Question

Find a particular solution to y"+25y=20sin(5t). I figured that Y = Asin(5t), so Y' = 5Acos(5t) and Y" = -25Asin(5t), but if you put Y and Y" in the equation, you get 0 = 20sin(5t)... so do you add a t to the initial Y?

Answer 1

hey

Answer 2

weren't you the one I showed that site to earlier?

Answer 3

if have the right side is sin( t) or cos(t ) or sin(t)+cos(t)
then the particular solution guess should be y=Asin(t)+Bcos(t)

so here you should have y=Asin(5t)+Bcos(5t)

Answer 4

Yes, and I've checked it but it really only has a few examples, such as y" - 4y' -12y = sin(2t).

Answer 5

Interesting... I wish my professor had gone more in depth with this type.

Answer 6

the table should say this though

Answer 7

and it does on row 3

Answer 8

Ah here we go, my table says \[e^(\alpha*t) (Q_n *t sinBt + R_ncosBt)\]. I didn't see all of it I suppose. Thanks!

Answer 9

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
I was talking about row 3 here:
g(t)=20sin(5t)
so
y_p=Acos(5t)+Bsin(5t)

Answer 10

I think, we can get the right hand side by differentiating either `sin` or `cosine` multiple times, so your guess must be a combination of both these functions

Answer 11

when the right hand side is sin or cosine or an exponential, we can express it as \(e^{ax}\) and then there is an easy direct formula for the particular solution :
\[y_p = \dfrac{e^{ax}}{p(a)}\]

Answer 12

don't bother about it if your professor haven't covered them yet..

Answer 13

It seems that it still comes out to 0. Y" = -25Asin(5t) -25Bcos(5t), Y = Asin(5t)+Bcos(5t). Multiply Y by 25, -25+25 = 0.

Answer 14

whats your homogeneous solution ?

Answer 15

ganeshie8, what does p(a) signify then?

Answer 16

oh you are right

Answer 17

so we have to adjust the particular solution

Answer 18

I think
(thinking)

Answer 19

So then maybe Y = At^2 sin(5t) + Bt^2cos(5t)?

Answer 20

\(y'' + 25y\)

can be expressed as \((D^2 + 25)y\)

\(p(D) = D^2+25\)

Hey don't bother about these, it is better you get these in class from your prof

Answer 21

lets focus on undetermined coefficients :)

Answer 22

Our homework is due tonight and he assigned these problems.

Answer 23

Do you know why you're stuck ?

Answer 24

Find the homogeneous solution and you will find the reason

Answer 25

Homogeneous as in the \[y_H\]?

Answer 26

oh... we need to multiply t to the particular guess we had right @ganeshie8 ?

Answer 27

Ah, so our c1 and c2 are imaginary numbers: +/- 5i.

Answer 28

So \[y_H = c_1e^(5t) + c_2e^(5t)?\]

Answer 29

when you have complex roots, \(r=\pm 5i\), the homogeneous solutoin can be written as :
\[y_H = c_1\cos(5t) + c_2\sin(5t)\]

Answer 30

Notice that your previous guess for particular solution is actually the solution to the homogeneous equaiton. So it never gonna work.

Answer 31

just multiiply `t` to the previous guess as freckles is saying and see...

Answer 32

I think I figured it out with multiplying by t, I just went back to the singular Y = Atsin(5t), and so my answer came out to A = 2tan(5t).

Answer 33

Still not coming up with the right answer though, my Y would then be \[2t \tan(5t)*\sin(5t)\]

Answer 34

thats the solution generated by wolfram using undetermined coefficients

you will be able to zoom in.. resolution is not an issue

Answer 35

Wow, wolfram alpha really does its job. I'll have to go over that and figure it out through the steps. Thanks very much!

Answer 36

the first half has steps to find the homogeneous solution

the second half is about the particular solution

Answer 37

the steps it has taken are exactly same as yours

Answer 38

Oh man, I had the exact same thing when I did Y = Atsin(5t) +Btcos(t)... I just didn't simplify.

Answer 39

Btcos(5t)*

Answer 40

I see.. happens haha!

Answer 41

It gives the exact reason for why we need to multiply our guess by "t" too...
thats the reason i like solutions generated by wolfram

but unfortunately it fails to generate solutions if the problems are a bit complicated..

Answer 42

Ah, I understand now. Some of this stuff is just small mistakes, and the problems take a pretty long time I suppose. Thanks again!

Answer 43

np