Help urgent integration

It's always safer to put the modulus sign, as in AQA, sometimes they let it slide, and other times they mark you down for it. So just to be on the cautious side, put the modulus in it. Since you can't input a negative number into lnx.

Also, regarding the +c. You can most definitely put +c instead of ln(k). In fact, most of the time with differential equations, it's nicer to put it in the form y = (a function). So doing +c is easier than ln(k).

Hope this helped!

It's always safer to put the modulus sign, as in AQA, sometimes they let it slide, and other times they mark you down for it. So just to be on the cautious side, put the modulus in it. Since you can't input a negative number into lnx.

Also, regarding the +c. You can most definitely put +c instead of ln(k). In fact, most of the time with differential equations, it's nicer to put it in the form y = (a function). So doing +c is easier than ln(k).

Hope this helped!

So if I put it as

lny = (function) + c
it would be fine for aqa

Yes. Although most of the time, they'd probably ask you to either work out this constant, or re-write the expression as y = function.

See attached to understand how to get y = function of x.
(You don't have to convert it into this form if they don't ask for it, but this form is a lot more useful when modelling differential equations) .

Agree with your comment about the
y = A e^(kx)
being more useful etc and you mention about negative values (y) in the previous post, but for a differential equation
dy/dx = k
y(0) = A (could be positive or negative), then doing a definite integral and seperation of variables you get to the usual
ln(|y|)-ln(|A\) = kx
or
ln(|y|/|A|) = ln(y/A) = kx
or
y = A e^kx
the sign of y (and A) is taken into consideration when you do the division (as both the numerator ..and denominator are the same sign) so taking exponetials (inverse logs) is ok for both positive and negative y.

If you did indefinite with + c, then youd get to the usual
ln(|y|) = kx + ln(|A|)
If y and A are positive then doing what you suggest is ok. If y and A are negative, youd have to write
ln(-y) = kx + ln(-A)
as this is what the equation says and taking exponentials
-y = -A e^kx
and obviously the - sign can be removed on both sides. Strictly speaking you cant just assume the A=e^c step works out when things are negative.

Realise the above may not even be worth a mark, but if you see a negative initial condition, be careful with the reasoning and treating it as a definite integral from 0..x and A..y. is probably safer.

Correct me if I'm misunderstanding your point, but isn't A just any constant? So it can refer to anything, regardless of its sign. For instance, if you had -e^c, can't you just call that entire term "A" instead of including the negative within the final expression?

I’m so lost 😭 which one is correct my a level exam is on Tuesday 😢😭😭😭

Correct me if I'm misunderstanding your point, but isn't A just any constant? So it can refer to anything, regardless of its sign. For instance, if you had -e^c, can't you just call that entire term "A" instead of including the negative within the final expression?

Not sure what you mean. Im trying to say that treating it as an indefinite up to
ln(|y|) = kx + c
is ok, and obviously if y is positive its straightforward. Similarly writing it as
y = A e^(kx)
is the "best", A represents y(0) so is effectively the constant of integration and this works for both positive and negative y and A.

If y and A are negative, then doing the indefinite gives
ln(|A|) = c
or |A| = e^c. As A is negative you have to remember to write it as
A = -e^c
if youve determined c in the first place. So you have to be careful when/where you remove the abs function.

To me, it seems fiddly and easier just to consider the definite integral from
0 to x and from A to y and it all pretty much works out as the abs gets removed when you do the division inside the log as y and A have the same sign.

Do you think AQA care about sign changes so long as you define a new constant? Because then, you don't have to worry about modulus or anything.

Do you think AQA care about sign changes so long as you define a new constant? Because then, you don't have to worry about modulus or anything.