anonymous
  • anonymous
Show how to find the particular solution of the differential equation, dy/dx+5y=9, x=0 when y=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I could use a LaPlace transform right?
anonymous
  • anonymous
Yeah.. and it's pretty easy.
anonymous
  • anonymous
Ok the only problem I'm having is, all I have to use is the formula L(y')=sL(y)-y(0) correct?

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anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
ok so I'm having trouble figuring out how to incorperate x=0 and y=1
anonymous
  • anonymous
x=0 when y=1 means that y(0)=1
anonymous
  • anonymous
Oh.. duh.
anonymous
  • anonymous
so I would just rewrite... \[s L(y)-y(0)+5L(y)=L(9)\]
anonymous
  • anonymous
OK.. what are the L(y) and L(9)?
anonymous
  • anonymous
Not 100% sure on L(y) but I know L(9)=9/s
anonymous
  • anonymous
Would it just be y/s ?
anonymous
  • anonymous
or no, I could solve for L(y)
anonymous
  • anonymous
I think I have it
anonymous
  • anonymous
Yeah.. you could do so, or you could write it as Y(s)
anonymous
  • anonymous
You know that you have to transform it back to the time domain.
anonymous
  • anonymous
L(y)=(9/s+1)/(s+5)
anonymous
  • anonymous
are you sure about the simplification you made?
anonymous
  • anonymous
oh I see it.. it's right.
anonymous
  • anonymous
I don't know how to do fractions notation using the equation thing or I would. :)
anonymous
  • anonymous
I think I need to grab a pen and a paper!
anonymous
  • anonymous
Well you so far have: \[Y(s)={9/s+1 \over s+5}\] you can multiply by s (since it's not 0).. you will get: \[{9+s \over s(s+5)}\]
anonymous
  • anonymous
Oh right to clean it up. Ok so now how do I evaluate for the particular solution?
anonymous
  • anonymous
\[{9+s \over s(s+5)}= {A \over s}+ {B \over s+5}\] how are you with partial fractions, can you find A and B?
anonymous
  • anonymous
not good at ALL with partial fractions...
anonymous
  • anonymous
it's easy.. multiply both sides of the last equation by s(s+5)
anonymous
  • anonymous
You mean A(s(s+5)/s+B(s(s+5)/(s+5) ?
anonymous
  • anonymous
Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.
anonymous
  • anonymous
don't*
anonymous
  • anonymous
A(s+5)+B(s)
anonymous
  • anonymous
err
anonymous
  • anonymous
that should equal to?
anonymous
  • anonymous
9+s/s(s+5)=A(s+5)+B(s)
anonymous
  • anonymous
nearly
anonymous
  • anonymous
you multiply both sides by s(s+5)
anonymous
  • anonymous
Oh right ok so its
anonymous
  • anonymous
9+s=A(s+5)+B(s)
anonymous
  • anonymous
Hey lokisan :).. sorry I took over. we seem to be useless wjen you're around :P
anonymous
  • anonymous
you're doing fine :D
anonymous
  • anonymous
I'm 'half' watching...
anonymous
  • anonymous
there is more than one way find A and B now.. take s to be zero and find the value of A.
anonymous
  • anonymous
If I did something wrong, Please let me know :)
anonymous
  • anonymous
you're on the right track
anonymous
  • anonymous
A=9/5
anonymous
  • anonymous
Then I put that back into the original for B?
anonymous
  • anonymous
or A and B for s...
anonymous
  • anonymous
Yeah.. or just plug s=-5
anonymous
  • anonymous
b=-4/5
anonymous
  • anonymous
now what does Y(s) equal to?
anonymous
  • anonymous
I'm kind of confused.. so I now put int he values of A and B back into 9+s=A(s+5)+B(s) to solve for the s?
anonymous
  • anonymous
Do you know why we used the partial fractions in the first place?
anonymous
  • anonymous
So we could evaluate what s was when x=0 and y=1
anonymous
  • anonymous
No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.
anonymous
  • anonymous
don't forget we are looking for the particular solution of the DE.
anonymous
  • anonymous
My instructor never used the notation as Y(s) so I think thats where I'm losing you.
anonymous
  • anonymous
Oh oh I think I understand... we're looking to change 9+s/s(s+5)
anonymous
  • anonymous
\[L(y)={9+5 \over s(s+5)}={9 \over 5}.{1 \over s}-{4 \over 5}.{1 \over s+5}\]
anonymous
  • anonymous
Isn't it 9+s
anonymous
  • anonymous
oh yeah 9+s sorry
anonymous
  • anonymous
now focus on the right part, and try to transform it back to y(x).
anonymous
  • anonymous
Ugh I understand what you mean, I just am unaware of the method. I don't do inverse laplace right?
anonymous
  • anonymous
You do nothing but the inverse Laplace.
anonymous
  • anonymous
So like... 1/s=1 etc?
