Question

I need help with a Differential Equations problem. Chapter on undamped forcing and resonance...

Answer 1

\[\frac{ d^2y }{ dt^2 } + 9y = 2 \cos(3t)\]

Answer 2

Compute the general solution of the given equation

Answer 3

I'm not sure if notations hold true in general
but I have an idea that \(\large y(t) = y_h(t)+y_p(t)\)

Answer 4

I think I got that ...
\(y_h(t) = k_1 cos(3t) + k_2 sin(3t)\)

Answer 5

anyways, I am stuck on getting \(y_p(t)\)
using complexification or otherwise .-.

Answer 6

Give me a sec ^.^

Answer 7

Use a calculator e.e

Answer 8

this is a practice problem for the quiz I'm gonna have to take later today - no calculators allowed.
I need to know the correct steps and processes and logic so I can do well on the quiz.

Answer 9

@.Sam.
@IrishBoy123
@iYuko

Answer 10

@Directrix @skullpatrol I require math help

Answer 11

I got the answer Jiggly :D

Answer 12

@.Sam. the smart one needs the smartest one. e.e

Answer 13

Ok Ok... but it's long .-.

Answer 14

@inkyvoyd

Answer 15

Wait you only want the yp(t) right?

Answer 16

yes

Answer 17

stop spamming, i doubt you can do this, if you are not going to help, leave @just_one_last_goodbye

Answer 18

^

Answer 19

I have so far attempted

\(\large y_p(t) = ke^{3it}\)
and
\(\large y_p(t) = a*cos(3t) + b*sin(3t)\)

both of which led to dead ends

Answer 20

Heres what you want Jiggly >.> http://prntscr.com/dh15cm

Answer 21

Let me know if my calculator is failing your needs .-. so I may improve it

Answer 22

I did that and everything on the left side seemed to cancel out

Answer 23

y_p(t), means a particular solution

Answer 24

yes

Answer 25

take y(t)=Acos(kt)

Answer 26

http://prntscr.com/dh16zg

Answer 27

leave @just_one_last_goodbye

Answer 28

I see one of my mistakes now...
brb I'ma recalculate

Answer 29

@ILovePuppiesLol leave me alone bruh ;-; im trying to help... not spam like you

Answer 30

I forgot a chain rule...

Answer 31

General solution.... \[y = c1\cos (3t) + c2\sin (3t) + \frac{ 1 }{ 3 } tsin (3t)\]

Answer 32

are you good now?

Answer 33

I told you know already knew the answer

Answer 34

I needed to know HOW to get it
so I'm still trying to work that out after finding my mistake, brb

Answer 35

._. my calculator has helped :D

Answer 36

Sadly Preetha rejected it :( after all my hard work trying to make it a part of OS...

Answer 37

I'm stuck at
-8at + 6b = 2
-6a - 8bt = 0

Answer 38

why is there a t

Answer 39

when using
\(\large y_p(t) = at*cos(3t) + bt*sic(3t)\)

Answer 40

*sin(3t)

Answer 41

dont use undetermined coeff the constants are cancelling out, they did it on purpose, u cant do it without some trig identities

Answer 42

use the general variation of parameter form

Answer 43

once you find your y_h you can find your y_p with this formula http://prntscr.com/dh1akg

Answer 44

interesting... I've never learned that

Answer 45

so you've only learn undetermined coefficients?

Answer 46

u can still do it that way, you just need to do a bit of algebra

Answer 47

and complexification

Answer 48

okok so then do this

Answer 49

\(\large y_p(t) = ke^{i \omega t} \)

Answer 50

I figured out how
-8at + 6b = 2
-6a - 8bt = 0
gets me the right answer .-.
not sure why I had to set t to 0
but ok