we're using Differential equations with boundary-value problems by Dennis Zill,the section about the integrating factor is 2.3. Or you can probably try googling solving a linear first-order equation. Basically you're given a formula and you need to identify parts of your given equation or arrange it into standard form to plug in to the exponent of e. In this video he gives us a solid example of when we actually get to use this formula.
ProjectDuck HQ could you point out a specific time stamp? If your talking about the integration of the left hand side of the equation, it’s just undoing the product rule of differentiation
Oh okay and in undoing the product rule, the (100+t)^2 * s’ becomes 0? I’m just a little confused on how (100+t)^2 s’ + 2(100+t)s became (100+t)^2 s since the first term of the left side disappeared.
ProjectDuck HQ let u = (100+t)^2 * s’ and let v = 2*(100+t)*s. You’ll see that the expression you wrote is equal to uv’ + u’v but that’s equal to (uv)’ and so the integral of that is just uv ( that’s what I mean by undoing the product rule) does that make sense?
Just for fun, I decided to work this problem but without supplying any values until the last step, so I created some general formulas for the tank problem. A couple interesting results: 1) In the solution in the video, that term (t + 100)^2 in the bottom ... ? That squaring comes from the ratio "Out / (In - Out)", as in, the 2 gal/min and the 3 gal/min rates. You'd come up with some crazy exponents if those two rates weren't integers right next to each other, for example if it were 3 gal/min and 5 gal/min. 2) If the In and Out rates are the same, the solution changes such that it's a simple exponential curve that asymptotically approaches the point where the water in the tank is indistinguishable from the water coming in. The interesting thing is if you also assume a non-zero amount of salt initially in the tank: then the solution is the sum of the asymptotic result I just described, PLUS an exponential decay of the salt that was initially in the tank. That makes me want to use tiny discrete particles instead of salt, in two different colors, so I could watch the initial particles disappear while the new particles would gradually take over.
We make a gross simplification by taking an assumption, that the solution of salt water mixes in the tank perfectly and instantly. Did you notice, that proper industrial chemical reactors are spheres? A more accurate solution to the problem would involve a diffusion equation in 3 dimensions of space and time. Symmetric boundary conditions help solve them and make for a better industrial process. In the simplified setup in an example at hand S is a concetration of salt in a tank as a function of time S(t), this is a changing value and we denote the rate of instantaneous change of S as S’. S’ is of course a function of t as well: S’(t) we are able to write our formula for S’(t) as [flow_in*concentration -flow_out*(S(t)/(100+time)] This is one level of abstraction up from your A level maths and there are still quite a number of these to go. Formulating problem properly, in other words, finding the right model, which is acceptably easy to solve, but close enough to reality, is the most important part of the overall task. Finally, did you write all the exercise on paper for yourself? Better still, take any free CAS and write it into this Caomputer Algebra System. Plot the solution and read some numerical values for different times.
Eren Karaca it’s fine. Just a different “+C”. It’s like saying the integral of x is 1/2x^2 + 1 + c. Reason why I have it the way I do is bc I used simple u substitution. Let u = 100+t. Then du = dx. And so rewriting the integral you get integral of 3u^2 du. Which is 3u^3 / 3 + c. = u^3 + c = (100+t)^3 + c Indefinite Integration is arbitrary up to a constant (the + c) so it doesn’t matter if you have any other constants in there - you’ll just end up with a different C value than me bc mine will be shifted by a constant that’s already present in the solution
im not seeing a mistake right now, but from what i see the "2" popped up to become a "squared" when calculating the integrating factor. Integrating both sides leaves you with the integrating factor * s on the LHS. so the 2 is gone. Maybe your confusion is coming from how I integrated both sides. Check out the 1st order non separable (method of integrating factors) videos that I have - that may clear up any confusion!
Nope! let u = (100+t)^2 and v = s. Product rule of differentiation is: (u*v)' = uv' + u'v. And so, we can say: the integral( u'v + uv' ) = integral( (uv)' ) = uv! So now, Substitute in for u and v now: and you have ( (100+t)^2 * s ) ' = (100+t)^2 * s' + 2*(100+t)*s which means the integral that you wrote above in your prev comment evaluates to u*v -- or -- (100+t)^2 * s.
