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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I could use a LaPlace transform right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. and it's pretty easy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok the only problem I'm having is, all I have to use is the formula L(y')=sL(y)y(0) correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so I'm having trouble figuring out how to incorperate x=0 and y=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=0 when y=1 means that y(0)=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I would just rewrite... \[s L(y)y(0)+5L(y)=L(9)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK.. what are the L(y) and L(9)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not 100% sure on L(y) but I know L(9)=9/s

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would it just be y/s ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or no, I could solve for L(y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. you could do so, or you could write it as Y(s)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know that you have to transform it back to the time domain.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure about the simplification you made?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh I see it.. it's right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know how to do fractions notation using the equation thing or I would. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I need to grab a pen and a paper!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh right to clean it up. Ok so now how do I evaluate for the particular solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{9+s \over s(s+5)}= {A \over s}+ {B \over s+5}\] how are you with partial fractions, can you find A and B?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not good at ALL with partial fractions...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's easy.. multiply both sides of the last equation by s(s+5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You mean A(s(s+5)/s+B(s(s+5)/(s+5) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that should equal to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.09+s/s(s+5)=A(s+5)+B(s)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you multiply both sides by s(s+5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey lokisan :).. sorry I took over. we seem to be useless wjen you're around :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm 'half' watching...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is more than one way find A and B now.. take s to be zero and find the value of A.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I did something wrong, Please let me know :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're on the right track

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then I put that back into the original for B?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. or just plug s=5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now what does Y(s) equal to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know why we used the partial fractions in the first place?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we could evaluate what s was when x=0 and y=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't forget we are looking for the particular solution of the DE.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My instructor never used the notation as Y(s) so I think thats where I'm losing you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh oh I think I understand... we're looking to change 9+s/s(s+5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[L(y)={9+5 \over s(s+5)}={9 \over 5}.{1 \over s}{4 \over 5}.{1 \over s+5}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now focus on the right part, and try to transform it back to y(x).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ugh I understand what you mean, I just am unaware of the method. I don't do inverse laplace right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You do nothing but the inverse Laplace.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So like... 1/s=1 etc?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not 1/s=1.. but the inverse laplace of 1/s=1 or more accurately u(x), (which is the unit step function)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do I solve the 9/5 and 4/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0laplace transform and its inverse are both linear operations, that's L[af(t)]=aL[f(t)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ugh I was understanding it and now I'm lost. I understand to use the table but this is just beyond me now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should not worry about these constants.. just keep them as they are.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Am I complicating things here? :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So just worry about the 1/s and 1/s+5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No you're doing good. I'm sure its just me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you already did 1/s.. what about 1/(s+5)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just think you're not familiar with notation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0e^at is the transform...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know what you need to do, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok so to clean it up I use use what I have left right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which basically will end up e^5t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your answer should have e^(5t) in it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm used to doing them in t...sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so final answer is: \[y(x)={9 \over 5} {4 \over 5} e ^{5x}\] right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We all are used to t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok I was rushing through on the simplification.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm 12 years old and what is this? \[\frac{\mathbb{d}y}{\mathbb{d}x} = 9  5y \implies \int \frac{1}{95y} \mathbb{d}y = \int \mathbb{d}x \implies c \frac{1}{5} \ln 9  5y = x\] x = 0 when y = 1 \[\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{95y} \implies y = \frac{94e^{5x}}{5}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He needed to use Laplace.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that just gave me the general solution right? Now I evaluate the particular?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He solved the problem without using the Laplace transform.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just kidding guys, no need to bully me :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0FWIW, he said ' I could us..', implying it wasn't necessary. And it isn't.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You spelt my name wrong. It's 'Isaac Newton'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im confused on what just happened.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. But you did spell my real name wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you just solved my problem doing tradidional integration then? Is that what I gather?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah.. He used separating variables method to solve the DE.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could have done it either way. You were on about Laplace, so Anwar answered as such.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^^ Thank you!! You saved me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see, its not solved all the way though right? I have C and X, I just need to use those to solve y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this a differential equation course or what exactly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Its a Calc II with LaPlace and Diff Eq in it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes.... so what do I need to do to finish using the method you did newton?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's OK, I finished it (I only posted the answer because you already had it).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, Scotty, are you okay with using Laplace now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes better. I need to practice a bit more I think :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0Damn! Was about to correct that. I'll get you next time...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
 5 years ago
Best ResponseYou've already chosen the best response.0You should also practice working them out yourself (i.e. that table entries).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0np. I'm off to bed. Good work, Anwar :)
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