Differential Equations Tank problem

[SpiffyEh]

Homework Statement

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.

Homework Equations

t:time
y: amount of salt in tank (lbs)
v:volume of water (lbs)
v(t) = 100+t

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

$$\mu$$(t) = $$e^{\int\frac{1}{100-t}dt}$$

$$e^{-ln|100-t|}$$ = $$e^{ln|100-t|^{-1}}$$ = $$(100-t)^{-1}$$

= $$\frac{1}{100-t}$$

so i multiply through and get

($$\frac{1}{100-t}$$ * y)' = $$\frac{1}{4(100-t)}$$
$$\frac{1}{100-t}$$ * y = $$\int$$$$\frac{1}{4(100-t)}$$dt
$$\frac{1}{100-t}$$ * 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.

 

[tiny-tim]

Hi SpiffyEh!

… v(t) = 100+t
…
a) for the equation dy/dt = 1/4 - y/(100-t) y(0) = 10
…
$$\frac{1}{100-t}$$ * y = ln|4(100-t)| +C

Technique looks fine, but

i] shouldn't it be 100+t ?

ii] your 4 should be outside the ln

 

[SpiffyEh]

oops, sorry yeah it should be plus t. Should it be outside the ln? or should i just multiply the values by 4?

 

[tiny-tim]

oops, sorry yeah it should be plus t. Should it be outside the ln? or should i just multiply the values by 4?

(just got up :zzz: …)

Not following you … aren't they the same?

 

[SpiffyEh]

(just got up :zzz: …)

Not following you … aren't they the same?

yeah, i just thought it would be easier to multiply it through

 

[SpiffyEh]

with the initital condition i got c = 1/10 - ln400
so.. the whole equation is...
y(t) = (100+t)*(ln|400+4t|+1/10-ln400)

does that look right?

 

[tiny-tim]

Nooo …

your integrating factor was for 100 - t.

Now it's 100 + t, the integrating factor is completely different.

 

[SpiffyEh]

Nooo …

your integrating factor was for 100 - t.

Now it's 100 + t, the integrating factor is completely different.

it was supposed to be 100+t, i accidently changed it in the middle of all the equations