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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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!
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: \[{9+s \over s(s+5)}\]
Oh right to clean it up. Ok so now how do I evaluate for the particular solution?
\[{9+s \over s(s+5)}= {A \over s}+ {B \over s+5}\] how are you with partial fractions, can you find A and B?
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) ?
Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.
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?
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?
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
No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.
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.
not 1/s=1.. but the inverse laplace of 1/s=1 or more accurately u(x), (which is the unit step function)
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)]
Ugh I was understanding it and now I'm lost. I understand to use the table but this is just beyond me now.
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.
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 \[\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{9-5y} \implies y = \frac{9-4e^{-5x}}{5}\]
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.
A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.
:( 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.
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.
*your
Damn! Was about to correct that. I'll get you next time...
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
haha
You should also practice working them out yourself (i.e. that table entries).
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
np. I'm off to bed. Good work, Anwar :)
...and Scotty :)
Oh thanks :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question