You're contagious! In an awesome way! Thank you for your enthusiasm and keep it up. Wish more teachers & professors shared the same passion for their subjects.
I hope that one day I can teach mathematics in this way. I wish he could produce more videos in pure mathematics courses such as manifolds, algebraic topology and so on.
I’ve taken calc 1 self study these last couple months and it’s gone pretty smoothly but currently doing separable differential equations has become quite the challenge! Your video here has helped me approach it from a fresh perspective! Thank you
Man, you saved my life. I missed two classes of Mathematical Analisis II, and really had troubles following the things. Thank you so, so much for your content
As a first year math student it's so interesting to see the many branches of mathematics. Pure math is still definitely more intimidating to me at this level than applied. Thank you for another awesome video!
I think i agree, pure math is such a new approach with theorems and proofs that the more applied math is familiar. However, once you get over those first exposures it really comes down to what you actually prefer doing. Excited to see you back here watching some of my 2nd and 3rd year level videos in a year or two:)
@@DrTrefor I will definitely be back. Your videos on the basics of linear algebra were a helpful place to start teaching myself matrix row operations and visualizing vectors.
Ah! I'm taking series and matrices this trimester and would have been so much excited to see any of that. Anyway, I'll come back for this series in the future.😄
Hey, thanks for the video. Can you explain what you mean by "defining dy to by dy/dt*dt"? How can we "define" dy, if it's already there in the equation?
Integral{u} [(1 / u) * du] u = y(t) du / dt = dy / dt du = (dy / dt) * dt Integral{t} [(1 / y) * (dy / dt) * dt] And realize the fist integral is equivalent to the last.
At the risk of sounding pedantic, I want to start by clarifying some things. Firstly, in the equation dy/dt = k·y, it is important to realize that, if you divide by y, then y must not be equal to 0 for any t, and since you cannot guarantee this a priori, you need to do some case analysis. If y = 0 on some interval, then dy/dt = k·0 = 0 on some nonempty open interval, and antidifferentiating thus implies y = C for some arbitrary C, for all t. Since y = 0 in some nonempty open interval, y = 0 for all t. This is one solution. With this, the remaining case is that there is no nonempty open interval where y = 0, hence you can safely divide by y, noting that y = 0 for all t is a solution. Secondy, at the extent that you antidifferentiate 1/y with respect to y, you technically need to consider this in separate cases, y < 0, and y > 0, because, as the integrand has a singularity at y = 0, the domain is disjoint, so different constants of integration are implied for both parts of the domain. If y < 0, then this results in ln(-y) + A = k·t, which can be simplified to -y = e^(-A)·e^(k·t). Thus y = -e^(-A)·e^(k·t). -e^(-A) is just an arbitrary negative real constant, so you can just recycle the name and call it A, acknowledging that A < 0, so y = A·e^(k·t) is a solution family. If y > 0, then ln(y) + B = k·t, and an analogous argument resuls in y = B·e^(k·t), with B > 0. With this, the three solution families are y = A·e^(k·t) with A < 0, y = 0, y = B·e^(k·t). As it happens, all three families can be expressed as special cases of the single solution family y = C·e^(k·t) with arbitrary real C. This does give the same result in the video, but this is a more careful derivation of it. Similar care should be taken in the general case dy/dt = f(t)·g(y)
Also, I want to add that it is possible to solve the equation y^2/2 + ln(|y|) = x^2/2 + C explicitly for y, as long as you are allowed to use the so-called Lambert W function. The idea is to multiply by 2, resulting in y^2 + 2·ln(|y|) = x^2 + A, and noting that 2·ln(|y|) = ln(y^2), so the equation simplifies to y^2 + ln(y^2) = x^2 + A. Exponentiating results in y^2·exp(y^2) = B·exp(x^2). In the assumption that B is nonnegative, you can now use the Lambert W function, which satisfies the property that W(z·exp(z)) = z whenever z·exp(z) is nonnegative. So this means y^2 = W[B·exp(x^2)], and this gives the solutions y = -sqrt(W[B·exp(x^2)]) and y = sqrt(W[B·exp(x^2)]). Interestingly, you can recover the solution y = 0 from either family by letting B = 0.
