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anonymous
 5 years ago
(Cosx )dY +(sinx)Y=1 anyone who can solve it?
anonymous
 5 years ago
(Cosx )dY +(sinx)Y=1 anyone who can solve it?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I first divide with cosx. So i get dY+ sinx/cosx Y =cosx..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[e ^{\int\limits_{}^{}sinx \div cosx} = e ^{\ln(cosx)} = e^{\ln(cosx)^{1}} = cosx ^{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1\div \cos x \times Y =\int\limits_{}^{} (1\div cosx) \times cox x\] \[1 \div cosx \times Y =\int\limits 1\] \[1 \div cosx \times Y =x +c\] \[Y= xcos(x) + Ccos(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyone no were i did the wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Umm...I'll write out a solution, scan and attach. You're on the right track, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you wait a bit? I want to make some tea.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okeej! i dont see... anywrong.. hm.. but.. were is the wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okej, so i can t do like i have done?..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, the symbol U is the as the integrating factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What i did, was that used U =e^ (int (Sinx\cosx))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or you could right int ( tan x) like you did.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0Then\[y=\frac{1}{\mu}\int\limits_{}{}\mu q(x) dx+\frac{c}{\mu}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Were you jumping to that formula to work it out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That was, what i was trying to do.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{1}{\sec x}\int\limits_{}{}\sec x \sec x dx+\frac{c}{\sec x}=\cos x \int\limits_{}{}\sec^2x dx+c \cos x\]\[=\cos x \tan x + c \cos x=\sin x + c \cos x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i must have done wrong with the e^∫sinx÷cosx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integration factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What happens when you try it again?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The forumla is not wrong? huh? I uses this formula, \[U \times Y =\int\limits_{}^{}U \times q(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where, DyDx + r(x)Y= q(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The formula will give you the same answer AS LONG AS you integrate the righthand side BEFORE dividing by the U on the lefthand 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes of course. that i now..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0imagine, 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
 5 years ago
Best ResponseYou've already chosen the best response.0What is your Integration factor ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't get the solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[U=\sec x = \frac{1}{\cos x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhh for god sake...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you sooooooooooooooooooooooooo much!!!!! appreciate the help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I saw that before  I thought you saw it too after the attachment was sent...lol, sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's fine. become a fan! ;)
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