Author Topic: Differential Equations (Read 10189 times) Tweet Share

Damo17 · « **Reply #15 on:** June 29, 2009, 12:17:56 pm »

Quote from: d0minicz on June 29, 2009, 12:07:01 pm

Find the general solution for:
$\frac {dy}{dx} = cos^2 y$
ty

$\frac{dy}{dx}= cos^2 y$

$\frac{dx}{dy}=\frac{1}{cos^2 y} = sec^2 y$

$x= \int sec^2 y \ dy = tan y +c$

$\therefore$ $y= tan^{-1} (x-c)$

dcc · « **Reply #16 on:** June 29, 2009, 01:12:47 pm »

Quote from: d0minicz on June 29, 2009, 12:07:01 pm

Find the general solution for:
$\frac {dy}{dx} = cos^2 y$

Or one could use seperation of variables to tackle this ( $\int \dfrac{dy}{\cos^2 y} = \int dx$ noting that we require $\cos y \neq 0$ ).

d0minicz · « **Reply #17 on:** June 29, 2009, 04:58:23 pm »

The rate of decay of a radioactive substance is proportional to the amoutn of Q of matter present at any time, t. The differntial equation for this situation is $\frac {dQ}{dt} = -kQ$ where k is a constant. If Q=50 when t=0 and Q=25 when t=10 , find the time taken for Q to reach 10.
thanks

Damo17 · « **Reply #18 on:** June 29, 2009, 05:18:52 pm »

Quote from: d0minicz on June 29, 2009, 04:58:23 pm

The rate of decay of a radioactive substance is proportional to the amoutn of Q of matter present at any time, t. The differntial equation for this situation is $\frac {dQ}{dt} = -kQ$ where k is a constant. If Q=50 when t=0 and Q=25 when t=10 , find the time taken for Q to reach 10.
thanks

$\frac{dt}{dQ}= -\frac{1}{kQ}$ $\therefore$ $t= -\frac{1}{k} log_e Q +c$

when $t=0$ , $Q=50$ $\therefore$ $c= \frac{1}{k} log_e 50$

$t= \frac{1}{k} log_e \frac{50}{Q}$

When $t=10$ , $Q=25$ $\therefore$ $10=\frac{1}{k} log_e 2$ $\implies$ $k= \frac{log_e 2}{10}$

$t= \frac{10log_e \frac{50}{Q}}{log_e 2}$

when $Q=10$

$t= \frac{10log_e 5}{log_e 2}=23.22s$

d0minicz · « **Reply #19 on:** June 29, 2009, 05:30:04 pm »

cheers damo
The number, n, of bacteria in a colony grows according to the law $\frac {dn}{dt} = kn$ , where k is a positive constant. If the number increases from 4000 to 8000 in four days, find, to the nearest hundred, the number of bacteria after three days more.
i need help interpreting this question.
thanks

Damo17 · « **Reply #20 on:** June 29, 2009, 05:43:54 pm »

Quote from: d0minicz on June 29, 2009, 05:30:04 pm

cheers damo
The number, n, of bacteria in a colony grows according to the law $\frac {dn}{dt} = kn$ , where k is a positive constant. If the number increases from 4000 to 8000 in four days, find, to the nearest hundred, the number of bacteria after three days more.
i need help interpreting this question.
thanks

np

$\frac{dt}{dn}= \frac{1}{kn}$ $\implies$ $t= \frac{1}{k}log_e n +c$

When $t=0$ , $n=4000$

put these into t and solve for c to get: $c= -\frac{1}{k} log_e 4000$

$t= \frac{1}{k} log_e \frac{n}{4000}$

When $t=4$ , $n=8000$

sub these into t and solve for k: $k= \frac{log_e 2}{4}$

$t= \frac{4}{log_e 2} log_e \frac{n}{4000}$

when $t= 7$ , find n: sub $t=7$ and solve for n:

