# Second Order Linear Equations? help

Find the solution of the given initial value problem. Sketch the graph of the solution and describe its behaviour as t increases?

$y'' + 4y' +3y = 0, y(0) = 3, y'(0)= -1$

I mean describing the behaviour as $t$ increases to infinity is no problem, nor is solving the initial value problem, The problem here is that I have no idea how to sketch something like that? any tips/advice? the only thing I can think of is to plot in values from $t$??

by the way the solution of the initial value problem is

$y=4e^{-1t} -e^{-3t}$ so I guess the question is how do I sketch that?

• Plotting the solution on wolframalpha or a graphing calculator might be a good start! – David H Mar 20 '14 at 22:08
• im not allowed to do that :O – gger234 Mar 20 '14 at 22:09
• well I guess since when t increases my equation goes to zero so, I guess it never intersect the x axis? that is something? maybe? – gger234 Mar 20 '14 at 22:13
• This is the difference of two exponentials. Interesting questions are "at what height does it start" and "where does the slope of the positive one exactly balance the slope of the negative one"? If relevant, "where are its critical points and are they maxima/minima/neither"? – Eric Towers Mar 20 '14 at 22:16
• Maybe start by finding where the first and second derivatives are zero? – Mike Mar 20 '14 at 22:16

An exponential of the form $e^{-t/\tau}$ has a so called characteristic time $\tau$. That is the the time in which the amplitude decreases to $1/e$ of its original value. After that it is considered less significant (in physics).

In your function the term $-e^{-3t}$ dominates (trying to slope up with slope +3) at first, since it has the shortest characteristic time. After that the term $4e^{-1t}$ dominates (sloping down with slope -4). Either way, you start off with a derivative of y'(0)=-1, meaning it slopes down with slope -1.

Good points to select for your graph are:

• $t=0$,
• $t=\frac 1 3$, which is the first characteristic time,
• $t=1$, which is the second characteristic time,
• $t=2$, which is just one more point to see how the function develops.

That should be enough to draw the graph.

• yeah I guess, I didnt understand much of the explanation for why you choose points like that would be great, my math is not that good neither is my english :P – gger234 Mar 20 '14 at 22:26
• At $t=\frac 1 3$ the term $e^{-3t}$ decreases to $\frac 1 e \approx \frac 1 3$ of its value at $t=0$, which is considered important. Similarly $e^{-t}$ decreases to $\frac 1 e$ of its original value at $t=1$. – Klaas van Aarsen Mar 20 '14 at 22:28
• hmm really? e^((-3)*(3)) = 1/e? – gger234 Mar 20 '14 at 22:30
• Already edited. – Klaas van Aarsen Mar 20 '14 at 22:31
• AHhh ok now I get it! thanks a lot man! and t=2 is just so see how the graph develops and and I guess t=0 is just standard? :P – gger234 Mar 20 '14 at 22:33