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Differential Eqautions

(-15)dV/dt=2V-450 Given that the volume initially is 300cm3 find the volume after 15seconds. Ok did the following: dt/dV=15/(450-2V) So c=15/2ln150 and when I do t=15/2ln150-15/2ln(450-2V) for t=15 I got V=214.8 but the answer say the actual expression should have been t=15/2ln150+15/2ln(450-2V) and the answer comes to 235. What did I do wrong I cant see is it which side I put the constant but that should not change the answer depending on what I have read....

Reply 1

IamDreaming

I got 235 ill show my working

Reply 2

IamDreaming

Original post by Merdan

I juat need to build up my rep so here you gp if you need anymore advise or help pm its not a problem look at the attachment for working

(edited 10 years ago)

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Reply 3

Merdan

Original post by IamDreaming

I juat need to build up my rep so here you gp if you need anymore advise or help pm its not a problem look at the attachment for working

When I did it I multiplied both sides by -1 to get 15dV/dt=450-2V I think that is the think that foul me cuz I got e^2(2V-450)= - 150. Algebraically I don't see an error in doing that but it doesn't give a write answer somehow....

Reply 4

IamDreaming

Okay theres an error I check if u x multiply by -1 you get -15/2ln(450-2v)
When u put it into c you get -15/2ln(-150) which is maths error or u could use it later technically if are good at algerbra you get 15/2ln(-150/450-2v) so it will reverse back to 15/2ln(150/2v-450) anyway

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Reply 5

Indeterminate

Original post by IamDreaming

I juat need to build up my rep so here you gp if you need anymore advise or help pm its not a problem look at the attachment for working

Note that, on this forum, we're not allowed to post full solutions (purely for the benefit of those we're helping).

Reply 6

Merdan

Original post by IamDreaming

Yes I see what you mean I assume it is an error for this particular question hence when I integrate it is actually -15/2ln|-150| modulus -150 which is 150 and in many questions in finding the areas under and whatever when you put in a value in ln() and get negative you always put it as positive due to modulus so I put it as -15/2ln150 which means when I do that I wont get 15/2ln(-150/450-2v) instead I will get 15/2(ln150/450-2v) which wont reverse to the correct form...

Reply 7

Merdan

Original post by Indeterminate

Note that, on this forum, we're not allowed to post full solutions (purely for the benefit of those we're helping).

I already had the solution in front of me before him posting and my question was not an full solution to the question(which I already had) but rather a reason why I went wrong while I was doing it without looking at the actual answer lol....

Reply 8

IamDreaming

When was that a rule this isnt exam anwsers ? Its like exams solutions should it be taken down , where does it say that ? Just wondering becuase I haven't heard of that rule

Reply 9

Merdan

Original post by IamDreaming

When was that a rule this isnt exam anwsers ? Its like exams solutions should it be taken down , where does it say that ? Just wondering becuase I haven't heard of that rule

When you are integrating partial fractions especially you usually get it in logarithm form such as ln|x| and not ln(x) . The difference is when you put your x value in when working out a specific solution to a question when you put your limits you sometimes get negative overall value in ln() and you don't ignore that term but rather take the positive of what you get that is the purpose of modulus otherwise there would be no need?

Reply 10

Indeterminate

Original post by Merdan

Regardless of whether you had the correct solution before he posted it, it is important for him to know this so he doesn't do it again in a situation where the person seeking help has no clue.

Reply 11

IamDreaming

Original post by Indeterminate

Regardless of whether you had the correct solution before he posted it, it is important for him to know this so he doesn't do it again in a situation where the person seeking help has no clue.

Ohhh now I get what you mean but yeah sometimes seeing the solution helps them understand the mistakes otherwise markschemes and past papers would be useless if the purpose wasnt to do the question and chexk the anwser to see if it was correct or how the mistake occured