(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The problem states that at time 0 the tank has 10 lb of salt dissolved in 100 gallons of water. The capacity is 200. Assume that water containing 1/8 lb of salt per gallon is entering the tank at a rate of 2 gal/min and the mizture is draining from the tank at a rate of 1 gal/min.

a) set up the initial value problem

b)solve using method of integrating factors.

2. Relevant equations

t:time

y: amount of salt in tank (lbs)

v:volume of water (lbs)

v(t) = 100+t

3. The attempt at a solution

a) for the equation dy/dt = 1/4 - y/(100-t) y(0) = 10

b) *this is where i think i'm messing up...

dy/dt +(1/(100-t))y = 1/4

[tex]\mu[/tex](t) = [tex]e^{\int\frac{1}{100-t}dt}[/tex]

[tex]e^{-ln|100-t|}[/tex] = [tex]e^{ln|100-t|^{-1}}[/tex] = [tex](100-t)^{-1}[/tex]

= [tex]\frac{1}{100-t}[/tex]

so i multiply through and get

([tex]\frac{1}{100-t}[/tex] * y)' = [tex]\frac{1}{4(100-t)}[/tex]

[tex]\frac{1}{100-t}[/tex] * y = [tex]\int[/tex][tex]\frac{1}{4(100-t)}[/tex]dt

[tex]\frac{1}{100-t}[/tex] * y = ln|4(100-t)| +C

then solving for y

y = (100-t)*(ln|4(100-t)| +C)

This doesn't seem right to me for some reason, I just want to make sure i'm doing this problem right. Can someone please let me know? Thank you.

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# Homework Help: Differential Equations Tank problem

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