for some reason, our math teacher thought it was a good idea to just throw how to solve 2nd order differential equations at our face without even introducing the big underlying ideas, like why we can choose y = ae^bx as a general form, or why there are 2 parts to a non-homogeneous diff equation. Thank you for this course, and I will learn about them from the start
Professor Bazett, thank you for a great explanation on The Separation of Variables in Introduction Differential Equations. Calculus is a strong tool for understanding this and future topics in Differential Equations.
At 3:25 isn't the reason we drop the absolute value because it is included in our new multiplicative C. Before that the abs(y) implied that we have two equations mirrored about the x axis.
dy=(dy/dt)dt. I see the reduction, but in reality it's not a (1/dt)dt division that is taking place. This is what people struggle with. Could you elaborate on the composition and chain rule that is taking place here. Thanks in advance
Unfortunately this is too simplistic and doesn't clarify what is really going on. The question: Why is this the case? That's the question people are having difficulties in comprehending. It's the shift from one dimension to another, hense the shift in domain and range. Could you please make a video of this? That's what people need to understand
@@Infinitesap This is not something you can make video about. You would need to dedicate an entire series to this topic, because the problem is that, to satisfactorily explain why dy is defined the way it is, you need to delve into differential geometry and multilinear algebra. Unfortunately, even an oversimplification would be too complicated for a calculus student. So you honestly just have to accept that this is the way the notation works. Is this itself satisfactory? No, but there is nothing that can change that, since there is no possible way to explain the topic accurately, concisely, satisfactorily, and intuitively, especially to a calculus student, let alone in a 10-minute video.
I know I'm a bit late but I've definitely been struggling to understand this and another mathematician told me that this is an example of a "1-form". Still doesn't make sense to me but hope that helps anyone else
great video as always Dr. Trefor! Excited to take DE during the summer. Curious about one aspect of the solution to the second example you walked us through. When we integrate both sides of the equation wouldn't we also have a constant on the left hand side of the equation from the y terms? Our implicit solution only contained a constant term on the right hand (x-side) of the equation. Hope to hear from you. Thanks!
Nice, the whole playlist should be out by then, hope it will help a lot! You are quite right, but if I have a +C on the left and a +D on the right, we may as well collect both of them into a single constant. The choice to leave it on the right as opposed to the left is completely arbitrary.
Nice explanation of the nature of the algebraic solution you found. I am teaching DE's this semester and will be following your videos, and adding links to your notes too ! My course is laid out pretty much along similar lines - with an emphasis of qualitative features of solutions.
Honourable Dr T.B. Would you please suggest a course, that full covers from basic to masters level mathematics on this subject of differential equations Thank you
2:18 Dr, nobody has ever explained this to me, but why formally you don't "cancel the differentials"? maths majors make fun of us engineers all the time for doing this 🙉
will you explain the relationship between math and physics?? That's will be a great course 😃😃 , but you are a great man all time keep up 😁😁 and thank you for your effort
Hi Thank you for the video it was so helpful! Can you just explain a little bit more what is the difference between implicit solution and non-implicit solution like dont both of them give you what x is if you know what y is and vice versa?
why is there not a plus c for the integrals on the left hand side and only for the right every time? i understand for indefinite integrals we need plus c but why do we not need it when differenting y? thanks
The reason is that it is redundant. If you want, you can put +C1 on the left, and +C2 on the right. Then, you can subtract C1 from the left, and you end up with (C2 - C1) on the right. Since C1 & C2 are both arbitrary constants, just consolidate them, and let C = C1 - C2. Generally, there are n independent constants that apply, for an nth order DiffEQ. If along the way, you produce more than n constants, there is a likely chance you can consolidate them to fewer constants. If along the way you produce fewer than n constants, chances are, you neglected to account for one.
