A 10 liter bucket is full of pure water. We start adding salt to the water at a

Question

A 10 liter bucket is full of pure water. We start adding salt to the water at a rate of 0.2 kilograms per minute. Assume also that every minute 1 liter of the salt water solution is leaking from the bucket, but that we are adding pure water to keep the bucket full. The salt water solution is always kept well mixed.   

Set up a differential equation modeling the amount of salt in the bucket.

What is the amount of salt in the bucket after 6.93147 minutes?

What is the amount of salt in the bucket after an extremely long time?

Explanation / Answer

let the concentration of salt at any time be C(t),

so, rate of accumulation = salt coming inside - salt going outside

a.. 10*d(C(t))/dt =0.2*1-1*C(t)

b. d(C(t))/dt= 0.02-0.1C(t)

d(C(t))/dt+0.1C(t)= 0.02

this differential equation becomes

dy/dx+py=q

so integrating it we get,

C(t)e0.1t=0.2e0.1t+ const

at t=0 , C(t)=0,

so, const= -0.2

so C(t)e0.1t=0.2e0.1t- 0.2

so at t= 6.93147

e0.1*6.93147 =2

C*2=0.2*2-0.2

C=0.1

part c

after long time c= 0.2

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