solving differential equation

a97775

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If you had a quantity h increasing by 125% every 5 hours, but 20 is subtracted every hour, how would you write the differential equation for the rate of change of h?
Is it

dh/dt = c*h -20*t where c is a constant and t is the number of hours

or is it just

dh/dt = c*h -20
?
And would you solve them using sep. of variables, integrating factor or another method?
Thank you for any help.
 
a97775 said:
If you had a quantity h increasing by 125% every 5 hours, but 20 is subtracted every hour, how would you write the differential equation for the rate of change of h?
Is it

dh/dt = c*h -20*t where c is a constant and t is the number of hours

or is it just

dh/dt = c*h -20
?
And would you solve them using sep. of variables, integrating factor or another method?
Thank you for any help.

20 what ? - 20 units - 20% ??

Assuming it is 20% - you'll need to compound it

So

say the rate of increase is (x*100)%/hour

then

P(1+x)[sup:10fu1c1r]5[/sup:10fu1c1r] = P*(1+1.25) ? x = 0.176079023

So now we see that in per hour basis

dh/dt = 0.176079023 - 0.20 = -0.023920977 ? it is decreasing
 
So how would the new differential eq. be with a constant being subtracted ratther than a %?
 
a97775 said:
So how would the new differential eq. be with a constant being subtracted ratther than a %?

It will depend on what "delta t" you are going to assume - 1 hour or 5 hours.
 
Then
\(\displaystyle \frac{dh}{dt} = 0.176079023 * h - 20\)

This is a "separable variable" problem.
 
would you solve it by dividing both sides by 0.176079023*h-20 and then integrating ?
 
a97775 said:
would you solve it by dividing both sides by 0.176079023*h-20 and then integrating ?

Review separable-variable method.
 
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