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
not 1/s=1.. but the inverse laplace of 1/s=1 or more accurately u(x), (which is the unit step function)
anonymous
  • anonymous
How do I solve the 9/5 and -4/5
anonymous
  • anonymous
laplace transform and its inverse are both linear operations, that's L[af(t)]=aL[f(t)]
anonymous
  • anonymous
Ugh I was understanding it and now I'm lost. I understand to use the table but this is just beyond me now.
anonymous
  • anonymous
You should not worry about these constants.. just keep them as they are.
anonymous
  • anonymous
Am I complicating things here? :(
anonymous
  • anonymous
So just worry about the 1/s and 1/s+5?
anonymous
  • anonymous
No you're doing good. I'm sure its just me.
anonymous
  • anonymous
yeah.
anonymous
  • anonymous
you already did 1/s.. what about 1/(s+5)?
anonymous
  • anonymous
I just think you're not familiar with notation.
anonymous
  • anonymous
e^-at is the transform...
anonymous
  • anonymous
e^-5t then right?
anonymous
  • anonymous
You know what you need to do, though.
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Me?
anonymous
  • anonymous
No, Scotty.
anonymous
  • anonymous
Exactly Scotty.
anonymous
  • anonymous
Sorry...
anonymous
  • anonymous
It's OK :)
anonymous
  • anonymous
Ok so to clean it up I use use what I have left right?
anonymous
  • anonymous
Which basically will end up e^5t
anonymous
  • anonymous
How?
anonymous
  • anonymous
Your answer should have e^(-5t) in it.
anonymous
  • anonymous
9/5*1-4/5*e^(-5t)
anonymous
  • anonymous
It's in x not in t.
anonymous
  • anonymous
Yes, Scotty.
anonymous
  • anonymous
But in x.
anonymous
  • anonymous
I'm used to doing them in t...sorry.
anonymous
  • anonymous
so e^-(5x)
anonymous
  • anonymous
so final answer is: \[y(x)={9 \over 5}- {4 \over 5} e ^{-5x}\] right?
anonymous
  • anonymous
We all are used to t.
anonymous
  • anonymous
perfect
anonymous
  • anonymous
Ok I was rushing through on the simplification.
anonymous
  • anonymous
I'm 12 years old and what is this? \[\frac{\mathbb{d}y}{\mathbb{d}x} = 9 - 5y \implies \int \frac{1}{9-5y} \mathbb{d}y = \int \mathbb{d}x \implies c -\frac{1}{5} \ln 9 - 5y = x\] x = 0 when y = 1 \[\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{9-5y} \implies y = \frac{9-4e^{-5x}}{5}\]
anonymous
  • anonymous
What?
anonymous
  • anonymous
He needed to use Laplace.
anonymous
  • anonymous
So that just gave me the general solution right? Now I evaluate the particular?
anonymous
  • anonymous
He solved the problem without using the Laplace transform.
anonymous
  • anonymous
Just kidding guys, no need to bully me :(
anonymous
  • anonymous
LOL
anonymous
  • anonymous
FWIW, he said ' I could us..', implying it wasn't necessary. And it isn't.
anonymous
  • anonymous
Hey Micheal!!
anonymous
  • anonymous
oh man...
anonymous
  • anonymous
You spelt my name wrong. It's 'Isaac Newton'
anonymous
  • anonymous
sure...
anonymous
  • anonymous
Oh yeah?
anonymous
  • anonymous
Im confused on what just happened.
anonymous
  • anonymous
No. But you did spell my real name wrong.
anonymous
  • anonymous
:(
anonymous
  • anonymous
Sorry Michael!
anonymous
  • anonymous
:D
anonymous
  • anonymous
...
anonymous
  • anonymous
So you just solved my problem doing tradidional integration then? Is that what I gather?
anonymous
  • anonymous
Yeah.. He used separating variables method to solve the DE.
anonymous
  • anonymous
You could have done it either way. You were on about Laplace, so Anwar answered as such.
anonymous
  • anonymous
^^ Thank you!! You saved me :)
anonymous
  • anonymous
I see, its not solved all the way though right? I have C and X, I just need to use those to solve y?
anonymous
  • anonymous
Is this a differential equation course or what exactly?
anonymous
  • anonymous
Its a Calc II with LaPlace and Diff Eq in it.
anonymous
  • anonymous
A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.
anonymous
  • anonymous
:( Oh dear.
anonymous
  • anonymous
Yes.... so what do I need to do to finish using the method you did newton?
anonymous
  • anonymous
It's OK, I finished it (I only posted the answer because you already had it).
anonymous
  • anonymous
So, Scotty, are you okay with using Laplace now?
anonymous
  • anonymous
Yes better. I need to practice a bit more I think :/
anonymous
  • anonymous
Sure.
anonymous
  • anonymous
It's a really useful method. You transform a differential equation into an algebraic one which can be easier to solve. Plus, you're initial conditions are taken into account in the process.
anonymous
  • anonymous
*your
anonymous
  • anonymous
Damn! Was about to correct that. I'll get you next time...
anonymous
  • anonymous
Yeah I can see how it would be useful, I just need a little more practice recognizing the transform tables and how to apply them. Also I guess my notation is different than the traditional method
anonymous
  • anonymous
haha
anonymous
  • anonymous
You should also practice working them out yourself (i.e. that table entries).
anonymous
  • anonymous
I wouldn't veen know how to start that, I'll just read some textbooks. Anyways off to the next problem on my review. Thanks guys
anonymous
  • anonymous
np. I'm off to bed. Good work, Anwar :)
anonymous
  • anonymous
...and Scotty :)
anonymous
  • anonymous
Oh thanks :)

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