You just defined dy/dt(dt)=dy while doing integration, but when we see differential equations in form of Mdx+Ndy=0, here what these symbol dx,dy means, here without Integral sign..please tell me.. waiting for your response..
For all the people disturbed by the dy/dx dx turning into just dy I don't think that saying that the dx "cancels out" or that we defined a new dy or moving stuff around is legal. I mean, it might somehow be, but I can't see why, I can only see that the exercises work. However it's not satisfying. Note that all we want and need to be satisfied is showing that ∫f(y) dy/dx dx = ∫f(y) dy If we can do that, we can then at least see why saying "cancel out the dx" or "substitute dy = dy/dx dx" would work. (Because we would then have proved the equivalence ∫f(y) dy/dx dx = ∫f(y) dy) Let us flip it for convenience: ∫f(y) dy = ∫f(y) dy/dx dx So let's do that! Let's prove it. Remember the chain rule: du/dx = (du/dy) * (dy/dx) where u is a function of y where y is a function of x (If you don't remember it look it up) Let our u = ∫f(y) dy Substitute u inside the chain rule equation d/dx (∫f(y) dy) = (d/dy (∫f(y) dy)) * (dy/dx) Kinda ugly in the comment, do it on paper if it's too weird. Now see that on the right side we are taking the derivative of a function we are integrating, and the two operations cancel out, let's simplify. d/dx (∫f(y) dy) = f(y) * (dy/dx) Now let's take the integral of both sides with respect to x. ∫ [d/dx (∫f(y) dy)] dx = ∫ [f(y) * (dy/dx)] dx Notice that on the left side we are taking the derivative of a function we are integrating (in respect to dx), the two operations cancel out, let's simplify. ∫f(y) dy = ∫ [f(y) * (dy/dx)] dx Oh heyyy, would you look at that, it's the same expression we wanted to prove! ∫f(y) dy = ∫f(y) dy/dx dx So we can at least informally accept we can "cancel out" the two dx and leave dy :)
Thanks! This really helped! Why does dy/dx dx not cancel into just dy? Does that have to do with what infinitesimally small amount we look at not cancelling and then by extension the philosophy and theory behind calc?
is there a method for systems of differential equations where the initial condition, itself depends on a n-1 system of reducible differential equations?
My teacher is a very smart guy, but he goes through this stuff so fast I'm left scratching my head. When the constant of integration morphed into the coefficient C~, that whole process went unexplained and I had absolutely no idea how it went from an additive constant to a multiplicative constant or why it was wearing that silly hat.
Thanks for your video series, also I love the textbook that you helped fork. Thanks for letting me download the pdf (not to be confused with probability distribution function, hehe)
2:14 How can you just "define" dy/dt * dt to be dy? like you cant just do that, You cant just call it dy and treat it like dy, it makes 0 sense. Hope someone has an answer would absolutely love to know the mechanics/thought process behind this, For everything else it was very easy to understand thank you Dr.Trefor!
We can define dy/dt multiplied by dt as dy, similar to a fraction. However, this approach does not apply to derivatives higher than the first derivative. It resembles a small change in distance, which equals velocity multiplied by a small change in time, denoted as "dt," where ds represents a small distance, and ds/dt represents speed. Therefore, ds/dt dt represents a differential variable in terms of a derivative multiplied by another differential variable. Although differential variables are infinitesimals , the ratio is kept the same.
