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# I'm so sorry - another C4 integration question watch

1. OK ive done something so wrong here bit i cant see it:

Find a particular solution:
(1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

So i did:
dy/dx = (1 + x)y/(1 - x^2)
INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
INT 1/y dy = INT 1/(1-x) dx
ln|y| = ln|1-x| + ln|k|
ln|y| = ln|k(1-x)|
y = k(1 - x)

6 = 0.5k
k = 12

y = 12(1-x)

What did i do wrong? (coz that is not the right answer by a long way)
2. (Original post by joker_900)
OK ive done something so wrong here bit i cant see it:

Find a particular solution:
(1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

So i did:
dy/dx = (1 + x)y/(1 - x^2)
INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
INT 1/y dy = INT 1/(1-x) dx
ln|y| = ln|1-x| + ln|k|
ln|y| = ln|k(1-x)|
y = k(1 - x)

6 = 0.5k
k = 12

y = 12(1-x)

What did i do wrong? (coz that is not the right answer by a long way)
int 1/(1-x) dx = - ln(1-x) not ln(1-x)
as d(1-x)/dx = -1 not 1
3. that would make the answer
y = k/(1-x)
find k... and u should get that answer u have..
4. ahhhhhhh thx a lot at least it wasnt some fundamental flaw

cheers
5. (Original post by joker_900)
ahhhhhhh thx a lot at least it wasnt some fundamental flaw

cheers
well you should always (in such cases) chech for the SIGN of x too!
6. (Original post by joker_900)
OK ive done something so wrong here bit i cant see it:

Find a particular solution:
(1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

So i did:
dy/dx = (1 + x)y/(1 - x^2)
INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
INT 1/y dy = INT 1/(1-x) dx
ln|y| = ln|1-x| + ln|k|
ln|y| = ln|k(1-x)|
y = k(1 - x)

6 = 0.5k
k = 12

y = 12(1-x)

What did i do wrong? (coz that is not the right answer by a long way)

Why did you do + ln(k) ??

isnt it just + k ?
7. (Original post by __WiZaRD__)
Why did you do + ln(k) ??

isnt it just + k ?
well lnk is a constant too.. but there is a ln in the integral... it's easier to make it lnk so as to put the k in the other ln.

ln|y| = ln|1-x| + ln|k|
ln|y| = ln|k(1-x)|

if it was
ln|y| = ln|1-x| + k
you will have to find k in the ln form. so putting the constant as lnk is easier to deal with...
8. but surely it would give you a completely different answer if youdo this? how can you just put it in ln form?
9. k is a constant that can be anything, so putting it as lnk doesnt make any difference, it just changes the value of k. it makes life much easier since your dealing with logs.

if you left it as
ln|y| = ln|1-x| + k:

'e' both sides
y= (1-x)*e^k
y= e^k - xe^k
= constant - (constant)x

compare this with
ln|y| = ln|1-x| + lnk = ln|k(1-x)|
y= k(1-x) = (constant) - (constant)x which is the same thing and arrived at much quicker.
10. (Original post by __WiZaRD__)
but surely it would give you a completely different answer if youdo this? how can you just put it in ln form?
no it would give u the same answer!
try it!

ln|y| = -ln|1-x| + ln|k|
y = k/(1-x)
6 = k/(0.5)
k = 3
y = 3/(1-x)

and

ln|y| = -ln|1-x| + C
ln6 = ln(1/0.5) + C
ln6 = ln2 + C
ln6 - ln2 = C
C = ln3

lny = ln(1/1-x) + ln3
lny = ln(3/(1-x))
y = 3/(1-x)

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Updated: June 13, 2006
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