Solving Differential Equations using A=+/- e^...

[acinod]

VCAA 2010 Exam 1 Question 2b:
Solve dv/dt =(v-4)/2 to find v as a function of t.

Now we the next step is dt/dv=2/(v-4) and then we can have v in t with an arbitary constant 'c' at the end. To solve the equation for v, I would always find v in terms of t so v=Ae^(t/2) where A=+/-e^(-c/2). In the end I would still get the right answer but after reading the Assessment Report, I'm not sure I should do it this way any more.

The Assessment Report says:

Some used A=e^-ct in their working and then found that A=-4, which is not consistent.

What does this mean? I found A=-4 but my A was +/-e^(c/2) not e^(-c/2). Could someone clarify this for me?

Thanks.

[dc302]

VCAA 2010 Exam 1 Question 2b:
Solve dv/dt =(v-4)/2 to find v as a function of t.

Now we the next step is dt/dv=2/(v-4) and then we can have v in t with an arbitary constant 'c' at the end. To solve the equation for v, I would always find v in terms of t so v=Ae^(t/2) where A=+/-e^(-c/2). In the end I would still get the right answer but after reading the Assessment Report, I'm not sure I should do it this way any more.

The Assessment Report says:

Some used A=e^-ct in their working and then found that A=-4, which is not consistent.

What does this mean? I found A=-4 but my A was +/-e^(c/2) not e^(-c/2). Could someone clarify this for me?

Thanks.

I'm not really sure what they're talking about either, but I assume it's because they had both the unknowns A and c in their solution, whereas you're only supposed to have one unknown.

edit: nvm, totally misread what vcaa said 

edit2: ok, I think I see their point. If you use A = e^whatever, and find that A = -4, then that is impossible because you can't get a negative number with an exponential (ie you would get loge(-4) = ... etc).

[acinod]

I'm not really sure what they're talking about either, but I assume it's because they had both the unknowns A and c in their solution, whereas you're only supposed to have one unknown.

edit: nvm, totally misread what vcaa said 

edit2: ok, I think I see their point. If you use A = e^whatever, and find that A = -4, then that is impossible because you can't get a negative number with an exponential (ie you would get loge(-4) = ... etc).

But my A could be positive or negative because I carried the +/- produced by the modulus of v-4. Is my solution still acceptable since VCAA might just be picking on those kids that ignored the modulus or something?

[Lord of Tzeentch]

Your probably better just solving it the long-winded way.
You don't want to make a careless error (could happen the long-winded way, but still).

[dc302]

I'm not really sure what they're talking about either, but I assume it's because they had both the unknowns A and c in their solution, whereas you're only supposed to have one unknown.

edit: nvm, totally misread what vcaa said 

edit2: ok, I think I see their point. If you use A = e^whatever, and find that A = -4, then that is impossible because you can't get a negative number with an exponential (ie you would get loge(-4) = ... etc).

But my A could be positive or negative because I carried the +/- produced by the modulus of v-4. Is my solution still acceptable since VCAA might just be picking on those kids that ignored the modulus or something?

Yeah your solution should be fine as long as you don't state something like -4 = e^c  etc.

[acinod]

Your probably better just solving it the long-winded way.
You don't want to make a careless error (could happen the long-winded way, but still).

I don't know how to do it the long-winded way.

Like when I get to the absolute sign, I don't know what to do with it. I might have an idea of whether to take the positive or negative but I don't know how to explain it and it just gets really messy. I've tried it and my teacher says what I wrote is mathematically incorrect and I might as well do my original way.