@@Mysoi123 thanks mate! makes more sense now. Will have to read it a couple of times to get the full understanding. Trying to get into differential equations is an adventure of its own😆
@@lusteen6318 You're welcome. I forgot to mention that this also doesn't work for partial derivatives. If your function has two or more variables, like f(y, x), then the derivative with respect to each variable is a partial derivative. This means we cannot integrate ∂f/∂x ∂x to obtain F(x, y). However, the differential form can be expressed as a multiplication of variables using the multivariable chain rule: ∂f = (∂f/∂x ∂x + ∂f/∂y ∂y) And for single-variable functions, it is: df = (df/dx) dx Divide both sides for “dt” and you get the good old chain rule df/dt = df/dx dx/dt
@@jeremykievit897 The equation "df = (df/dx) dx" only works when you relate two differential variables with a function in between, where that function is the derivative of "f." For example, a tiny change in distance (ds) equals velocity as a function of time multiplied by a tiny change in time (dt), expressed as ds = v(t) dt = (ds/dt) dt. In this context, an integral is akin to an infinite continuous sum of tiny distances, providing a helpful perspective when studying motion using calculus.
There are two related perspectives on this. One is that we are DEFINING dy to be dy/dt*dt. The second is thinking of this as a change of variables (i.e a u-sub with u=y) and then applying chain rule. Basically differentials are always defined to be obeying chain rule, and chain rule is what makes "cancelling the dt on top and bottom" turn from a convenient fiction to something that actually works.
But why is this the case? That's the question people are having difficulties in comprehending. It's the shift from one dimension to another, hense the shift in domain and range.
@@Infinitesap Actually, this has nothing to do with "shifting dimensions." The dimensionality of the problem is the exact same. Also, I already replied to your objection elsewhere.
@@angelmendez-rivera351 Please help me here then, because if you shift from (dy/dt)*dt to dy then there has been a tour across the dy/dt and as I see it this is indeed a change in range (of dy) to another domain (of dy/dt).
@@Infinitesap Yes, a change in range, and in domain, but not a change in dimensionality. A change in dimensionality would be like if the domain went from R to R^2.
but integrating each side means we add more information! because now the area under the curve of the LHS must be equals to the area under the curve on the RHS. When dealing with x^2 = x it becomes quite obvious, that this is not true. The area on the LHS would be 1/3 * x^3. whereas the area on the RHS would be 1/2 * x^2. Why is this allowed?
In this example, x^2=x is ONLY true at 0 and 1, so integrating it and considering other points doesn't work. But if f(x) =g(x) for all the x values in a domain, you CAN integrate both sides and it will be equal.
It doesn't necessarily do that in all cases. That's simply a given, for this particular family of problems. It would come from the application of the differential equation. One such application is Newton's law of cooling, where the rate at which a body cools is proportional to the difference between its temperature and its environment. Consider a body at a positive temperature T, put in a 0 Celsius environment. Its temperature would be governed by: dT/dt = -k*T The solution would be: T = T0*e^(-k*t) The constant in front of the variable that changes is positive for exponential growth, and negative for either exponential approach or exponential decay. Side note: if its environment were any temperature (Tbg) other than zero, then T would be replaced with (T - Tbg), and the solution would be: T = (T0 - Tbg)*e^(-k*t) + Tbg.
Pretty sure I solved the DE but can someone verify? I multiird everything by 2 and put the 2 in the exponent of the ln. Then just raise to e and w lambert and sqrt to get y=+/-sqrt(w(e^(2x+2c))). Again, not sure if this works.
"dropped absolute sign because exponential is always positive" BIG blunder here, the absolute value of y is equal to the exponential hence y itself could be equal to the negative of an exponential for example if y = -e^x then |y|=e^x but |y|=e^x DOES NOT imply y = e^x instead: |y|=e^x implies y=plus or minus e^x HOWEVER since you wrote C tilde infront of the exponential you somewhat saved it because C tilde can also represent a negative constant
Bc if u add a constant, let’s say C1 to the right and a constant C2 to the left, just subtract C1 from both sides. On the left you end up with C2 - C1. This is a constant minus a constant, which is just another constant. After all this, you end up with just one constant on the right, so they only add it on the right in the first place.
