The Student Room Group

Differential equation help

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
1D61C4E2-4307-4254-A7CC-79B01E883D6A.jpeg
(edited 2 years ago)

Scroll to see replies

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
1D61C4E2-4307-4254-A7CC-79B01E883D6A.jpeg

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.....
Reply 2
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?
(edited 2 years ago)
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
Original post by mathperson
(1/3)e^3y = (1/4)e^4x + C


EDIT your post - we only give hints and not solutions
Reply 5
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().
(edited 2 years ago)
Reply 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(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
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)
Reply 9
Original post by mathperson
You've misread what I have written.
I used ln(ab) = ln(a) + ln(b)

line 1 to line 2?
Original post by mqb2766
line 1 to line 2?

What about it?

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

Example:
ln[(1/3)e^3y] = ln(1/3) + ln(e^3y)
Reply 11
Original post by mathperson
What about it?

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

Example:
ln[(1/3)e^3y] = ln(1/3) + ln(e^3y)

Yea your solution gives the mark scheme's answer
Original post by mathperson
What about it?

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.
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().

try reading it again
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.
Please reread my solution CAREFULLY.
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.
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
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.
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?
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.

Quick Reply

Latest