At first the final answer confused me: how could it be that the amount of salt is increasing at all times, and doesn't asymptotically approach a limit? Well the answer is, there is more water going into the tank than coming out, so that's where the increase comes from. If the deal were that 2 gallons were being pumped in and 2 gallons were being removed, then that changes the initial differential equation so that the volume is a constant "100" rather than "100 + t", which drastically alters the math, and indeed does give us the asymptotic behavior I was (mistakenly) expecting.
You forget you're teaching DE to people that want to learn people that are trying very hard to understand the subject and when you start skipping steps you sabotage yourself and confuse everyone else. We are not stupid and we understand when the explanation is clear and obvious but it is almost as if you need to show off at a cost of confusing everyone else consequently you get all these angry comments which should be a hint to you that something is a miss and people watching this channel are dissatisfied with your tutoring. Do everyone a favor by not skipping steps it's stupid to tutor this way, do it right the first around and explain the logic behind the numbers. I'm so disappointed!
i tried many tutorials and videos but non of them were clear as this... This series helped me alot ..
I appreciate you breaking everything down into simple terms i.e. dimensional analysis. Very easy to follow and comprehend.
you are a gift from God, please know you are one of the reasons I am going to pass diff eq :)
I finally understand🙏🏻🙏🏻 Thank You. Hope both sides of ur pillow are cold when you go to sleep
what an amazing video!! thank you I understood all
You're an artist!
6:50, why are you taking the exponential to the integral of that? Is that Mew(t) and P(t)
we're using Differential equations with boundary-value problems by Dennis Zill,the section about the integrating factor is 2.3. Or you can probably try googling solving a linear first-order equation. Basically you're given a formula and you need to identify parts of your given equation or arrange it into standard form to plug in to the exponent of e. In this video he gives us a solid example of when we actually get to use this formula.
question: What happened to the (100+t)^2 s' when integrating? does that become 0? if so is it because of the integral of s'?
ProjectDuck HQ could you point out a specific time stamp? If your talking about the integration of the left hand side of the equation, it’s just undoing the product rule of differentiation
Oh okay and in undoing the product rule, the (100+t)^2 * s’ becomes 0? I’m just a little confused on how (100+t)^2 s’ + 2(100+t)s became (100+t)^2 s since the first term of the left side disappeared.
ProjectDuck HQ let u = (100+t)^2 * s’ and let v = 2*(100+t)*s. You’ll see that the expression you wrote is equal to uv’ + u’v but that’s equal to (uv)’ and so the integral of that is just uv ( that’s what I mean by undoing the product rule) does that make sense?
The Lazy Engineer OH okay, that makes more sense. Thank you for the help!
you are saving my life!
you > my diff eq prof
Just for fun, I decided to work this problem but without supplying any values until the last step, so I created some general formulas for the tank problem. A couple interesting results:
1) In the solution in the video, that term (t + 100)^2 in the bottom ... ? That squaring comes from the ratio "Out / (In - Out)", as in, the 2 gal/min and the 3 gal/min rates. You'd come up with some crazy exponents if those two rates weren't integers right next to each other, for example if it were 3 gal/min and 5 gal/min.
2) If the In and Out rates are the same, the solution changes such that it's a simple exponential curve that asymptotically approaches the point where the water in the tank is indistinguishable from the water coming in. The interesting thing is if you also assume a non-zero amount of salt initially in the tank: then the solution is the sum of the asymptotic result I just described, PLUS an exponential decay of the salt that was initially in the tank. That makes me want to use tiny discrete particles instead of salt, in two different colors, so I could watch the initial particles disappear while the new particles would gradually take over.
I use de UV method because yours too advance xd. Thants for your explanation.
thanks i can understand this now YAY
how do we know it is rate coming in - rate coming out? Isn't rate of change final minus initial?