Hi Trefor please help me and let me know what you think about it. I know and my family knows I am very smart, as a 7th grader I do calculus 2 right now and understand everything really quick. My parent is strict about grades and keep pressuring me about bad grades in middle school will not get you to good high school. And yes my grade is very bad and I don’t know why I’m not doing so well in as online school, for some reasons my motivations towards listening well to teachers, doing work in time and ect.., I don’t want to do work hard as I used to. For first quarter grades, I got 5 A’s and 2 B’s, second quarter I got 4 A’s and 3 B’s. Now in third quarter I now have 2 A’s 2 D’s and 5 C’s. Mostly the B, C, grade is form science, health, art, social studies. I want to be an inventor, most my interests is about teleportation. Is it still possible for me to do well now from on and actually see my self graduating one of the top college and succeed in life? Or just I just give up at this point. Please let me know.
David K You can do everything.. you are capable of everything, don't be disappointed 🙁 maybe the online school isn't the best for you.. Try learning new things or take a rest every once in a while .. don’t care about grade coz if you did u wouldn’t get good grades.. that’s what I believe in .. hope I could help u 💖
I don't like it. dy is what? It's notation. You're manipulating notation as if were a real object with a real definition. I think the dy is just notation used in conjunction with the integral sign that means, integrate with respect to y. It's not a thing.
You are half right. Integral in itself is not something completely abstract, it rather represents an infinite sum, hence an integral sign is a deformed letter S. dx on the other hand stands for infinitely small change of x, so integrating is actually adding up small pieces of (f(x) times dx), written down as [Integral sign]f(x) dx. When we write derivative this way: dy/dx, notation here also represents infinitely small values. In some way, the definition of limit is built inside the d(something) so we can actually work with it.
You're contagious! In an awesome way! Thank you for your enthusiasm and keep it up. Wish more teachers & professors shared the same passion for their subjects.
Wow, thank you!
@@DrTrefor You're very welcome! :D
I hope that one day I can teach mathematics in this way. I wish he could produce more videos in pure mathematics courses such as manifolds, algebraic topology and so on.
I’ve taken calc 1 self study these last couple months and it’s gone pretty smoothly but currently doing separable differential equations has become quite the challenge! Your video here has helped me approach it from a fresh perspective! Thank you
Man, you saved my life. I missed two classes of Mathematical Analisis II, and really had troubles following the things. Thank you so, so much for your content
As a first year math student it's so interesting to see the many branches of mathematics. Pure math is still definitely more intimidating to me at this level than applied. Thank you for another awesome video!
I think i agree, pure math is such a new approach with theorems and proofs that the more applied math is familiar. However, once you get over those first exposures it really comes down to what you actually prefer doing. Excited to see you back here watching some of my 2nd and 3rd year level videos in a year or two:)
@@DrTrefor I will definitely be back. Your videos on the basics of linear algebra were a helpful place to start teaching myself matrix row operations and visualizing vectors.
Great video as always Dr.! Must say again I love how you edit all the math onto the screen so clear!
Taking an ODE's class at the exact same time as this series, nice videos to have to supplement the course, thank you!
Thank you for the reminder on exponent rules that results in ce^kt 3:08
Ah! I'm taking series and matrices this trimester and would have been so much excited to see any of that. Anyway, I'll come back for this series in the future.😄
Nice! You're in for a fun trimester:)
Plzz Make complete videos about Engineering mathematics sequencely . So from this I feel so glad to learn with u . Thank you very much
Hey, thanks for the video.
Can you explain what you mean by "defining dy to by dy/dt*dt"? How can we "define" dy, if it's already there in the equation?
I think it's because in both cases it represents a change in y?
Integral{u} [(1 / u) * du]
u = y(t)
du / dt = dy / dt
du = (dy / dt) * dt
Integral{t} [(1 / y) * (dy / dt) * dt]
And realize the fist integral is equivalent to the last.
Sir your explanation is really wonderful!!
I needed videos on this course as this is what is going to be taken up this semester in our college.
Sir, I am your fan now... The way you teach is soooooo good and your excitement for teaching excites me to understand the concept...
