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# Integration+Rate of change+Differential equation

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1. Integration+Rate of change+Differential equation
HI there can someone help me solve this problem:

The Rate of Change of dy/dt of a variable x at time t is proportional to 4t-3. Given that at t=2, x is 4 more than its original value and that at t=3, x is seven times its original value, find x in terms of t. Find the time at which x has its lowest value and find this lowest value.

Last edited by tyre; 05-05-2012 at 15:25.
2. Re: Integration+Rate of change+Differential equation
So my working is as follows:

Since dy/dt is proportional to 4t-3

->That dy/dt=k(4t-3)

Next making dy subject of formulae

->dy= k(4t-3) x dt

Next integrating both sides

->integrate(1 dy)= integrate (k(4t-3))
-> x=2kt^2-3kt+c

only that i came up with the rest i don't know wht to do??
Last edited by tyre; 05-05-2012 at 15:25.
3. Re: Integration+Rate of change+Differential equation
Are you sure you mean dy/dx ? What is y?
4. Re: Integration+Rate of change+Differential equation
Sorry an error has occurred now the question is set right..
5. Re: Integration+Rate of change+Differential equation
(Original post by tyre)
So my working is as follows:

Since dy/dt is proportional to 4t-3

->That dy/dt=k(4t-3)

Next making dy subject of formulae

->dy= k(4t-3) x dt

Next integrating both sides

->integrate(1 dy)= integrate (k(4t-3))
-> x=2kt^2-3kt+c

only that i came up with the rest i don't know wht to do??
"Given that at t=2, x is 4 more than its original value". This means that at t=2, 2kt^2-3kt+c is 4 more than when t=0. Can you carry on from here?
6. Re: Integration+Rate of change+Differential equation
How did you know that x is more than 4 of its original when t=0?? Please tell me maybe then i can continue
7. Re: Integration+Rate of change+Differential equation
(Original post by tyre)
How did you know that x is more than 4 of its original when t=0?? Please tell me maybe then i can continue
It's not completely clear but "original value" implies the value at the start of whatever's happening i.e. when the time starts (t=0).
8. Re: Integration+Rate of change+Differential equation
(Original post by tyre)
How did you know that x is more than 4 of its original when t=0?? Please tell me maybe then i can continue
if it says original or initial they mean when t = 0
9. Re: Integration+Rate of change+Differential equation
ok thnks yu guys i will try it now and notice yu back if anything goes wrong ok??
10. Re: Integration+Rate of change+Differential equation
Hi i am stuck again because when t=0, from x=2kt^2-3kt+C
->x=C

when t=2 , X=C+4 and when t=3, X=7C

now what??
11. Re: Integration+Rate of change+Differential equation
by the way the answer is x=4t^2-6t+3 -> x in terms of t
12. Re: Integration+Rate of change+Differential equation
(Original post by tyre)
Hi i am stuck again because when t=0, from x=2kt^2-3kt+C
->x=C

when t=2 , X=C+4
This is all correct but if you look at the bold part above, you have that when t=2, x=8k-6k+C = 2k+C

So 2k+C=C+4

Can you finish it from here?
13. Re: Integration+Rate of change+Differential equation
Therefore

k=2

giving,

x=4t^2-6t+c

when t=3, x=7c and x=9k+c
->9k+c=7c

since k=2

->from 9k+c=7c implies 6c=18 therefore, c=3

substituting k=2 and c=3

-> x=4t^2-6t+3 which is the answer therefore

Congrats to notnek and the bear for helping me without you another copybook would have finished with trial and error of the problems

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Last updated: May 5, 2012
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