We make a gross simplification by taking an assumption, that the solution of salt water mixes in the tank perfectly and instantly. Did you notice, that proper industrial chemical reactors are spheres? A more accurate solution to the problem would involve a diffusion equation in 3 dimensions of space and time. Symmetric boundary conditions help solve them and make for a better industrial process.
In the simplified setup in an example at hand S is a concetration of salt in a tank as a function of time S(t), this is a changing value and we denote the rate of instantaneous change of S as S’. S’ is of course a function of t as well: S’(t)
we are able to write our formula for S’(t) as [flow_in*concentration -flow_out*(S(t)/(100+time)]
This is one level of abstraction up from your A level maths and there are still quite a number of these to go.
Formulating problem properly, in other words, finding the right model, which is acceptably easy to solve, but close enough to reality, is the most important part of the overall task.
Finally, did you write all the exercise on paper for yourself? Better still, take any free CAS and write it into this Caomputer Algebra System. Plot the solution and read some numerical values for different times.
what is mew and why do you multiply by equation
What happened to 3 when you were integrating (100+t)^2 ?.
When you integrate u^2, the result is 1/3 u^3. So the 3 canceled out with the 1/3.
how the exponential is integral the whole thing
Umm how is the integral of 3(100+t)^2 is equal to (100+t)^3?? Thats not how the integral works, the answer should be 30000t+300t^2 +t^3 +C
Eren Karaca it’s fine. Just a different “+C”.
It’s like saying the integral of x is 1/2x^2 + 1 + c.
Reason why I have it the way I do is bc I used simple u substitution. Let u = 100+t. Then du = dx. And so rewriting the integral you get integral of 3u^2 du. Which is 3u^3 / 3 + c. = u^3 + c = (100+t)^3 + c
Indefinite Integration is arbitrary up to a constant (the + c) so it doesn’t matter if you have any other constants in there - you’ll just end up with a different C value than me bc mine will be shifted by a constant that’s already present in the solution
What happened to the 2 infront of (100+ t)s??
im not seeing a mistake right now, but from what i see the "2" popped up to become a "squared" when calculating the integrating factor. Integrating both sides leaves you with the integrating factor * s on the LHS. so the 2 is gone. Maybe your confusion is coming from how I integrated both sides. Check out the 1st order non separable (method of integrating factors) videos that I have - that may clear up any confusion!
Integral(100+t)^2 s' + 2(100+t)s dt you have 2 in front of (100+t)s don't you need to algebraically remove that first?
Nevermind, yeah you differentiate it. You are right no problem. Thanks
Nope! let u = (100+t)^2 and v = s. Product rule of differentiation is: (u*v)' = uv' + u'v. And so, we can say: the integral( u'v + uv' ) = integral( (uv)' ) = uv! So now, Substitute in for u and v now: and you have ( (100+t)^2 * s ) ' = (100+t)^2 * s' + 2*(100+t)*s which means the integral that you wrote above in your prev comment evaluates to u*v -- or -- (100+t)^2 * s.
The Lazy Engineer got it. I got confused there for a bit. Had to watch the previous video about product rule. Thanks for clarifying. You are awesome!
show each integration....separately.....
Thank you
At first the final answer confused me: how could it be that the amount of salt is increasing at all times, and doesn't asymptotically approach a limit? Well the answer is, there is more water going into the tank than coming out, so that's where the increase comes from. If the deal were that 2 gallons were being pumped in and 2 gallons were being removed, then that changes the initial differential equation so that the volume is a constant "100" rather than "100 + t", which drastically alters the math, and indeed does give us the asymptotic behavior I was (mistakenly) expecting.
could you do an example where you explain why and what you’re doing please because the steps don’t make sense
You forget you're teaching DE to people that want to learn people that are trying very hard to understand the subject and when you start skipping steps you sabotage yourself and confuse everyone else. We are not stupid and we understand when the explanation is clear and obvious but it is almost as if you need to show off at a cost of confusing everyone else consequently you get all these angry comments which should be a hint to you that something is a miss and people watching this channel are dissatisfied with your tutoring. Do everyone a favor by not skipping steps it's stupid to tutor this way, do it right the first around and explain the logic behind the numbers. I'm so disappointed!