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differential equations

mickymouse1995

hey how do you differentiate this ??

dn/dt = (1/5000)*(n(500-N))

THANK YOU XXX

Reply 1

BabyMaths

Original post by mickymouse1995

hey how do you differentiate this ??

dn/dt = (1/5000)*(n(500-N))

THANK YOU XXX

You don't differentiate. You integrate.

\int \frac{5000}{n(500-n)} dn = \int dt

I'm assuming n and N were supposed to both be n.

Next use partial fractions.

Reply 2

mickymouse1995

i havent done partial fractions could you run through this question with me thanks
yes im sorry typo!!

Reply 3

Indeterminate

Original post by mickymouse1995

hey how do you differentiate this ??

dn/dt = (1/5000)*(n(500-N))

THANK YOU XXX

To generalise the above, if you have

\dfrac{dy}{dx} =f(x)\cdot g(y)

,

with f and g both non-zero, you can solve this by the method of separating variables.

We get

\displaystyle \int \dfrac{1}{g(y)} dy = \int f(x) dx

Reply 4

mickymouse1995

it isnt showing what you get!!

Reply 5

BabyMaths

Original post by mickymouse1995

i havent done partial fractions could you run through this question with me thanks
yes im sorry typo!!

Not really. I'm not going to try to teach from scratch on an internet forum.

We can easily give hints and little reminders but it sounds like you need to talk to your teacher. Maybe ask why they set work needing partial fractions when they haven't taught you about partial fractions.

On the other hand, it will be explained well in your textbook. You could teach yourself.

ok thank you xx

Original post by mickymouse1995

i havent done partial fractions could you run through this question with me thanks
yes im sorry typo!!

\dfrac{5000}{n(500-n)} = \dfrac{A}{n} + \dfrac{B}{500-n}

Now add the fractions on the RHS and equate the numerators (5000=...) Then substitute, in turn, values of n that would make each of n and 500-n go to zero. This will give you A and B.