Solving First Order Diff EQ: Salt Tank Example

[adl2114]

Homework Statement

Salt water pours into a 10 liter tank at a rate of 4 l/min. Its concentration is 2 g/l. The brine in the tank is well mixed and it drains out at a rate of 4 l/min. Call y the grams of salt in the tank at time t. The tank is initially full of fresh water. Solve the differential equation that models the salt in the tank.

Homework Equations

dy/dx = rate in - rate out
y(t)= grams of salt (y) at time (t)

The Attempt at a Solution

I worked the problem and got the differential equation y'=8-(y/10)4 now I don't know how to solve that to get a particular solution assuming the initial condition y(0)=0

 

[hgfalling]

Your equation is fine. Now you can separate the variables (y and t) and integrate both sides.

$$\frac{dy}{dt} = 8 - \frac{y}{2.5}$$

When you are ready to integrate, this will look like:

something with only ys dy = something with only ts dt

Then use the initial condition to solve for your constant of integration.

 

[rock.freak667]

Actually, I think that DE will require an integrating factor.EDIT: Nevermind...constants are making me overkill it.

 

[adl2114]

ok so here's what i got so far:

dy/dt=8-(4y/10)
dy/dt=(80-4y)/10
dy/(80-4y)=dt/10
integrate to get
(-1/4)ln(80-4y)=t/10 + C

what do you think?

 

[hgfalling]

That looks okay to me, now solve for y.

(you can do this with an integrating factor, too, but I don't see that's it's necessary).

 

[adl2114]

Ok if what I have so far is correct solving for y gets me:

y(t)=20 + e^(-2t/5)C

then assuming the inital condition of y(0)=0 I get C= -20

so the final answer would be y(t)=20-20e^(-2t/5)

can someone verify this is correct?

 

[hgfalling]

You can verify it easily yourself, right? Take the derivative of y, plug into the original equation, make sure it works out.

Yes, it does. But you should still do it yourself! This is one of the best things about differential equations. It's usually really easy to check your work, because taking derivatives of functions is easy.

 

[Mark44]

Ok if what I have so far is correct solving for y gets me:

y(t)=20 + e^(-2t/5)C

then assuming the inital condition of y(0)=0 I get C= -20

so the final answer would be y(t)=20-20e^(-2t/5)

can someone verify this is correct?

You can. Check that
1) y(0) = 0, using your solution.
2) y' = 8 - 2y/5, using your solution.

That's all you need to do to verify your solution.

 

[adl2114]

perfect! I know I did check it myself but I am very poor in math skills I like to have another set of eyes thanks so much for your quick responses