# Maths Question

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#1
when t=0 h=0.6

solve the differential equation from part (a) to show that
t=A ln (B/(5-4h)) where A and B are constants to be found

part (a) was a show that which I managed to solve with the answer 40dh/dt = 5-4h

I have solved differential equations to give 2h^2=t/8 +c
and using the information at the start of the question found that c=0.72 but I'm not sure where to go from here.

Any help would be greatly appreciated
0
1 year ago
#2
Could you post your working pls?
(Original post by yellow sunshine)
when t=0 h=0.6

solve the differential equation from part (a) to show that
t=A ln (B/(5-4h)) where A and B are constants to be found

part (a) was a show that which I managed to solve with the answer 40dh/dt = 5-4h

I have solved differential equations to give 2h^2=t/8 +c
and using the information at the start of the question found that c=0.72 but I'm not sure where to go from here.

Any help would be greatly appreciated
0
#3
unfortunately I can't seem to upload the photo as it keeps coming up with error
(Original post by mqb2766)
Could you post your working pls?
0
#4
I think this should work now?
0
1 year ago
#5
I got c to be -10ln(2.6) and then A = 10 and B = 2.6, but I'm not entirely sure either
0
#6
how did you manage to get c in terms of ln?
(Original post by FlawlessChicken)
I got c to be -10ln(2.6) and then A = 10 and B = 2.6, but I'm not entirely sure either
0
1 year ago
#7
40 dh/dt = 5-4h
40 dh = (5-4h) dt
(40/(5-4h)) dh = 1 dt {divided both sides by 5-4h}

integrate both sides:
-10 ln |5-4h| = t + c

sub in t= 0 and h = 0.6
-10 ln |5-0.6*4| = c
c = -10 ln2.6

you can then sub it back into the equation to get t in terms of h and ln in the form they want using log rules and stuff

(Original post by yellow sunshine)
how did you manage to get c in terms of ln?
Last edited by username4591252; 1 year ago
0
#8
omg that makes so much sense! Thankyou so much

(Original post by FlawlessChicken)
40 dh/dt = 5-4h
40 dh = (5-4h) dt
(40/(5-4h)) dh = 1 dt {divided both sides by 5-4h}

integrate both sides:
-10 ln |5-4h| = t + c

sub in t= 0 and h = 0.6
-10 ln |5-0.6*4| = c
c = -10 ln2.6

you can then sub it back into the equation to get t in terms of h and ln in the form they want using log rules and stuff
Last edited by yellow sunshine; 1 year ago
0
1 year ago
#9
No worries! It was actually really good differential equations practice!
(Original post by yellow sunshine)
omg that makes so much sense! Thankyou so much
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