Question

Two second order De problem due tomorrow, Help me now!

Answer 1

\[3y \prime \prime + 2y \prime - y =4\sin 5x\]

Answer 2

\[x \prime \prime + 16x =3\cos t\]

Answer 3

I cannot find the PI for these two equations

Answer 4

what does PI mean?

Answer 5

Oh no! It should be 3 cos 4t in the second problem

Answer 6

PI stands for "particular integral"

Answer 7

the first one first, what is your characteristic equation for homogenous part? don't forget it's second ODE, not first ODE, no need to find PI

Answer 8

my AQE is \[3u ^{2}+2^{u}-1=0\]

Answer 9

why 2^u ? but 2u?

Answer 10

so CF should be \[y=Ae ^{\frac{ x }{ 3 }}+Be ^{-x}\]

Answer 11

yes, my typo. Obviously it's 2u

Answer 12

but the general solution is supposed to be CF + PI right?

Answer 13

oh, I got you , PI is partial solution, comes from nonhomogeneous part.

Answer 14

yes, it has formula to get,

Answer 15

right, my CF is correct , but I can't solve for PI

Answer 16

want to take my note and find out by yourself?

Answer 17

I let my PI to be\[y=\lambda \sin 5x + \mu \cos 5x\]

Answer 18

this lead to the equations\[-76\lambda-10\mu=4\]
and\[-76\mu-10\lambda=0\]

Answer 19

The solution is weird, so maybe there is something wrong

Answer 20

Is my equation correct?

Answer 21

the whole solution is add them together, I mean homo and nonhomo solution parts
hey, your prof didn't teach you that way??

Answer 22

my answers are :
\[\lambda=-\frac{ 76 }{ 1469 }\]
\[\mu=-\frac{ 10 }{ 1469 }\]

Answer 23

I mean my solution is weird, look at the numbers above, they are so strange!

Answer 24

No I am going to solve the constant in the PI, not in the CF

Answer 25

so, I am useless for you. Sorry , I waste your time. Do you want me clear up or just close this post and open a new one to get other's help?

Answer 26

You are helpful if you can just check my answer is correct or not.
Here is my final solution:
\[y=Ae ^{\frac{ 3 }{ x }}+Be ^{-x}-\frac{ 76 }{ 1469 }\sin 5x -\frac{ 10 }{ 1469 }\cos 5x\]

Answer 27

but how ?! what's wrong with my equation?

Answer 28

ok, show me your work from \(y=\lambda \sin 5x + \mu \cos 5x\)
step by step, don't say "which leads to...."

Answer 29

how do you get the next equation for \(\lambda ~~and~~\mu\)

Answer 30

\[y=\lambda \sin 5x +\mu \cos 5x\]\[y \prime=5\lambda \cos 5x -5\mu sin 5x\]

Answer 31

\[y \prime \prime = -25\lambda \sin 5x - 25 \]

Answer 32

Am I right up until now?

Answer 33

y" is weird

Answer 34

the last term, where are your \(\mu\) and sin 5x from y'?

Answer 35

If you substitute all of them to the original equation. \[-75\lambda \sin 5x -75mucos5x + 10 \lambda \cos5x -10 \mu \sin 5x - \lambda \sin 5x - \mu \cos 5x =4 \sin 5x\]

Answer 36

OMG typo again!

Answer 37

the above equation should have " sin 5x " at the end, but they disappeared.

Answer 38

then you compare coefficients and you should arrive at my equations for mu and lambda

Answer 39

ok, combine the like term first, respect to cos and sin

Answer 40

it should be \[(-76 \lambda -10\mu) \sin 5x + ( -76\mu + 10 \lambda) \cos 5x = 4 \sin 5x \]

Answer 41

yup

Answer 42

thats why -76 l -10 m =4 and -76 m + 10 l =0
l is lambda and m is mu

Answer 43

so how do you get your answer?

Answer 44

My Ti 83 said that \(\lambda = -\dfrac{76}{1469}\) and \(\mu=-\dfrac{10}{1469}\)

Answer 45

yes, that's my answer

Answer 46

ok, we did all our best, if it's not correct, no regret

Answer 47

aha! wolfram alpha says my answer is correct!

Answer 48

hehehe unfortunately we are not allowed to use that tool on teeeeeests

Answer 49

the second equation you solve by yourself, I don't have time to play with you anymore. good luck