# Differential equation help

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#1
Can somebody plz explain how I get from what I've done below to the yellow answer?
I took ln of both sides but don't think I'm doing it right.
Thanks
Last edited by Rhys_M; 1 month ago
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1 month ago
#2
(Original post by Rhys_M)
Can somebody plz explain how I get from what I've done below to the yellow answer?
I took ln of both sides but don't think I'm doing it right.
Thanks
On the line where you have (1/3)e^3y = (1/4)e^4x + C (the line second from bottom), try taking natural logarithms (ln) the get rid of the exponentials.....
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#3
(Original post by mathperson)
On the line where you have (1/3)e^3y = (1/4)e^4x + C (the line second from bottom), try taking natural logarithms (ln) the get rid of the exponentials.....
But doesn't that give 1/3(3y) = 1/4(4x) + c So y=x+c?
Or am I taking ln wrong - do you have to do something with the coefficients too?
Last edited by Rhys_M; 1 month ago
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1 month ago
#4
(Original post by Rhys_M)
But doesn't that give 1/3(3y) = 1/4(4x) + c So y=x+c?
Or am I taking ln wrong - do you have to do something with the coefficients too?
(1/3)e^3y = (1/4)e^4x + C

ln[(1/3)e^3y] = ln[(1/4)e^4x] + ln[C]

ln(1/3) + ln(e^3y) = ln(1/4) + ln(e^4x) + ln(C)

ln(1/3) + 3y = ln(1/4) + 4x + ln(C)

3y = 4x + [ln(C) + ln(1/4) - ln(1/3)]

3y = 4x + C

y = (4/3)x + C
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1 month ago
#5
(Original post by mathperson)
(1/3)e^3y = (1/4)e^4x + C
EDIT your post - we only give hints and not solutions
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1 month ago
#6
(Original post by mathperson)
(1/3)e^3y = (1/4)e^4x + C

ln[(1/3)e^3y] = ln[(1/4)e^4x] + ln[C]
...
ln(a+b)
is not
ln(a)+ln(b).
on the right hand side. You can't simplify it down to a linear relationship, the answer y= ... will involve ln() and exp().
Last edited by mqb2766; 1 month ago
1
#7
(Original post by mathperson)
(1/3)e^3y = (1/4)e^4x + C

ln[(1/3)e^3y] = ln[(1/4)e^4x] + ln[C]

ln(1/3) + ln(e^3y) = ln(1/4) + ln(e^4x) + ln(C)

ln(1/3) + 3y = ln(1/4) + 4x + ln(C)

3y = 4x + [ln(C) + ln(1/4) - ln(1/3)]

3y = 4x + C

y = (4/3)x + C
Thank you so much
1
1 month ago
#8
You realize that is incorrect
https://www.wolframalpha.com/input/?...exp%284x-3y%29
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1 month ago
#9
(Original post by mqb2766)
ln(a+b)
is not
ln(a)+ln(b).
on the right hand side. You can't simplify it down to a linear relationship, the answer y= ... will involve ln() and exp().
You've misread what I have written.
I used ln(ab) = ln(a) + ln(b)
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1 month ago
#10
(Original post by mathperson)
You've misread what I have written.
I used ln(ab) = ln(a) + ln(b)
line 1 to line 2?
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1 month ago
#11
(Original post by mqb2766)
line 1 to line 2?

I used ln(ab) = ln(a) + ln(b)

Example:
ln[(1/3)e^3y] = ln(1/3) + ln(e^3y)
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#12
(Original post by mathperson)

I used ln(ab) = ln(a) + ln(b)

Example:
ln[(1/3)e^3y] = ln(1/3) + ln(e^3y)
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1 month ago
#13
(Original post by mathperson)

I used ln(ab) = ln(a) + ln(b)

Example:
ln[(1/3)e^3y] = ln(1/3) + ln(e^3y)
log(x + c)
is not
ln(x) + ln(c)
That occurs before the 1/3 and 1/4 multipliers.
0
1 month ago
#14
(Original post by mqb2766)
ln(a+b)
is not
ln(a)+ln(b).
on the right hand side. You can't simplify it down to a linear relationship, the answer y= ... will involve ln() and exp().
1
1 month ago
#15
(Original post by mqb2766)
log(x + c)
is not
ln(x) + ln(c)
That occurs before the 1/3 and 1/4 multipliers.
Oh for goodness sake, am I really having to explain this to you?!

I did not say that ln(a+b) = ln(a) + ln(b).
I used ln(ab) = ln(a) + ln(b).

I am not going to go around in circles with you for the rest of the day.
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1 month ago
#16
(Original post by mqb2766)
log(x + c)
is not
ln(x) + ln(c)
That occurs before the 1/3 and 1/4 multipliers.
you do know how to read, correct? His solution literally STARTS with the 1/3 and 1/4 multipliers, so how can he make an error before them?

Also, he never once converts log(x+c) into log(x) + log(c). Someone also confirmed the mark scheme gives the exact same answer.
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1 month ago
#17
Not my thread, but I saw the question and thought I'd give it a go! I'm confused as to where the ln(C) comes in in line 2?

(Original post by mathperson)
(1/3)e^3y = (1/4)e^4x + C

ln[(1/3)e^3y] = ln[(1/4)e^4x] + ln[C]

ln(1/3) + ln(e^3y) = ln(1/4) + ln(e^4x) + ln(C)

ln(1/3) + 3y = ln(1/4) + 4x + ln(C)

3y = 4x + [ln(C) + ln(1/4) - ln(1/3)]

3y = 4x + C

y = (4/3)x + C
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1 month ago
#18
(Original post by user342)
Not my thread, but I saw the question and thought I'd give it a go! I'm confused as to where the ln(C) comes in in line 2?
So essentially you have the +C term from the original integration, and then a load of constants (natural logarithm of a number), so you collect all those together and then just make the term equal to 'C' - some constant.
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1 month ago
#19
(Original post by mathperson)
So essentially you have the +C term from the original integration, and then a load of constants (natural logarithm of a number), so you collect all those together and then just make the term equal to 'C' - some constant.
Why don't you take ln of the whole of the RHS?
1
1 month ago
#20
(Original post by user342)
Why don't you take ln of the whole of the RHS?
because you will then have what we were arguing about earlier, a log(x+c) term which you really do not want. Your end goal is to get an equation like y=mx+c - something ideally without an "unsimplifiable" log term. log(x+c) is really useless to have and unless theres no other way, you should avoid them.
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