how do you solve the initial value problem of 4y(3)+2y=0;y(0)=2y(0)=5y(0)=6

- anonymous

how do you solve the initial value problem of 4y(3)+2y=0;y(0)=2y(0)=5y(0)=6

- jamiebookeater

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- anonymous

i got the wrong answer :(

- anonymous

Can you retype the question? It didn't come out right? What are those 0s? Are they exponents or numbers?

- anonymous

they are numbers

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## More answers

- anonymous

But they are 0s?

- lalaly

What course are studying?

- anonymous

Here's is the problem again: 4y'''+2y''=0
y(0)=2
y'(0)=5
y''(0)=6

- anonymous

yes they are zeros

- anonymous

calc?

- lalaly

differential eqns:D

- anonymous

and im taking differential equations

- anonymous

yes lol

- anonymous

the 3 in the problem represents the third derivative im sorry the original post looks weird

- anonymous

its supposed to be 4y'''+2y''=0 as in to the third derivative and the second derivative

- lalaly

\[4r^3+2r^2=0\]\[2r^2(r+2)=0\]\[r=0,0,-1/2\]so \[y=c_1e^{-0.5x}+c_2x+c3\]

- lalaly

\[y'=-0.5c_1e^{-0.5x}+c_2\]

- lalaly

\[y''=0.25c_1e^{-0.5x}\]

- anonymous

yes i got that but i think i did something wrong after that

- lalaly

just plug in the values of x and y, nothing'd go wrong

- anonymous

i did but the answer is wrong! i did y(0)=2=c_1 +c_3

- anonymous

y'(0)=5=c2-(1/2)c3e^(-1/2x)

- lalaly

no the e part cancels coz it equals 1 when x =0

- lalaly

y'(0)=c2-1/2c3

- anonymous

yea i know and thats how i solved for c2=12

- lalaly

wait let me do it

- lalaly

\[y(0)=2\]\[c_1+c_3=2\]
\[y'(0)=5\]\[-0.5c_1+c_2=5\]
\[y''(0)=6\]\[0.25c_1=6\]\[c_1=24\]
\[24+c_3=2\]\[c_3=-22\]\[-0.5(24)+c_2=5\]\[c_2=17\]

- anonymous

I got c2=-7 and thats where i went wrong! thanks soooo much!

- lalaly

:D:D its ok, its confusing i know lol,
ur welcome

- anonymous

i frogot .5 was negative

- anonymous

Thanks again! :)

- lalaly

anytime:)

- anonymous

can I ask another question?

- lalaly

sure:D

- anonymous

8y''+4y'+4y=0, where y(0)=4 and y'(0)=1

- lalaly

u want the answers or the whole solution?

- anonymous

the solution please

- anonymous

but if you cant you can give the answer and I can try to work backwards

- lalaly

ok give me 5 min let me work it out

- anonymous

thanks sooooo much

- lalaly

\[y=c_1e^{-\frac{x}{4}}\cos(\frac{\sqrt 7}{4}x)+c_2e^{-\frac{x}{4}}\sin(\frac{\sqrt 7}{4}x)\]
i got \[c_1=4\]\[c_2=3.02\]

- anonymous

ok let me check thanks again

- lalaly

:):)

- anonymous

woohoooo its correct! thanks sooo much

- lalaly

hehe yaaay :D

- anonymous

ok i have one last question but if you dont want to do it its totally fine! i had a bunch of roblems to do and these were the three i couldnt get so if its ok can i ask you another one? i feel bad for asking!

- lalaly

lol dont feel bad, id love to help, why else would i be on openstudy

- anonymous

true lol thanks soo much! here is the question:

- anonymous

Find a linear homogeneous constant-coefficient differential equation with the general solution:
y(x)=c1e^(2x) + c2Cos(9x) +c3Sin(9x)
that has the form y'''+_________y''+_____y'+____y=0
the underscores are supposed to represent blanks lol

- lalaly

i got off here by mistake :S let me do it again

- anonymous

okie dokes! :)

- lalaly

the solutions are
2,+/-9i

- lalaly

so
(r-2)(r^2+81)=0

- lalaly

r^3-2r^2+81r-162=0

- lalaly

y'''-2y''+81y'-162y=0

- anonymous

how did you get the solutions?

- lalaly

do u know how u find the general solution like when u get r=2,3 the general solution is y=c1e^2x+c_2e^3x and when u get complex solutions a+bi and a-bi
the general solution is y=e^ax(c1cos(bx)+c2sin(bx)...
u just do it back wards

- anonymous

oh ok I got it! thnkssss sooo much!!! i really appreciate all yr help!! I also noticed that you joioned 3arabi open stidy group! r u taking arabic classes?

- anonymous

study*

- lalaly

lol no im half arab :D:D i speak arabic

- anonymous

oh lol I was going to ask if you need any help because I also speak arabic and it's my minor

- lalaly

oh wow really :D:D where are u from hehe

- anonymous

I'm palestinian! hbu?

- lalaly

o ana kmaaaaaaaaaaaaaan :D

- anonymous

but if you're referring to city, I'm from Dearborn lol

- anonymous

omg nooo waaayyy!!! how cool!

- lalaly

hehe im from hebron hehe

- anonymous

oh by dearborn i meant city in the US haha I'm from beit safafa in palestine

- anonymous

do you live in palestine?

- lalaly

oh lol i dont know abt US lol, ana 3ayshe bi newzealand

- anonymous

ooooo keef new zealns? keef al; taqs hunak?

- anonymous

zealand*

- lalaly

hehe its summer so yaaaaaaaaay !! :D newzealand btjanen bmoot feha hehe, 3njad sodfe helwee:D:D iza bedik more help ask me :D

- anonymous

3njad inti bit mooti feeha? lol hon it's winter bes ana behab al pelletaa3! bhab al thalj! and shukran bes khalasit wajabatee! thanks soo much! i really appreciate your help!

- anonymous

i was supposed to say pelletaa3 not pelletaa3 lol i have no idea hot i typed that instead! lol sorry for the confusion

- anonymous

pelletaaaa3 as in winter* i dont know why it keeps sending it as something else

- lalaly

if u catch me on here at anytime ill be glad to help u
o sara7a u made my day hehe i was a lil upset, bas u made me smile :D
hehe Fhemt-ha

- anonymous

thanks so much! and same to you! if you have any questions I'll be more than glad to help to the best of my ability! and I'm glad to have made your day! a smile is always better than a frown! :)

- lalaly

hehe nice to meet you, esme lana btw

- anonymous

nice to meet you lana! ismee Ranya

- lalaly

oooo kman my favorite name ya salam hehe

- anonymous

haha shukran!

- anonymous

btasif bes lazim arooh leana lazam arooh ijeeb igrath min al mahalat! bye! and talk to you soon inshallah! :)Have a great day!! :)

- lalaly

:D:D okay take care,, bye byee

- anonymous

bye :)

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