You are absolutely right
= rate coming in - rate coming out
= 1*0.004-3y/200
=(-0.015)(y-4/15)
=
take integral of both side
=
=
I'm having a bit of trouble with some questions, I'm not sure whether it is from deficiancies in D.E or integration and would like some help if possible.
so here's an example.
A tank with 200L of pure water initially, has 4grams of salt per litre run into the tank at 1L per min. A well mixed solution is drained out at 3L per min.
I am to work out the amount of salt in the tank after t mins. Here's what im doing:
V=200-2t--------volume function of time
Dx/Dt=4-(3x/200-2t)------transpose
Dx/dt+(3x/200-2t)=4
integrating factor= e^((integral of)3/200-2t)
=(200-2t)^(-3/2)
Equation then becomes
x(200-2t)^(-3/2)={integral of}4((200-2t)^(-3/2))
Transposing
x =4(200-2t)^(-.5)+c
This equation doesn't seem to be realistic as shouldn't the amount of salt in the container look like a parabola or trig(sin/cos) function?
Thanks in advance