1. ## differential equation

recipe is based on 75% chocolate and 25% fudge by volume.Chocolate is pumped into a vat at 10litres per minute,into a vat with 30litres of melted fudge.mixture is pumped out at same rate.Write a differential equation for amount of chocolate in the vat,hence find how many seconds it takes for the mixture to be right recipe.

ok so far all i can really put down is dv/dt =10, stuck at starting it! help,thanks.

2. The tank initially has 30 liters of just fudge. Chocolate is pumped in and out at the same rate, so it will be 75% chocolate when there are 22.5 liters of chocolate in the tank.

$\frac{dy}{dt}=\text{rate in - rate out}$

$\text{rate in}=10 \;\ \frac{L}{min}$

$\text{rate out}=\frac{y(t)}{30}\cdot{10}=\frac{y(t)}{3} \;\ \frac{L}{min}$

$\frac{dy}{dt}+\frac{y}{3}=10$

y(0)=0 because there is no chocolate in the vat initially.

Can you finish now?.

3. thanks,im still a bit confused, 1st time ive seen this sort of problem.

id be eternally grateful if you could give me an outline of the method for all general questions like this.heres another problem i have:

a mixing machine initially contains 10l of orange,machine switched on apple juice enters at a rate of 5l per min, and mixed leaves at same rate.write a differential equation for apple juice in the machine, calculate how many seconds after switch on that the machine contains equal amounts of apple and orange.

4. Originally Posted by skystar
thanks,im still a bit confused, 1st time ive seen this sort of problem.

id be eternally grateful if you could give me an outline of the method for all general questions like this.heres another problem i have:

a mixing machine initially contains 10l of orange,machine switched on apple juice enters at a rate of 5l per min, and mixed leaves at same rate.write a differential equation for apple juice in the machine, calculate how many seconds after switch on that the machine contains equal amounts of apple and orange.
Galactus gave you a good out line. Lets use it to analyze this problem.

Let A be the amount of apple juice in the machine. We know at the beginning the machine has no apple juice in it so A(0)=0. We also know that Apple juice is flowing into the take at a constant rate of 5L per min. Now we just need to figure out how fast it is leaving the machine.

The volume of the machine is fixed at 10L and we know that the mixed product is leaving the the machine at 5L per min.

Now we need to compute how much of the mix is apple juice. We now that A is the amount of apple juice in the tank at any time so if we divide A by the volume of the machine we will get the part of the mix that is apple juice.
$\frac{A}{10}$ but this mix is flowing out at a rate of $\frac{5L}{min}$

So the ammount of Apple juice leaving at any time is $\frac{A}{10}\cdot \frac{5L}{min}=\frac{A \\\ L}{2 \\\ min}$

Now the rate of change of Apple juice is given by

$\frac{dA}{dt}=\mbox{rate in } -\mbox{ rate out}=5-\frac{A}{2}$

Now we just need to solve

$\frac{dA}{dt}=5-\frac{A}{2} \\\ A(0)=0$

I hope this helps.

Good luck.

5. EmptySet gave you some good pointers. I may as well follow through a little more on the "Oompa-Loompa" problem.

Now that we have our DE, we can find the integrating factor.

$\frac{dy}{dt}+\frac{y}{3}=10$

The integrating factor is arrived at by $e^{\int\frac{y}{3}dt}=e^{\frac{t}{3}}$

This gives us:

$\frac{d}{dt}(e^{\frac{t}{3}}y)=10e^{\frac{t}{3}}$

Integrate both sides:

$e^{\frac{t}{3}}y=30e^{\frac{t}{3}}+C$

$y=30+Ce^{\frac{-t}{3}}$

Now, you can solve for C by using the initial condition. Then you pretty much have it.
Set equal to 22.5 and solve for t.

6. "Oompa-Loompa" problem.
That is awesome!!
I think we need more more Oompa-Loompa's in ODE's