Heads Off!!!
At the risk of sounding pedantic, I want to start by clarifying some things.
Firstly, in the equation dy/dt = k·y, it is important to realize that, if you divide by y, then y must not be equal to 0 for any t, and since you cannot guarantee this a priori, you need to do some case analysis. If y = 0 on some interval, then dy/dt = k·0 = 0 on some nonempty open interval, and antidifferentiating thus implies y = C for some arbitrary C, for all t. Since y = 0 in some nonempty open interval, y = 0 for all t. This is one solution. With this, the remaining case is that there is no nonempty open interval where y = 0, hence you can safely divide by y, noting that y = 0 for all t is a solution.
Secondy, at the extent that you antidifferentiate 1/y with respect to y, you technically need to consider this in separate cases, y < 0, and y > 0, because, as the integrand has a singularity at y = 0, the domain is disjoint, so different constants of integration are implied for both parts of the domain. If y < 0, then this results in ln(-y) + A = k·t, which can be simplified to -y = e^(-A)·e^(k·t). Thus y = -e^(-A)·e^(k·t). -e^(-A) is just an arbitrary negative real constant, so you can just recycle the name and call it A, acknowledging that A < 0, so y = A·e^(k·t) is a solution family. If y > 0, then ln(y) + B = k·t, and an analogous argument resuls in y = B·e^(k·t), with B > 0. With this, the three solution families are y = A·e^(k·t) with A < 0, y = 0, y = B·e^(k·t). As it happens, all three families can be expressed as special cases of the single solution family y = C·e^(k·t) with arbitrary real C. This does give the same result in the video, but this is a more careful derivation of it.
Similar care should be taken in the general case dy/dt = f(t)·g(y)
Being pedantic is definitely good, these details don't always make it into the videos can sometimes be quite important!
Also, I want to add that it is possible to solve the equation y^2/2 + ln(|y|) = x^2/2 + C explicitly for y, as long as you are allowed to use the so-called Lambert W function.
The idea is to multiply by 2, resulting in y^2 + 2·ln(|y|) = x^2 + A, and noting that 2·ln(|y|) = ln(y^2), so the equation simplifies to y^2 + ln(y^2) = x^2 + A. Exponentiating results in y^2·exp(y^2) = B·exp(x^2). In the assumption that B is nonnegative, you can now use the Lambert W function, which satisfies the property that W(z·exp(z)) = z whenever z·exp(z) is nonnegative. So this means y^2 = W[B·exp(x^2)], and this gives the solutions y = -sqrt(W[B·exp(x^2)]) and y = sqrt(W[B·exp(x^2)]).
Interestingly, you can recover the solution y = 0 from either family by letting B = 0.
Been looking for the right video for the past two days and i finally got one. Thanks s lot Sir
for some reason, our math teacher thought it was a good idea to just throw how to solve 2nd order differential equations at our face without even introducing the big underlying ideas, like why we can choose y = ae^bx as a general form, or why there are 2 parts to a non-homogeneous diff equation. Thank you for this course, and I will learn about them from the start
Professor Bazett, thank you for a great explanation on The Separation of Variables in Introduction Differential Equations. Calculus is a strong tool for understanding this and future topics in Differential Equations.
At 3:25 isn't the reason we drop the absolute value because it is included in our new multiplicative C. Before that the abs(y) implied that we have two equations mirrored about the x axis.
dy=(dy/dt)dt. I see the reduction, but in reality it's not a (1/dt)dt division that is taking place. This is what people struggle with. Could you elaborate on the composition and chain rule that is taking place here.
Thanks in advance
One way I like to think at it is we are DEFINING dy in such a way that the “fiction” of cancelation of the dt is actually true via chain rule.
Unfortunately this is too simplistic and doesn't clarify what is really going on. The question: Why is this the case? That's the question people are having difficulties in comprehending.
It's the shift from one dimension to another, hense the shift in domain and range.
