Show that the origin is a critical point (linear differential equation)

[JJBladester]

1. The problem statement, all variables and given/known data

Show that the origin (0,0) is a critical point. Write the linear differential equation in operator format and solve.

2. Relevant equations

x'' + 10x' + 25x = 0

3. The attempt at a solution

I am not sure how to show that the origin is a critical point (without using a graph).

As for solving the 2nd-order linear differential equation, here's what I did:

x'' + 10x' + 25x = 0

Auxillary equation: m2 + 10m + 25 = 0
Roots: m1 = m2 = -5

General solution: x = c1e-5x + c2xe-5x

I believe my general solution is correct, but am not sure if I solved it by "operator method".

[tiny-tim]

Hi JJBladester! :wink:
Show that the origin (0,0) is a critical point. Write the linear differential equation in operator format and solve.

x'' + 10x' + 25x = 0
…
General solution: x = c1e-5x + c2xe-5x

I believe my general solution is correct, but am not sure if I solved it by "operator method".

Yes, that's the correct general solution :smile: …

I think by operator format they just mean D2 + 10D + 25 = 0 :wink:

But I don't quite understand what is meant by (0,0) :confused: …

does that mean that there is an initial condition of x' = 0 at x = 0?

if so, (0,0) is a critical point just by looking at x'' + 10x' + 25x = 0 … you don't need to solve it. :wink:

[JJBladester]

I don't quite understand what is meant by (0,0) :confused: …

does that mean that there is an initial condition of x' = 0 at x = 0?

if so, (0,0) is a critical point just by looking at x'' + 10x' + 25x = 0 … you don't need to solve it. :wink:

My professor said I should:

"Change the 2nd order DEQ into a system of first order DEQ's.
Basically...Let y=dx/dt and then substitute.... When you get these two equations, set both derivatives equal to zero and solve for the x and y value that solves the system."

I'm still a little confused what that means.

Here is my attempt:

Let y = x', y'=x''

So x'' + 10x' + 25x = 0 becomes y' + 10y + 25x = 0.

-10y - 25x = 0
y = 0
-25x = 0
x = 0

[tiny-tim]

hmm … I've no idea what he means :redface: …

unless maybe he means put y = x' + 5x …

then y' + 5y = 0 …

dunno … as I said in my last post, I don't really understand the question :confused:

[HallsofIvy]

Let y= x'. Then x'' + 10x' + 25x = 0 becomes y'+ 10y+ 25x= 0 and we have the system of equations x'= y, y'= -10x-25y. A "critical point" is (x,y) such that x'= 0 and y'= 0.

[tiny-tim]

Let y= x'. Then x'' + 10x' + 25x = 0 becomes y'+ 10y+ 25x= 0 and we have the system of equations x'= y, y'= -10x-25y. A "critical point" is (x,y) such that x'= 0 and y'= 0.

why can't we just put x'' = x' = 0 in the original equation, then, and forget about y? :confused: