Question

Use a characteristic equation to find the particular solution of
\[\large \frac{ d^{2} x}{ dt^{2} }-4\frac{ dx }{ dt }-32x=0\]
given that x(0)=5 and x'(0)=16

Answer 1

so far worked out that the homogeneous comes to\[\huge Ae^{-4x}+Be^{8x}\]
Can someone please check that and help me with the rest?

Answer 2

You seem familiar, have we encountered each other on here before?

Answer 3

hahahha sarcasm? :P

Answer 4

Decide for yourself.
Right now however, we are faced with a rather daunting differential equations problem.
Well, YOU'RE faced with it, I may well be only here to watch you... >:)

Answer 5

well I'm hoping you remember me :O
yes rather daunting :/
that is true :(

Answer 6

Well, I remember people based on feats... you managed a fiasco with Pan involving complex roots... am I right?

Answer 7

ah yes you are very right!

Answer 8

Now this appears to be nothing more than an initial value problem, are you really daunted by this? :D

Answer 9

I have a large disliking for differential equations requiring this :(

Answer 10

I see. What was your name again?

Answer 11

Ray

Answer 12

Oh.. right, silly me :)
TJ. Terry. Terence.
Pleasure's mine.

Now, let's get to work...
I want to make sure you knew what you were doing here, would you mind telling me how you arrived at that general solution you posted immediately after your question?

Answer 13

I remember you saying to call you TJ.! :)
Glad to be talking!
well first I went to find the homogeneous solution by solving the first eqution;

Answer 14

equation*

Answer 15

Go on...

Answer 16

\[\large \lambda^{2}-4\lambda-32=0\] then it can be found that \[\large \lambda =-4\] or \[\large \lambda =8\]

Answer 17

then I looked at the table and the example I have is similar, using the homogeneous solution as \[\large Ae^{\alpha*x}+Be^{\beta*x}\]

Answer 18

I see. Well I can tell you that the general solution you posted is incorrect, however :)

Answer 19

Much as I'd like to see you flail about in search of an error, even I'm not THAT sadistic.
The error was minor, but an error nonetheless:
\[\huge x(t)=Ae^{-4\color{red}t}+Be^{8\color{red}t}\]

Answer 20

I see :/ could you please walk me through where I went wrong? I'm not to bright with using the table for this :(

Answer 21

Relax. Minor error. Refer to previous post ^

Answer 22

Oh!!! Right! Of course! thank you :)

Answer 23

I didn't see that myself

Answer 24

Now, to the "hard" part, the initial values.
First, the initial values not only involve x(t) but also x'(t)

Care to find x'(t) ?

Answer 25

this is where I seem to get lost, in previous questions, the right hand side had another expression

Answer 26

this one has a zero

Answer 27

Again, relax. It's not nearly as hard as you think.
\[\huge x(t)=Ae^{-4t}+Be^{8t}\]\[\huge x'(t) = \frac{d}{dt}x(t)\]
in other words, just differentiate the bloody solution XD

Answer 28

ohhh differentiate the homogeneous equation! :O

Answer 29

Oh, the sound of realisation... I get that often ;)

Answer 30

I am very well aware you do! :P
\[\large 8*Be^{8t}-4*Ae^{-4t}\]

Answer 31

Now, let me recap for you, ok?
\[\Large x(t) = Ae^{-4t}+Be^{8t}\]\[\Large x'(t) = 8Be^{8t}-4Ae^{-4t}\]

catch me so far?

Answer 32

yes I sure do

Answer 33

Good. This entire thing is just a fancy way to present a system of linear equations.
I trust you'd have no problem with that?
....
I didn't think you would.

NOW.
x(0) = 5
How do you put that to use?

Answer 34

oh I see, so now it's just the simple use of:
\[\large 5=Ae^{-4*0}+Be^{8*0}\]

Answer 35

\[\large 5=Ae^{0}+Be^{0}\]

Answer 36

And even simpler,
A + B = 5
Good.
Now put x'(0) = 16 to use.

Answer 37

\[\large -4A+8B=16\] also \[\large -4\]

Answer 38

don't worry about the second negative four

Answer 39

A+B = 5
-4A + 8B = 16

Hardly daunting anymore, aye?

Answer 40

not at all anymore!
solve simultaneously :D

Answer 41

I shall wait.

Answer 42

work out A and B and then substitute them into the equation! :D
\[\large 2e^{-4t}+3e^{8t}\]

Answer 43

That wasn't too bad, now was it?

Answer 44

It made so much sense when you helped me see that it's a simple question just differently worded, thank you !!! :D

Answer 45

I am...
TERENZ

LOL


Signing off for now ^_^
---------------------------
Terence out

Answer 46

Ter-iffic!!