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

- anonymous

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

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- 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

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

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- 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

;)

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