(y-1) dx + x(x+1) dy = 0 stuck

[J4e8a16n]

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

[SL4P]

Aren't you supposed to observe the (precedence) !?

[onlooker]

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).

[J4e8a16n]

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

[JacquesBBB]

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

[J4e8a16n]

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)

[JacquesBBB]

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 ?

[J4e8a16n]

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
?

[Andy Watson]

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)).

[IanB]

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)

[J4e8a16n]

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.

[IanB]

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?

[J4e8a16n]

Well, both are equal to C(x+1),  but I am not supposed to know the answer.  ;o)


Jean

[IanB]

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.

[J4e8a16n]

Right

[JacquesBBB]

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.

[J4e8a16n]

thank you so much for your patience.

I get it.

partial fraction