Could you please make a video of this?
That's what people need to understand
@@Infinitesap This is not something you can make video about. You would need to dedicate an entire series to this topic, because the problem is that, to satisfactorily explain why dy is defined the way it is, you need to delve into differential geometry and multilinear algebra. Unfortunately, even an oversimplification would be too complicated for a calculus student. So you honestly just have to accept that this is the way the notation works. Is this itself satisfactory? No, but there is nothing that can change that, since there is no possible way to explain the topic accurately, concisely, satisfactorily, and intuitively, especially to a calculus student, let alone in a 10-minute video.
I know I'm a bit late but I've definitely been struggling to understand this and another mathematician told me that this is an example of a "1-form". Still doesn't make sense to me but hope that helps anyone else
@@TheMasterLich139 In fact it is the chain rule that is applied here. If you need more info let me know. I can explain it to you.
Dr. Trefor you are doing great tutorial...i love your work. Keep making more videos.
Glad you find them helpful!
great video as always Dr. Trefor! Excited to take DE during the summer. Curious about one aspect of the solution to the second example you walked us through. When we integrate both sides of the equation wouldn't we also have a constant on the left hand side of the equation from the y terms? Our implicit solution only contained a constant term on the right hand (x-side) of the equation. Hope to hear from you. Thanks!
Nice, the whole playlist should be out by then, hope it will help a lot! You are quite right, but if I have a +C on the left and a +D on the right, we may as well collect both of them into a single constant. The choice to leave it on the right as opposed to the left is completely arbitrary.
@@DrTrefor Awesome. With your help I've attained clarity once more :)
Why do you add a constant to the integration on the RHS but not the LHS?
This guy is a legend. Single-handedly saving my math grades in college.
Nice explanation of the nature of the algebraic solution you found. I am teaching DE's this semester and will be following your videos, and adding links to your notes too ! My course is laid out pretty much along similar lines - with an emphasis of qualitative features of solutions.
Can u please make complete videos about topology? You are the best 💖
That's the plan!
Honourable Dr T.B.
Would you please suggest a course, that full covers from basic to masters level mathematics on this subject of differential equations
Thank you
Thank you so much! This course is such a help for my DEQ class!!
2:18 Dr, nobody has ever explained this to me, but why formally you don't "cancel the differentials"? maths majors make fun of us engineers all the time for doing this 🙉
will you explain the relationship between math and physics?? That's will be a great course 😃😃 , but you are a great man all time keep up 😁😁 and thank you for your effort
Thank you sir. Love from India
Hi Thank you for the video it was so helpful! Can you just explain a little bit more what is the difference between implicit solution and non-implicit solution like dont both of them give you what x is if you know what y is and vice versa?
The integration for y^2+1 is supposed to be y^3 /3 +y right?
High-quality content!! Great job!
why is there not a plus c for the integrals on the left hand side and only for the right every time? i understand for indefinite integrals we need plus c but why do we not need it when differenting y? thanks
The reason is that it is redundant. If you want, you can put +C1 on the left, and +C2 on the right. Then, you can subtract C1 from the left, and you end up with (C2 - C1) on the right. Since C1 & C2 are both arbitrary constants, just consolidate them, and let C = C1 - C2.
Generally, there are n independent constants that apply, for an nth order DiffEQ. If along the way, you produce more than n constants, there is a likely chance you can consolidate them to fewer constants. If along the way you produce fewer than n constants, chances are, you neglected to account for one.
You just defined dy/dt(dt)=dy while doing integration, but when we see differential equations in form of Mdx+Ndy=0, here what these symbol dx,dy means, here without Integral sign..please tell me.. waiting for your response..
For all the people disturbed by the dy/dx dx turning into just dy
I don't think that saying that the dx "cancels out" or that we defined a new dy or moving stuff around is legal.
I mean, it might somehow be, but I can't see why, I can only see that the exercises work.
However it's not satisfying.
