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I could use a LaPlace transform right?

Yeah.. and it's pretty easy.

Ok the only problem I'm having is, all I have to use is the formula L(y')=sL(y)-y(0) correct?

Yeah.

ok so I'm having trouble figuring out how to incorperate x=0 and y=1

x=0 when y=1 means that y(0)=1

Oh.. duh.

so I would just rewrite...
\[s L(y)-y(0)+5L(y)=L(9)\]

OK.. what are the L(y) and L(9)?

Not 100% sure on L(y) but I know L(9)=9/s

Would it just be y/s ?

or no, I could solve for L(y)

I think I have it

Yeah.. you could do so, or you could write it as Y(s)

You know that you have to transform it back to the time domain.

L(y)=(9/s+1)/(s+5)

are you sure about the simplification you made?

oh I see it.. it's right.

I don't know how to do fractions notation using the equation thing or I would. :)

I think I need to grab a pen and a paper!

Oh right to clean it up. Ok so now how do I evaluate for the particular solution?

not good at ALL with partial fractions...

it's easy.. multiply both sides of the last equation by s(s+5)

You mean A(s(s+5)/s+B(s(s+5)/(s+5) ?

don't*

A(s+5)+B(s)

err

that should equal to?

9+s/s(s+5)=A(s+5)+B(s)

nearly

you multiply both sides by s(s+5)

Oh right ok so its

9+s=A(s+5)+B(s)

Hey lokisan :).. sorry I took over. we seem to be useless wjen you're around :P

you're doing fine :D

I'm 'half' watching...

there is more than one way find A and B now.. take s to be zero and find the value of A.

If I did something wrong, Please let me know :)

you're on the right track

A=9/5

Then I put that back into the original for B?

or A and B for s...

Yeah.. or just plug s=-5

b=-4/5

now what does Y(s) equal to?

Do you know why we used the partial fractions in the first place?

So we could evaluate what s was when x=0 and y=1

don't forget we are looking for the particular solution of the DE.

My instructor never used the notation as Y(s) so I think thats where I'm losing you.

Oh oh I think I understand... we're looking to change 9+s/s(s+5)

\[L(y)={9+5 \over s(s+5)}={9 \over 5}.{1 \over s}-{4 \over 5}.{1 \over s+5}\]

Isn't it 9+s

oh yeah 9+s sorry

now focus on the right part, and try to transform it back to y(x).

Ugh I understand what you mean, I just am unaware of the method. I don't do inverse laplace right?

You do nothing but the inverse Laplace.

So like... 1/s=1 etc?

Yeah.

How do I solve the 9/5 and -4/5

laplace transform and its inverse are both linear operations, that's L[af(t)]=aL[f(t)]

You should not worry about these constants.. just keep them as they are.

Am I complicating things here? :(

So just worry about the 1/s and 1/s+5?

No you're doing good. I'm sure its just me.

yeah.

you already did 1/s.. what about 1/(s+5)?

I just think you're not familiar with notation.

e^-at is the transform...

e^-5t then right?

You know what you need to do, though.

Yes

Me?

No, Scotty.

Exactly Scotty.

Sorry...

It's OK :)

Ok so to clean it up I use use what I have left right?

Which basically will end up e^5t

How?

Your answer should have e^(-5t) in it.

9/5*1-4/5*e^(-5t)

It's in x not in t.

Yes, Scotty.

But in x.

I'm used to doing them in t...sorry.

so e^-(5x)

so final answer is:
\[y(x)={9 \over 5}- {4 \over 5} e ^{-5x}\]
right?

We all are used to t.

perfect

Ok I was rushing through on the simplification.

What?

He needed to use Laplace.

So that just gave me the general solution right? Now I evaluate the particular?

He solved the problem without using the Laplace transform.

Just kidding guys, no need to bully me :(

LOL

FWIW, he said ' I could us..', implying it wasn't necessary. And it isn't.

Hey Micheal!!

oh man...

You spelt my name wrong. It's 'Isaac Newton'

sure...

Oh yeah?

Im confused on what just happened.

No. But you did spell my real name wrong.

:(

Sorry Michael!

:D

...

So you just solved my problem doing tradidional integration then? Is that what I gather?

Yeah.. He used separating variables method to solve the DE.

You could have done it either way. You were on about Laplace, so Anwar answered as such.

^^
Thank you!! You saved me :)

I see, its not solved all the way though right? I have C and X, I just need to use those to solve y?

Is this a differential equation course or what exactly?

Its a Calc II with LaPlace and Diff Eq in it.

:( Oh dear.

Yes.... so what do I need to do to finish using the method you did newton?

It's OK, I finished it (I only posted the answer because you already had it).

So, Scotty, are you okay with using Laplace now?

Yes better. I need to practice a bit more I think :/

Sure.

*your

Damn! Was about to correct that. I'll get you next time...

haha

You should also practice working them out yourself (i.e. that table entries).

np. I'm off to bed. Good work, Anwar :)

...and Scotty :)

Oh thanks :)