I'm having a bit of difficulty with these. I'm sort of new to this so I'm somewhat confused, maybe someone could lighten a bit of my confusion.(adsbygoogle = window.adsbygoogle || []).push({});

Here's the first problem:

http://www.synthdriven.com/images/deletable/img01.jpg [Broken]

This is what I've done, please let me know if I'm doing this correctly...

[tex]\frac{dy}{dt}=[/tex]rate in - rate out

rate in =[tex](0kg/G)(5G/hr)=0[/tex]

rate out =[tex](\frac{y(t)}{10G})(5G/hr)=\frac{5y(t)}{10}=\frac{1}{2}y(t)[/tex]

[tex]\frac{dy}{dt}=-\frac{1}{2}y(t)[/tex]

[tex]\int \frac{1}{y}=-\int \frac{1}{2}dt[/tex]

[tex]\ln{y}=-\frac{1}{2}t[/tex]

This is where I've ended up... Am I to find y and then solve for y=(1/2)10=5?? If so, that'd give me the following:

[tex]\ln{5}=-\frac{1}{2}t[/tex]

[tex]t=-2\ln{5}[/tex]

Is this correct??

The second problem is the following:

http://www.synthdriven.com/images/deletable/img02.jpg [Broken]

For this, I got the following:

[tex]\frac{dy}{dt}=[/tex]rate in - rate out

rate in=[tex](0.15\frac{ft^3}{min})(0.06\frac{C0}{ft^3})[/tex]

rate out=[tex](\frac{y(t)CO}{1800ft^3})(\frac{0.15ft^3}{min})=\frac{0.15y}{1800}CO/min[/tex]

[tex]\frac{dy}{dt}=0.009-\frac{0.15y}{1800}[/tex]

[tex]\frac{dy}{dt}=-(\frac{0.15}{1800}-0.009)[/tex]

[tex]\int{\frac{1800}{0.15y}-\frac{1}{0.009}dy}=-\int{dt}[/tex]

[tex]\frac{1800}{0.15}\ln{y}-\frac{y}{0.009}=-t[/tex]

I would then plug in 0.00018 for y, right?

When I do that, I get 51735.34 min... Is this correct...?

Any pointers?

Thanks

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# Homework Help: Diff Equations, Mixing

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