EEVblog® Electronics Community Forum
Electronics => Beginners => Topic started by: J4e8a16n on July 08, 2015, 09:36:42 pm
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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 = 0
x(x+1) dy = - (y-1) dx = (1-y) dx
Exact? no pcq partial (y-1) dx/partial dy = 0
1/x(x+1) dx + 1/(y-1) dy = 0
1/x * 1/(x+1) dx + 1/(y-1) dy = 0
Separable?
1/x * 1/(x+1) dx = - 1/(y-1) dy
1/(x+1) dx = - x * dy/(y-1)
1/(x+1) dx = x * dy/(y+1)
u1 = x+1
du1 = 1 dx
u2 = y+1
du2 = 1 dy
du1/u1 = x* du2/u2
ln|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
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Aren't you supposed to observe the (precedence) !?
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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).
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answer : xy+1 = C (x+1)
C(x+1) = x(y-1)
C(x+1) = xy-y
For the minus sign
I thought
- 1/(y-1) dy
1/(y+1) dy
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Just separate the variables
dx/(x(x+1)) + dy/(y-1 ) = 0
then 1/(x(x+1)) = 1/x-1/(x+1)
thus by integrating
Log(y-1) = C+Log(x+1) - Log(x)
and
y = C(1+1/x) +1
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Just separate the variables
dx/(x(x+1)) + dy/(y-1 ) = 0
then 1/(x(x+1)) = 1/x-1/(x+1)
thus by integrating
Log(y-1) = C+Log(x+1) - Log(x)
and
y = C(1+1/x) +1
I don't get that x-1 then 1/(x(x+1)) = 1/x-1/(x+1)
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I don't get that x-1 then 1/(x(x+1)) = 1/x-1/(x+1)
[/quote]
You 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.
and
1/(x*(x+1)) = 1/x - 1/(x+1)
to verify it, just compute by reduction to the same denominator
1/x - 1/(x+1) = (x+1 -x)/(x*(x+1))
Is it more clear ?
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1/(x*(x+1)) = 1/x - 1/(x+1)
why is there a negative sign?
1/x * 1 / x+1
do you mean
ln|x| + ln|x+1| +C
ln|x| = - ln|x+1| +C
?
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1/(x*(x+1)) = 1/x - 1/(x+1)
why is there a negative sign?
It'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)).
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answer : xy+1 = C (x+1)
C(x+1) = x(y-1)
C(x+1) = xy-y?
You really need to pay attention to details.
x(y-1) = xy - x
For the minus sign
I thought
- 1/(y-1) dy
1/(y+1) dy
Why would you think that?
- (y-1) = -y - (-1) = -y + 1 = (1-y)
Therefore:
- 1/(y-1) = 1/(1-y)
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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, Bernouilli
I think we have taken of Exact and separable.
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Thanks. So we are there..
answer : xy+1 = C (x+1)
my answer:x(1-y) = C(x+1)
You need to show that "xy + 1 = C(x+1)" is the same as "x(1-y) = C(x+1)". Can you do that?
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Well, both are equal to C(x+1), but I am not supposed to know the answer. ;o)
Jean
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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.
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Right
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Thanks. So we are there..
answer : xy+1 = C (x+1)
my answer:x(1-y) = C(x+1)
I 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''-1
You obtain
xy+1 = C''(x+1)
which is the other solution you mention. They are all solutions, and just correspond to different values of the constant.
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thank you so much for your patience.
I get it.
partial fraction