Differential solving

a97775

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Here's another difficult one.
I have a quantity of drug in the body, d, and every four hours a dosage, m, is injected. Taking into account a rate of elimination by the kidneys of 5% every hour, this dosage m needs to be such that every four hours the quantity of drug d is 90 g.
So when m=90, t=0, 4, 8, etc

Would the differential equation be
dd/dt=(m+d)-d/20 ?
 
a97775 said:
Here's another difficult one.
I have a quantity of drug in the body, d, and every four hours a dosage, m, is injected. Taking into account a rate of elimination by the kidneys of 5% every hour, this dosage m needs to be such that every four hours the quantity of drug d is 90 g.
So when m=90, t=0, 4, 8, etc

Would the differential equation be
dd/dt=(m+d)-d/20 ?

\(\displaystyle \frac{dd}{dt} = -0.05 \cdot d\)
 
Wouldn't there be an m in the equation, since it is what I need to find?
And where does the 0.5 come from?
 
a97775 said:
Wouldn't there be an m in the equation, since it is what I need to find?

m will come in as your initial condition (d = m + 90 at t = 0)

And where does the 0.5 come from? <<<< That was typo - fixed it - thanks for catching it.
 
S. Khan, I need help with solving it.
This is what I did

dd/dt=-0.05d
1/d dd=-0.5 dt
ln d=-0.05t+c
d=Ae^(-0.05t)
m+90=A

m+90=(m+90)e^(-0.05*4)
m= -90
can't be negative.
 
a97775 said:
S. Khan, I need help with solving it.
This is what I did

dd/dt=-0.05d
1/d dd=-0.5 dt
ln d=-0.05t+c
d=Ae^(-0.05t)
m+90=A

m+90=(m+90)e^(-0.05*4)
m= -90 ...... How is that ???can't be negative.

at t= 4 you should have d = 90

90 = (m+90)e^(-0.2)

e^(0.2) = 1 + (m/90)

m = 90*0.221402758 = 19.92624823

and that's it.....
 
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