tags: linear, order
Share this thread:
Apteronotus
#1
Jan17-13, 12:28 PM
P: 203
Hi,

When solving a 2nd order Linear DE with constant coefficients ([itex]ay''+by'+cy=0[/itex]) we are told to look for solutions of the form [itex]y=e^{rt}[/itex] and then the solution (if we have 2 distinct roots of the characteristic) is given by
[itex]y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}[/itex]

This is clearly a solution, but how do we know there are no other solutions?
That is, how do we know this is the general solution?
Phys.Org News Partner Science news on Phys.org
Apple to unveil 'iWatch' on September 9
NASA deep-space rocket, SLS, to launch in 2018
Study examines 13,000-year-old nanodiamonds from multiple locations across three continents
tiny-tim
#2
Jan17-13, 02:33 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi Apteronotus!
Quote Quote by Apteronotus View Post
how do we know there are no other solutions?
It's easy to prove for the first-order case

if y' - ry = 0, put y = zert, then (z' + rz)ert = rzert

so ert = 0 (which is impossible),

or z' + rz = rz, ie z' = 0, ie z is constant
and now try (y' - ry)(y' - sy) = 0, using the same trick twice


Register to reply

Related Discussions
Difference between linear and non-linear first order DE Calculus & Beyond Homework 0
Help with solving a first order linear and first order non-linear Differential Equations 4
Second Order non-linear ODE Differential Equations 4
First order linear PDE Differential Equations 8
Reduction of order (2nd order linear ODE homogeneous ODE) Calculus & Beyond Homework 10