Question

@Psymon So is what you basically do substitute v like you do u in integrating?

Answer 1

Im used to u, but yeah, youre subbing in a way as to make the equation linear in another variable. You have that isolated y term of apower higher than 1 on the right:

\[y' + P(x)y + Q(x)y^{n}\]

The two things you need is u= y^(1-n) and then the derivative of that, du = (1-n)y^-n. YOu would then multiply through by du and as a result, you would have a remaining
\[u' +P(x)u = Q(x) \],which you would then treat like any other linear equation, integrating factor,e tc.

Answer 2

Multiply by du? I've been thinking you multiply by u. Dang no wonder it looks weird. Ok thanks!

Answer 3

Yep. Once you multiply by du, assumign you isolated y' before, you would automatically be left with u', sometimes times u and then= blah blah. Of course once youget it linear, find your integrating factor etc, dontforget to sub u back in the end xD

Answer 4

Oh boy. Now I'll review compartment analysis before going back to how to force 2 discontinuous solutions to become continuous.

Answer 5

yeah, theyre not bad, trust me :P if you got an example problem can go through it really quick to show ya

Answer 6

Gotcha. I'll look for examples.

Answer 7

This might be long @Psymon

Answer 8

well, leaving and entering at the same rate helps. Okay, so we want the rate of change of the salt basically, and thats dependent on flow in times concentration minus flow out times concentration. \[\frac{ dA }{ dt }=R1Q1 - R2Q2\] So r is flow rate and q is concentration. You pretty much always are going to need to find that Q2. So for the firstpart, we have a rate in of 2 gallons a min times a concentration of 3 lbs of salt. So R1Q1 is just 6 (if you need theunits then keep em).
R2 is just that equal flow rate out of 2. Now Q2 is always going to be the mixture divided by the volume in the tank at any time. So usually something like this:

\[Q_{2} = \frac{ A }{ volume + (r_{2}-r_{1})t }\] So in our case the volume is 100 gallons and since r1 and r2 are equal, Q2 is just A/100. Multiplying that by R2, we just have:

\[\frac{ dA }{ dt }= 6 - \frac{ 2A }{ 100 } \implies \frac{dA}{dt} + \frac{ A }{ 50 }=6\]Thatlast step I did was set itup like a linear equation. So justtreat it as such, get your integrating factor and go : ) Oh, and the.5 lbsof salt is an initial condition at t = 0.

Answer 9

Right. So I find integrating factor using 1/50 and so on, right? I think I can manage this. The hard part is if the outflow is different than the inflow. What about B though? What is it asking for?

Answer 10

@John_ES Help? :3

Answer 11

I just need to know what B asks for. :) Thanks

Answer 12

Do you have the solution of the differential equation of the problem?

Answer 13

No not yet, but can you tell me what it asks for? I think if I know what I'm looking for I can do it. Would the limiting amount be when the concentration inside the tank remains constant?

Answer 14

If I'm correct, you should obtain a funciton of the form,
\[f(t)=C_1+C_2e^{-\alpha t}\] When time "goes" to infinity then you see that this function has a clear tendency, C1. This is the value they ask for (assuming C is the amount of salt).

Answer 15

And yes, it should be something like what you said.

Answer 16

Ahhhh I get it. I think from how you write it f(t) is the mass of salt and the C's are constants/ At least that's what I got from other examples we did in class.

Answer 17

Yes, :)

Answer 18

Thanks a lot John! Sorry I can't give another medal for this :/

Answer 19

Don't worry, the problem seems very interesting ;).

Answer 20

I have one more if you're interested and have time to burn.

Answer 21

You can find the solution of the differential equation,
\[\frac{dy}{dx}=\sqrt{y-5}\Rightarrow \int\frac{dy}{\sqrt{y-5}}=\int dx \\
\frac{1}{2}\sqrt{y-5}=x+C\]Now you put the initial conditions. For example, for (1),
\[\frac{1}{2}\sqrt{9-5}=0+C\Rightarrow C=-1\]
\[\frac{1}{2}\sqrt{y-5}=x-1\]

Answer 22

Ohh I see. And I do this for a and b? And I'm guessing for c I'm supposed to combine a and b?

Answer 23

That is something I'm not sure how to do.

Answer 24

No needed. The solution to (c) can be find without solve the differential equation. As the function must be constant (check the derivative and you'll see that when y=5, y'=0), you only need to choose y=5.

For (b) the procedure I wrote before is needed.
\[\frac{1}{2}\sqrt{y-5}=x\]And for (d), well this must be done with the theorem.

Answer 25

Hmm I will have to try and understand that part for c. But thanks John! I have to go to class now. Wish me luck!

Answer 26

Good luck ;)