Differential equations (1 Viewer)

[Affinity]

I asked my teacher if the following would be ok in exams, he answered: it might cost me marks. Could anyone give their opinion?

given dN/dt= k(N-B) find N in terms of t.

dN/dt= k(N-B)

1/(N-B) dN = k dt

/
| 1/(N-B) dN =
/

/
| k dt
/


ln | N - B | = kt + C

N - B = e^(kt + C)
N = Ae^kt + B For some constant A.

 

[Lazarus]

You'd only lose marks if the question had specifically asked you to employ some other method. I can't see anything wrong with what you've done; it's the standard process for backtracking from a differential equation.

 

[underthesun]

thats a pretty nice method, i'm going to ask the same thing to my teacher as well.

 

[kini mini]

That seems perfectly valid to me, in fact it's the usual on as Laz said.

Why would you lose marks for it? Did your teacher explain?

 

[Harimau]

Originally posted by Affinity
I asked my teacher if the following would be ok in exams, he answered: it might cost me marks. Could anyone give their opinion?

given dN/dt= k(N-B) find N in terms of t.

dN/dt= k(N-B)

1/(N-B) dN = k dt

/
| 1/(N-B) dN =
/

/
| k dt
/


ln | N - B | = kt + C

N - B = e^(kt + C)
N = Ae^kt + B For some constant A.

My tutor taught me the same thing, and i think it might actually even be considered superior compared to the other method of guessing the answer and differentiating it...

 

[Affinity]

Thanks everyone

 

[fatmuscle]

I'll try post up the other method, if I find it