(y-1) dx  +  x(x+1) dy = 0   stuck - Page 1
Forum section: electronics / beginners

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 08, 2015, 09:36:42 pm
content: Hi,Is there someone to solve this?answer : xy+1 = C (x+1)my answer C(x+1) = x(y+1)(y-1) dx + x(x+1) dy = 0x(x+1) dy = - (y-1) dx = (1-y) dxExact? no pcq partial (y-1) dx/partial dy = 01/x(x+1) dx + 1/(y-1) dy = 01/x * 1/(x+1) dx + 1/(y-1) dy = 0Separable?1/x * 1/(x+1) dx = - 1/(y-1) dy1/(x+1) dx = - x * dy/(y-1)1/(x+1) dx = x * dy/(y+1)u1 = x+1du1 = 1 dxu2 = y+1du2 = 1 dydu1/u1 = x* du2/u2ln|u1| + ln|c| = x * ln|u2| + ln|c|ln|u1| + ln|c| = x * ln|u2|C(x+1) = x(y+1)Non separable?Thanks,JP

author: SL4P
author_rank: Super Contributor
timestamp: July 08, 2015, 09:54:35 pm
content: Aren't you supposedto observe the (precedence) !?

author: onlooker
author_rank: Frequent Contributor
timestamp: July 08, 2015, 11:40:14 pm
content: Your C(x+1) = x(y+1) should really be C(x+1) = x(y-1). Then, it is the same as xy+1 = C (x+1), except your C is not their C (differ by 1).

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 01:04:16 am
content: answer : xy+1 = C (x+1)C(x+1) = x(y-1)C(x+1) = xy-yFor the minus signI thought- 1/(y-1) dy1/(y+1) dy

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: July 09, 2015, 08:25:23 am
content: Just separate the variablesdx/(x(x+1)) + dy/(y-1 ) = 0then 1/(x(x+1)) = 1/x-1/(x+1)thus by integratingLog(y-1) = C+Log(x+1) - Log(x)andy = C(1+1/x) +1

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 12:02:17 pm
content: Quote from: JacquesBBB on July 09, 2015, 08:25:23 amI don't get that x-1Quote

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: July 09, 2015, 12:32:32 pm
content: Quote from: J4e8a16n on July 09, 2015, 12:02:17 pmI don't get that x-1QuoteYou have to integrate dx/(x(x+1))for this the classical way is to decompose in simple elements. Here it is easy as we have only simple poles.and1/(x*(x+1)) = 1/x - 1/(x+1)to verify it, just compute by reduction to the same denominator1/x - 1/(x+1) = (x+1 -x)/(x*(x+1))Is it more clear ?

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 12:58:48 pm
content: 1/(x*(x+1)) = 1/x - 1/(x+1)why is there a negative sign?1/x * 1 / x+1do you meanln|x| + ln|x+1| +Cln|x| = - ln|x+1| +C?

author: Andy Watson
author_rank: Super Contributor
timestamp: July 09, 2015, 01:27:13 pm
content: Quote from: J4e8a16n on July 09, 2015, 12:58:48 pmIt's also know as partial fractions. Put 1/x and -1/(x+1) over a common denominator and add them together - you should arrive at the original equation 1/(x(x+1)).

author: IanB
author_rank: Super Contributor
timestamp: July 09, 2015, 02:50:11 pm
content: Quote from: J4e8a16n on July 09, 2015, 01:04:16 amYou really need to pay attention to details.x(y-1) = xy -xQuoteWhy would you think that?- (y-1) = -y - (-1) = -y + 1 = (1-y)Therefore:- 1/(y-1) = 1/(1-y)

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 07:25:03 pm
content: Thanks. So we are there..answer : xy+1 = C (x+1)my answer:x(1-y) = C(x+1)The idea is to find wich one of the methods solves the equation:Linear, Exact, Separable, Integrating factor, Continuous, BernouilliI think we have taken of Exact and separable.

author: IanB
author_rank: Super Contributor
timestamp: July 09, 2015, 07:56:45 pm
content: Quote from: J4e8a16n on July 09, 2015, 07:25:03 pmYou need to show that "xy + 1 = C(x+1)" is the same as "x(1-y) = C(x+1)". Can you do that?

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 09:52:43 pm
content: Well, both are equal to C(x+1), but I am not supposed to know the answer. ;o)Jean

author: IanB
author_rank: Super Contributor
timestamp: July 09, 2015, 09:59:58 pm
content: If your answer is different from the correct answer, then your answer is not correct.If your answer is not correct then you have not solved the problem.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 10, 2015, 12:42:59 am
content: Right

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: July 10, 2015, 08:44:19 am
content: Quote from: J4e8a16n on July 09, 2015, 07:25:03 pmI gave you the proper answer above (y = C(1+1/x) +1) and the way to compute it.this is equivalent (for x non zero ) to(y-1)x = C(1+x)Now you can take any value of the constant C. If we use C=-C',we obtain(1-y)x = C'(1+x)which is one of the solution you find.If you take C=C''-1You obtainxy+1 = C''(x+1)which is the other solution you mention. They are all solutions, and just correspond to different values of the constant.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 10, 2015, 01:10:40 pm
content: thank you so much for your patience.I get it.partial fraction