sinxcosx(dy/dx)+y=sinx how to find y?

- anonymous

sinxcosx(dy/dx)+y=sinx how to find y?

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- anonymous

you mean y' right?

- anonymous

y' = cosx ^_^

- anonymous

wait a min, it's:
y' + y = sinx

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## More answers

- anonymous

right?

- anonymous

if so, then y= sinx - y'

- anonymous

y'=(sinx-y)/(sinxcosx)
?
i dont fully understand the question DX

- anonymous

lol no, the question is :
y' + y = sinx

- anonymous

whats the sinxcosx?

- anonymous

wait....it's the derivative of sinxcosx + y = sinx

- anonymous

so y = sinx + sin^x + cos^2x
= sinx + 1
= sinx

- anonymous

sin^2x*

- anonymous

there's something wrong ._.

- anonymous

whats wrong?

- anonymous

mowazzem can you please restate the question

- anonymous

in a more understandable way

- anonymous

sinxcosxy'+y=sinx

- anonymous

find y' or y?

- anonymous

its a derivative, i need to solve this

- anonymous

that's mean y

- anonymous

you want to derive y = sinx - sinxcosx?

- anonymous

linear homogeneous

- anonymous

again, mowa is this the question you want to derive:
y = sinx - sinxcosx?

- anonymous

no this is not a question. its a DERIVATIVE FORM sinxcosx(y')+y=sinx

- anonymous

and you want to find y?

- anonymous

yeah

- anonymous

then y = sinx - sinxcosxy'

- anonymous

as far as I have understood from the problem

- anonymous

wait, do you want to integrate?!

- anonymous

right hand side has also a derivative of y'. so your suggestion might be worng

- anonymous

lol! so you DO want to integrate ^_^

- anonymous

what class is this? calculus or diff equas?

- anonymous

I think it's calculus since she wants to get y

- anonymous

diff equas

- anonymous

only way to get rid of y' is to integrate

- anonymous

oh, no wonder, diff euas (at that level) are just beyond me

- anonymous

just a little*

- anonymous

mowa, did you take integration by parts?

- anonymous

sstaica what do you do?

- anonymous

did you take integration?

- anonymous

do you know bernoulli equation

- anonymous

heard of it

- anonymous

y'+p(x)y=r(x)

- anonymous

general form of bernoullis equation

- anonymous

its semelar form of previous equation

- anonymous

hmm, the bernoulli that I know comes from physics P + 1/2pV^2 + pgh

- anonymous

are undergraduate student

- anonymous

you mean a college student?

- anonymous

yeah

- anonymous

yep I am, first year ^_^

- anonymous

or graduate student

- anonymous

why? :)

- anonymous

ok, probably you don't know

- anonymous

are you a graduate?

- anonymous

its higher class math

- anonymous

I am doing PhD

- anonymous

I see, then you are right, I prolly don't know it :) still taking calculus II ^_^

- anonymous

whoa! GOOD LUCK :)

- anonymous

i know that why you usde very simple rule

- anonymous

lol, still learning ^^"
anyways, sorry for not being able to help , I wish I could help though, you can try google? or youtube ^_^

- anonymous

and I wish you the best of luck :)

- anonymous

its very simple but I forgot. I done it about 4 yrs ago.

- anonymous

you can try google though , the only way I know right now is integration by parts

- anonymous

anyaway thanks

- anonymous

np ^_^

- anonymous

haha yup diff equas are some crazy fun stuff XD

- anonymous

mowazzem, can i ask you a question?

- anonymous

sorry i am going out. What you wanna ask?

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