Question

Derive \(y''+4y=cos \omega t\).
Second order ODE problem.

Answer 1

@amistre64 help please?

Answer 2

Solve the homogeneous problem first (complex roots) and then make a guess for the function on the right hand side.

I would try it with:
\[\Large A\sin(\omega t) + B\cos ( \omega t) \]

Answer 3

I just had a lecture on this but he was very unclear on this topic, basically he told us to replace \(y''+4y\) with an equation like \(Ly\) and then he lost me :(

Answer 4

The LaPlace transform?

Answer 5

there are many methods that can be used; so unless youre tied to a method ....

Answer 6

sorry but the only thing I know was that we know
\((Ly_p=g\) and \(L(c_1 y_1+c_2 y_2 )=0)\) and we want \(L(y_p+c_1 y_1+c_2 y_2 )=g\)

Answer 7

im unsure about the method presented then. It does look laplacey, but its not ringing any bells for me.

Answer 8

can you give me an example with the LaPlace transform? If there were any easier equations to solve with this method, I would gladly take it. I just want to understand this before we move on in class

Answer 9

if it is laplace, then i fond pauls site to do a splendid job at being readable and concise.

http://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx

Personally, I would have gone with Limbuses suggestion for a solution.

Answer 10

Thanks you! I will check that out.

Answer 11

this is linear operator

read more about linear operator. first find the complementary solution and general solution.

Answer 12

solve the homogeneous equation

Answer 13

and guess a general function that would give you a form of cosine when you take the first and the second derivative.

Answer 14

hmmm, getusel is making more sense :) The "L" will later be used in LaPlace transforms, but at the moment it is most likely a generic operating notation.

Answer 15

laplace is not represented by capital l for most cases
rather by some distorted l

Answer 16

ok so I am confused again. Do I solve \(y''+4y=0\) first?

Answer 17

L is one of the linear operators like laplace.

Answer 18

yes

Answer 19

go for the homogenous solution first

y'' + 4y = 0

this will be the yc, or yh depending on textbook

Answer 20

solve it

Answer 21

using characterstics equation

Answer 22

ok.
\[\lambda ^{2} + 4\lambda =0\]
so \(\lambda = 2i\) and
\(y=c_{1}cos2t+c_{2}sin2t\)
right?

Answer 23

and then i can set \(c_1cos2t+c_2sin2t=cos \omega t\)?

Answer 24

yes, but the way i was taught it was to consider that ca and c2 as functions; and get it down to the y'' again

Answer 25

well, almost right; yp''+4yp = cos wt

Answer 26

\[y_p=A(t)cos(2t)+B(t)sin(2t)\]
\[y'_p=A'cos(2t)-A~2sin(2t)+B'sin(2t)+B~2cos(2t)\]
but the A',B' terms equate to the homogenous part, and zero out leaveing us with

\[y'_p=-A~2sin(2t)+B~2cos(2t)\]
\[y''_p=-A'~2sin(2t)+B'~2cos(2t)-A~4cos(2t)-B~4sin(2t)\]

Answer 27

now we have all the parts needed to express this in terms of y''p + 4yp= cos wt

\[~~~~~y''_p=-A'~2sin(2t)+B'~2cos(2t)-A~4cos(2t)-B~4sin(2t)\\~+4y_p=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A~4cos(2t)+B~4sin(2t)\\---------------------------\\cos(wt)=-A'~2sin(2t)+B'~2cos(2t)\]

Answer 28

what this does is give us 2 equations in 2 unknowns
\[A'cos(2t)+B'sin(2t)=0\\-2A'sin(2t)+2B'cos(2t)=cos(wt)\]

solve for A' adn B' and integrate them back into A and B

Answer 29

This is very helpful. Thank you so much!

Answer 30

theres a method called the wronskian what shortens this up alittle, since the repetitious stuff can be formulized into a matrix