ODE homogeneous equations w/constant coefficients

Click For Summary
SUMMARY

The discussion focuses on solving the homogeneous second-order linear differential equation y'' + 3y' + 2y = 0. The characteristic equation is derived as r^2 + 3r + 2 = 0, yielding roots r_1 = -2 and r_2 = -1. The general solution is expressed as y(t) = c_1e^(-2t) + c_2e^(-t). Participants clarify that the order of terms in the solution does not affect its correctness, although some conventions prefer the slower-decreasing terms to be listed first.

PREREQUISITES
  • Understanding of differential equations, specifically homogeneous equations.
  • Familiarity with characteristic equations and their roots.
  • Knowledge of exponential functions and their properties.
  • Basic algebra skills for solving quadratic equations.
NEXT STEPS
  • Study the method of solving higher-order linear differential equations.
  • Explore conventions in writing solutions for differential equations.
  • Learn about the implications of root multiplicity in differential equations.
  • Investigate applications of homogeneous equations in physics and engineering.
USEFUL FOR

Students studying differential equations, educators teaching mathematical concepts, and anyone seeking to understand the conventions in presenting solutions to linear differential equations.

mmont012
Messages
39
Reaction score
0

Homework Statement


Find the general solution
y"+3y'+2y=0

Homework Equations


y(t) =c_1e^r_1t + c_2e^r_2t

The Attempt at a Solution


a=1 b=3 c=2
r^2+3r+2=0
(r+2)(r+1)=0
r_1=-2
r_2=-1

General solution: y(t) =c_1e^(-2t)+c_2e^(-t)I was wondering if the order mattered. The answer in the book is y(t)=c_1e^(-t)+c_2e^(-2t)

I don't think that it does, but I want to make sure before continuing and submitting my hw.
If the order DOES matter, how can I get the correct order?

Thank you.
 
Physics news on Phys.org
mmont012 said:

Homework Statement


Find the general solution
y"+3y'+2y=0

Homework Equations


y(t) =c_1e^r_1t + c_2e^r_2t

The Attempt at a Solution


a=1 b=3 c=2
r^2+3r+2=0
(r+2)(r+1)=0
r_1=-2
r_2=-1

General solution: y(t) =c_1e^(-2t)+c_2e^(-t)I was wondering if the order mattered. The answer in the book is y(t)=c_1e^(-t)+c_2e^(-2t)
No, the order doesn't matter
mmont012 said:
I don't think that it does, but I want to make sure before continuing and submitting my hw.
If the order DOES matter, how can I get the correct order?
By the way, what you wrote as a relevant equation confused me for a while.
c_1e^r_1t + c_2e^r_2t
For the two functions, are they ##e^{r_1}t## and ##e^{r_2}t## or are they ##e^{r_1t}## and ##e^{r_2t}##?
I know what you mean, but if you write them inline as you did, at the least use parentheses around the exponent, such as e^(r_1t).
 
Thank you, and sorry about the confusion. I'm using my phone and its acting up. Next time I'll be sure to add ().
 
mmont012 said:

Homework Statement


Find the general solution
y"+3y'+2y=0

Homework Equations


y(t) =c_1e^r_1t + c_2e^r_2t

The Attempt at a Solution


a=1 b=3 c=2
r^2+3r+2=0
(r+2)(r+1)=0
r_1=-2
r_2=-1

General solution: y(t) =c_1e^(-2t)+c_2e^(-t)I was wondering if the order mattered. The answer in the book is y(t)=c_1e^(-t)+c_2e^(-2t)

I don't think that it does, but I want to make sure before continuing and submitting my hw.
If the order DOES matter, how can I get the correct order?

Thank you.

Either order is correct because there are not really any rules about that.

However, sometimes authors adopt certain conventions, such as having the most slowly-decreasing functions first, followed by more rapidly-decreasing functions. In that convention, ##e^{-t}## decreases more slowly than ##e^{-2t}##, so would come first. In the opposite case of increasing functions, people sometimes want the most rapidly-increasing functions to come first, in part because they govern the asymptotic large-##t## behavior. So, if you had increasing functions ##e^{t}## and ##e^{2t}##, some people might write the ##e^{2t}## first.

However, as I said, there really are no rules, and not everyone subscribes to the type of conventions I have mentioned.
 

Similar threads

Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
1K