$n= 4000e^\frac{7log_e 2}{4} = 13,500$

d0minicz · « **Reply #21 on:** June 29, 2009, 06:01:22 pm »

thanks again. sorry more q's, im just really bad at differentials.
A town had a population of 10 000 in 1990 and 12 000 in 2000. If the population is N at a time t years after 1990, find the predicted population in the year 2010 assuming;
a) $\frac {dN}{dt} = \frac {k}{N}$

thanks in advance !!!

edit: solved b)

d0minicz · « **Reply #22 on:** June 29, 2009, 06:23:19 pm »

wait do you nmean 13 711 ? thanks for the workings =]

A tank intially contains 200L of pure water. A salt solution containing 5kg of salt per litre is added at the rate of 10 L/min, and the mixed solution is drained simultaenously at the rate of 12 L/min. There is m kg of salt in the tank after t mins. Find $\frac {dm}{dt}$
thanks again

Damo17 · « **Reply #23 on:** June 29, 2009, 06:28:27 pm »

Quote from: d0minicz on June 29, 2009, 06:01:22 pm

thanks again. sorry more q's, im just really bad at differentials.
A town had a population of 10 000 in 1990 and 12 000 in 2000. If the population is N at a time t years after 1990, find the predicted population in the year 2010 assuming;
a) $\frac {dN}{dt} = \frac {k}{N}$

thanks in advance !!!

edit: solved b)

$<br />t= \frac{N^2}{2k} +c$

t=0, N=10000 : solve for c

$t= \frac{N^2}{2k} - \frac{(10000)^2}{2k}$

t=10, N=12000 : solve for k
$<br />t= \frac{N^2 - 10^8}{4.4 \times 10^6}$

sub in t=20 and you get N= 13711

pHysiX · « **Reply #24 on:** June 29, 2009, 06:29:35 pm »

oops made a silly mistake sorry

Damo17 · « **Reply #25 on:** June 29, 2009, 06:37:59 pm »

Quote from: d0minicz on June 29, 2009, 06:23:19 pm

wait do you nmean 13 711 ? thanks for the workings =]

A tank intially contains 200L of pure water. A salt solution containing 5kg of salt per litre is added at the rate of 10 L/min, and the mixed solution is drained simultaenously at the rate of 12 L/min. There is m kg of salt in the tank after t mins. Find $\frac {dm}{dt}$
thanks again

$V(t)= 200+10t-12t= 200-2t$

$IR= 10L/min \times 5kg/L= 50kg/min$

$OR= \frac{m}{V(t)} \times 12 = \frac{6m}{100-t}$

$\frac{dm}{dt} = IR-OR = 50 - \frac{6m}{100-t}$

EDIT: Restriction on t: $0 \le t < 100$

d0minicz · « **Reply #26 on:** June 30, 2009, 11:11:31 am »

A partially filled tank contains 200L of water in which 1500g of salt have been dissolved. Water is poured into the tank at a rate of 6L/min. The mixture, which is kept uniform by stirring, leaves the tank through a hole at a rate of 5L/min. There are x grams of salt in the tank after t mins. Find $\frac {dx}{dt}$
thank you

edit: solved ; but thanmks for any attemps

d0minicz · « **Reply #27 on:** June 30, 2009, 12:48:13 pm »

A country's population N at time t years after 1 Jan 2000 changes according to the differential equation $\frac {dN}{dt} = 0.1N - 5000$ .

(Five thousand people leave the country every year and there is a 10% growth rate).

a) Given that the population was 5 000 000 at the start of 2000, find N in terms of t.

b)In which year will the country have a population of 10 million?

Basically need help with a) mostly. Need the equation before i can do b).

thanks alot

kamil9876 · « **Reply #28 on:** June 30, 2009, 12:53:38 pm »

Use the same process as here:

http://vcenotes.com/forum/index.php/topic,13737.msg151711.html#msg151711

or even better, TT's post just below it.

d0minicz · « **Reply #29 on:** June 30, 2009, 12:55:19 pm »

thanks man but can soemone please show the working

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