# Differential equation help

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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

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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#2

(Original post by

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

**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

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(Original post by

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.....

**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.....

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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#4

(Original post by

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?

**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?

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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#5

(Original post by

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

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

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#6

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

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(Original post by

(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

**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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#8

You realize that is incorrect

https://www.wolframalpha.com/input/?...exp%284x-3y%29

https://www.wolframalpha.com/input/?...exp%284x-3y%29

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#9

(Original post by

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().

**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().

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

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#11

(Original post by

line 1 to line 2?

**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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(Original post by

What about it?

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

Example:

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

**mathperson**)What about it?

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

Example:

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

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#13

**mathperson**)

What about it?

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

Example:

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

is not

ln(x) + ln(c)

That occurs before the 1/3 and 1/4 multipliers.

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#14

**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().

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#15

(Original post by

log(x + c)

is not

ln(x) + ln(c)

That occurs before the 1/3 and 1/4 multipliers.

**mqb2766**)log(x + c)

is not

ln(x) + ln(c)

That occurs before the 1/3 and 1/4 multipliers.

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.

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#16

**mqb2766**)

log(x + c)

is not

ln(x) + ln(c)

That occurs before the 1/3 and 1/4 multipliers.

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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#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

(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

**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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#18

(Original post by

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?

**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?

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#19

(Original post by

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.

**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.

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#20

(Original post by

Why don't you take ln of the whole of the RHS?

**user342**)Why don't you take ln of the whole of the RHS?

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