Show how to find the particular solution of the differential equation, dy/dx+5y=9, x=0 when y=1

- anonymous

Show how to find the particular solution of the differential equation, dy/dx+5y=9, x=0 when y=1

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

I could use a LaPlace transform right?

- anonymous

Yeah.. and it's pretty easy.

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Yeah.

- anonymous

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

- anonymous

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

- anonymous

Oh.. duh.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Would it just be y/s ?

- anonymous

or no, I could solve for L(y)

- anonymous

I think I have it

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

are you sure about the simplification you made?

- anonymous

oh I see it.. it's right.

- anonymous

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

- anonymous

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

- 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

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

- 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

not good at ALL with partial fractions...

- anonymous

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

- anonymous

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

- anonymous

Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.

- anonymous

don't*

- anonymous

A(s+5)+B(s)

- anonymous

err

- anonymous

that should equal to?

- anonymous

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

- anonymous

nearly

- anonymous

you multiply both sides by s(s+5)

- anonymous

Oh right ok so its

- anonymous

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

- anonymous

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

- anonymous

you're doing fine :D

- anonymous

I'm 'half' watching...

- anonymous

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

- anonymous

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

- anonymous

you're on the right track

- anonymous

A=9/5

- anonymous

Then I put that back into the original for B?

- anonymous

or A and B for s...

- anonymous

Yeah.. or just plug s=-5

- anonymous

b=-4/5

- anonymous

now what does Y(s) equal to?

- 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

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

- anonymous

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

- anonymous

No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Isn't it 9+s

- anonymous

oh yeah 9+s sorry

- anonymous

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

- anonymous

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

- anonymous

You do nothing but the inverse Laplace.

- anonymous

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

- anonymous

Yeah.

- 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

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

- anonymous

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

- 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

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

- anonymous

Am I complicating things here? :(

- anonymous

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

- anonymous

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

- anonymous

yeah.

- anonymous

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

- anonymous

I just think you're not familiar with notation.

- anonymous

e^-at is the transform...

- anonymous

e^-5t then right?

- anonymous

You know what you need to do, though.

- anonymous

Yes

- anonymous

Me?

- anonymous

No, Scotty.

- anonymous

Exactly Scotty.

- anonymous

Sorry...

- anonymous

It's OK :)

- anonymous

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

- anonymous

Which basically will end up e^5t

- anonymous

How?

- anonymous

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

- anonymous

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

- anonymous

It's in x not in t.

- anonymous

Yes, Scotty.

- anonymous

But in x.

- anonymous

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

- anonymous

so e^-(5x)

- anonymous

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

- anonymous

We all are used to t.

- anonymous

perfect

- anonymous

Ok I was rushing through on the simplification.

- 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

What?

- anonymous

He needed to use Laplace.

- anonymous

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

- anonymous

He solved the problem without using the Laplace transform.

- anonymous

Just kidding guys, no need to bully me :(

- anonymous

LOL

- anonymous

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

- anonymous

Hey Micheal!!

- anonymous

oh man...

- anonymous

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

- anonymous

sure...

- anonymous

Oh yeah?

- anonymous

Im confused on what just happened.

- anonymous

No. But you did spell my real name wrong.

- anonymous

:(

- anonymous

Sorry Michael!

- anonymous

:D

- anonymous

...

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

^^
Thank you!! You saved me :)

- 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

Is this a differential equation course or what exactly?

- anonymous

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

- anonymous

A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.

- anonymous

:( Oh dear.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Sure.

- 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

*your

- anonymous

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

- 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

haha

- anonymous

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

- 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

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

- anonymous

...and Scotty :)

- anonymous

Oh thanks :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.