Math Is Fun Forum

glenn101

Member

Registered: 2008-04-02

Posts: 108

c constant with applications of differential equations

Hi all, I'm getting frustrated with dealing with applications of differential equations.
With regards to when to add in the c constant, and how you are to go about manipulating it.
Take for example this applications question;

A city with population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a) i) Set up a DE to describe this situation;
Let P= Population
    t= time (years)
dP/dt proportional to p
hence Dp/dt=kP where k>0.

ii) Solve to obtain a general solution.
dP/dt=kP
dt/dP=1/kP
t=∫1/kP dP
this is where the problem arises;I want to solve the integral for P which will help with later questions.
My teacher goes about it this way.

t= 1/k integral(1/P) dP
tk= loge|P|+c      oh and if you solve for t here would you get; t=1/kloge|P|+c/k?
tk-c=loge|P|
e^(tk-c)=P

Alternatively I tried this approach;
t=1/k∫(1/P) dP
t=1/kloge|P|+c
t-c=1/kloge|P|
k(t-c)=loge|P|
e^k(t-c)=P
which is correct? and why?

I have solved for P because the next question asks to find the population after x years given other data.

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: c constant with applications of differential equations

the two equations are exactly the same in the end; both C and K are both constants, so the product CK is also a constant.

thus e^k(t-c) is essentialy exactly the same as e^(tk-c), only that c has a different value because it is not multiplied by k

You could say that your teachers solution is simplified.

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In the same way, you might end up with an integration where when you manipulate it you end with a constant like:

I(x) = f(x) + 4.5C

which ofcourse, you can equally just write as

I(x) + f(x) + C'

the C is written with the apostrophe here just to explicitly show that it is not the same constant as the first equation, but both are correct

Last edited by luca-deltodesco (2009-07-17 20:48:14)

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glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: c constant with applications of differential equations

the two equations are exactly the same in the end; both C and K are both constants, so the product CK is also a constant.

thus e^k(t-c) is essentialy exactly the same as e^(tk-c), only that c has a different value because it is not multiplied by k

You could say that your teachers solution is simplified.

--

In the same way, you might end up with an integration where when you manipulate it you end with a constant like:

I(x) = f(x) + 4.5C

which ofcourse, you can equally just write as

I(x) + f(x) + C'

the C is written with the apostrophe here just to explicitly show that it is not the same constant as the first equation, but both are correct

Thanks for the reply luca,
it seems you are right that they end to be the same, however, I discovered that plugging values into e^(tk-c) and e^k(t-c) gave different results, they only came out to be the same when I solved for P after finding values for k and c. Any ideas?

for e^k(t-c) I have found that k= 1/2loge(11/10) and c= -6loge(10)/loge(11/10)
for e^(tk-c) I have found k=1/2loge(11/10) and c=-3loge(10)
actually never mind your absolutely right.
Thanks ever so much:D

Last edited by glenn101 (2009-07-18 17:07:13)

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"If your going through hell, keep going."