(Cosx )dY +(sinx)Y=1 anyone who can solve it?

- anonymous

(Cosx )dY +(sinx)Y=1 anyone who can solve it?

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

I first divide with cosx. So i get dY+ sinx/cosx Y =cosx..

- anonymous

Then i Take U as... \[U=e ^{(sinx \div cosx)} ..... \]
\[e ^{\int\limits_{}^{}(sinx \div cosx)} Y \int\limits_{}^{}e ^{\int\limits_{}^{}(sinx \div cosx)} \times Cos x\]

- anonymous

\[e ^{\int\limits_{}^{}sinx \div cosx} = e ^{-\ln(cosx)} = e^{\ln(cosx)^{-1}} = cosx ^{-1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

\[1\div \cos x \times Y =\int\limits_{}^{} (1\div cosx) \times cox x\]
\[1 \div cosx \times Y =\int\limits 1\]
\[1 \div cosx \times Y =x +c\]
\[Y= xcos(x) + Ccos(x)\]

- anonymous

Anyone no were i did the wrong?

- anonymous

Yes

- anonymous

Umm...I'll write out a solution, scan and attach. You're on the right track, though.

- anonymous

Thank you!

- anonymous

Can you wait a bit? I want to make some tea.

- anonymous

Np!

- anonymous

Scanning now...

- anonymous

okeej! i dont see... anywrong.. hm.. but.. were is the wrong.

- anonymous

##### 1 Attachment

- anonymous

If it's unfamiliar, it's because I just derive the integrating factor from first principles. It's just one of those differential equations where you use an integrating factor, as you have done.

- anonymous

Okej, so i can t do like i have done?..

- anonymous

I'm not too sure what you've done (on the face of it). Did you attempt to use an integrating factor, or are you looking at a book and seeing the form of your equation in there and following the book's description on how to solve it?

- anonymous

Yes, the symbol U is the as the integrating factor

- anonymous

brb

- anonymous

ok

- anonymous

What i did, was that used U =e^ (int (Sinx\cosx))

- anonymous

or you could right int ( tan x) like you did.

- anonymous

Yes, I saw.

- anonymous

The method of integrating factors boils down to this: for a differential equation \[y'+p(x)y=q(x)\]then the solution y is found by solving\[\frac{d(\mu y)}{dx}=\mu q(x) \rightarrow \mu y= \int\limits_{}{}\mu q(x) dx + c\]where c is a constant and\[\mu = e^{\int\limits_{}{} p(x)dx}\]

- anonymous

Then\[y=\frac{1}{\mu}\int\limits_{}{}\mu q(x) dx+\frac{c}{\mu}\]

- anonymous

Were you jumping to that formula to work it out?

- anonymous

Yes Exactly

- anonymous

That was, what i was trying to do.

- anonymous

\[y=\frac{1}{\sec x}\int\limits_{}{}\sec x \sec x dx+\frac{c}{\sec x}=\cos x \int\limits_{}{}\sec^2x dx+c \cos x\]\[=\cos x \tan x + c \cos x=\sin x + c \cos x\]

- anonymous

okeey, i see ..

- anonymous

i must have done wrong with the
e^âˆ«sinxÃ·cosx

- anonymous

The integration factor

- anonymous

What happens when you try it again?

- anonymous

The forumla is not wrong? huh? I uses this formula, \[U \times Y =\int\limits_{}^{}U \times q(x)\]

- anonymous

Where, DyDx + r(x)Y= q(x)

- anonymous

The formula will give you the same answer AS LONG AS you integrate the right-hand side BEFORE dividing by the U on the left-hand side.
If you don't do this, you won't get full use of the constant of integration. It forms part of the solution.

- anonymous

For example

- anonymous

Yes of course. that i now..

- anonymous

imagine, you divide through by U now. Then you'd have\[Y=\frac{1}{\sec x}\int\limits_{}{}\sec x \sec x dx=\cos x \tan x +c = \sin x +c\]

- anonymous

Okay

- anonymous

What is your Integration factor ?

- anonymous

sec x

- anonymous

same as before

- anonymous

Equals = 1/ cosx

- anonymous

as my.

- anonymous

You don't get the solution.

- anonymous

\[U=\sec x = \frac{1}{\cos x}\]

- anonymous

I found my wrong... I ve right... , at first when i divided with Cosx to get Dydx free... i, wrote dY+ sinx/cosx Y =cosx. it should be dY+ sinx/cosx Y =1\cosx.

- anonymous

Yes

- anonymous

ahhhh for god sake...

- anonymous

Thank you sooooooooooooooooooooooooo much!!!!! appreciate the help

- anonymous

Sorry, I saw that before - I thought you saw it too after the attachment was sent...lol, sorry.

- anonymous

that's fine. become a fan! ;)

- anonymous

Already one..

- anonymous

nice

- anonymous

;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.