Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Anyone no were i did the wrong?

Answer 6

Yes

Answer 7

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

Answer 8

Thank you!

Answer 9

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

Answer 10

Np!

Answer 11

Scanning now...

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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?

Answer 16

Yes, the symbol U is the as the integrating factor

Answer 17

brb

Answer 18

ok

Answer 19

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

Answer 20

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

Answer 21

Yes, I saw.

Answer 22

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\[\mu = e^{\int\limits_{}{} p(x)dx}\]

Answer 23

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

Answer 24

Were you jumping to that formula to work it out?

Answer 25

Yes Exactly

Answer 26

That was, what i was trying to do.

Answer 27

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

Answer 28

okeey, i see ..

Answer 29

i must have done wrong with the
e^∫sinx÷cosx

Answer 30

The integration factor

Answer 31

What happens when you try it again?

Answer 32

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

Answer 33

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

Answer 34

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.

Answer 35

For example

Answer 36

Yes of course. that i now..

Answer 37

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

Answer 38

Okay

Answer 39

What is your Integration factor ?

Answer 40

sec x

Answer 41

same as before

Answer 42

Equals = 1/ cosx

Answer 43

as my.

Answer 44

You don't get the solution.

Answer 45

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

Answer 46

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.

Answer 47

Yes

Answer 48

ahhhh for god sake...

Answer 49

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

Answer 50

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

Answer 51

that's fine. become a fan! ;)

Answer 52

Already one..

Answer 53

nice

Answer 54

;)