anonymous
  • anonymous
(Cosx )dY +(sinx)Y=1 anyone who can solve it?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I first divide with cosx. So i get dY+ sinx/cosx Y =cosx..
anonymous
  • 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
  • anonymous
\[e ^{\int\limits_{}^{}sinx \div cosx} = e ^{-\ln(cosx)} = e^{\ln(cosx)^{-1}} = cosx ^{-1}\]

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anonymous
  • 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
  • anonymous
Anyone no were i did the wrong?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Umm...I'll write out a solution, scan and attach. You're on the right track, though.
anonymous
  • anonymous
Thank you!
anonymous
  • anonymous
Can you wait a bit? I want to make some tea.
anonymous
  • anonymous
Np!
anonymous
  • anonymous
Scanning now...
anonymous
  • anonymous
okeej! i dont see... anywrong.. hm.. but.. were is the wrong.
anonymous
  • anonymous
1 Attachment
anonymous
  • 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
  • anonymous
Okej, so i can t do like i have done?..
anonymous
  • 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
  • anonymous
Yes, the symbol U is the as the integrating factor
anonymous
  • anonymous
brb
anonymous
  • anonymous
ok
anonymous
  • anonymous
What i did, was that used U =e^ (int (Sinx\cosx))
anonymous
  • anonymous
or you could right int ( tan x) like you did.
anonymous
  • anonymous
Yes, I saw.
anonymous
  • 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
  • anonymous
Then\[y=\frac{1}{\mu}\int\limits_{}{}\mu q(x) dx+\frac{c}{\mu}\]
anonymous
  • anonymous
Were you jumping to that formula to work it out?
anonymous
  • anonymous
Yes Exactly
anonymous
  • anonymous
That was, what i was trying to do.
anonymous
  • 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
  • anonymous
okeey, i see ..
anonymous
  • anonymous
i must have done wrong with the e^∫sinx÷cosx
anonymous
  • anonymous
The integration factor
anonymous
  • anonymous
What happens when you try it again?
anonymous
  • anonymous
The forumla is not wrong? huh? I uses this formula, \[U \times Y =\int\limits_{}^{}U \times q(x)\]
anonymous
  • anonymous
Where, DyDx + r(x)Y= q(x)
anonymous
  • 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
  • anonymous
For example
anonymous
  • anonymous
Yes of course. that i now..
anonymous
  • 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
  • anonymous
Okay
anonymous
  • anonymous
What is your Integration factor ?
anonymous
  • anonymous
sec x
anonymous
  • anonymous
same as before
anonymous
  • anonymous
Equals = 1/ cosx
anonymous
  • anonymous
as my.
anonymous
  • anonymous
You don't get the solution.
anonymous
  • anonymous
\[U=\sec x = \frac{1}{\cos x}\]
anonymous
  • 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
  • anonymous
Yes
anonymous
  • anonymous
ahhhh for god sake...
anonymous
  • anonymous
Thank you sooooooooooooooooooooooooo much!!!!! appreciate the help
anonymous
  • anonymous
Sorry, I saw that before - I thought you saw it too after the attachment was sent...lol, sorry.
anonymous
  • anonymous
that's fine. become a fan! ;)
anonymous
  • anonymous
Already one..
anonymous
  • anonymous
nice
anonymous
  • anonymous
;)

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