Note that all we want and need to be satisfied is showing that ∫f(y) dy/dx dx = ∫f(y) dy
If we can do that, we can then at least see why saying "cancel out the dx" or "substitute dy = dy/dx dx" would work.
(Because we would then have proved the equivalence ∫f(y) dy/dx dx = ∫f(y) dy)
Let us flip it for convenience:
∫f(y) dy = ∫f(y) dy/dx dx
So let's do that! Let's prove it.
Remember the chain rule:
du/dx = (du/dy) * (dy/dx)
where u is a function of y
where y is a function of x
(If you don't remember it look it up)
Let our u = ∫f(y) dy
Substitute u inside the chain rule equation
d/dx (∫f(y) dy) = (d/dy (∫f(y) dy)) * (dy/dx)
Kinda ugly in the comment, do it on paper if it's too weird.
Now see that on the right side we are taking the derivative of a function we are integrating, and the two operations cancel out, let's simplify.
d/dx (∫f(y) dy) = f(y) * (dy/dx)
Now let's take the integral of both sides with respect to x.
∫ [d/dx (∫f(y) dy)] dx = ∫ [f(y) * (dy/dx)] dx
Notice that on the left side we are taking the derivative of a function we are integrating (in respect to dx), the two operations cancel out, let's simplify.
∫f(y) dy = ∫ [f(y) * (dy/dx)] dx
Oh heyyy, would you look at that, it's the same expression we wanted to prove!
∫f(y) dy = ∫f(y) dy/dx dx
So we can at least informally accept we can "cancel out" the two dx and leave dy :)
Thanks! This really helped! Why does dy/dx dx not cancel into just dy? Does that have to do with what infinitesimally small amount we look at not cancelling and then by extension the philosophy and theory behind calc?
I watched it once.
Pls tell when do you start pde ?
Going to be a couple months, going to get through ODEs first before doing some PDEs
Please I want to ask if you divided ysqaured by the y on the bottom in the example you gave
at 2:38 why we didnt add a constant at the left side after integration?
We could have, but then we can combine both constants into a single one, so I skipped it.
at 3:33 the reason you gave to remove absolute value sing is wrong, we can remove abs value sing if we assume that y>0
Why was C on only one side of the equation? Since they were both indefinite integrals.
What topic am I missing?
That's fine too, but you can combine the two constants into just one.
is there a method for systems of differential equations where the initial condition, itself depends on a n-1 system of reducible differential equations?
My teacher is a very smart guy, but he goes through this stuff so fast I'm left scratching my head. When the constant of integration morphed into the coefficient C~, that whole process went unexplained and I had absolutely no idea how it went from an additive constant to a multiplicative constant or why it was wearing that silly hat.
Glad it helped!
If there is a 5% interest rate, should not the actual capital be 100*e^((ln1.05)*t)?
What is the app you were using to draw the graph
Desmos I think.
Super Sir
Thanks for your video series, also I love the textbook that you helped fork. Thanks for letting me download the pdf (not to be confused with probability distribution function, hehe)
Very very appreciated.
2:14 How can you just "define" dy/dt * dt to be dy? like you cant just do that, You cant just call it dy and treat it like dy, it makes 0 sense. Hope someone has an answer would absolutely love to know the mechanics/thought process behind this, For everything else it was very easy to understand thank you Dr.Trefor!
We can define dy/dt multiplied by dt as dy, similar to a fraction. However, this approach does not apply to derivatives higher than the first derivative. It resembles a small change in distance, which equals velocity multiplied by a small change in time, denoted as "dt," where ds represents a small distance, and ds/dt represents speed. Therefore, ds/dt dt represents a differential variable in terms of a derivative multiplied by another differential variable.
Although differential variables are infinitesimals , the ratio is kept the same.
@@Mysoi123 thanks mate! makes more sense now. Will have to read it a couple of times to get the full understanding. Trying to get into differential equations is an adventure of its own😆
@@lusteen6318 You're welcome.
I forgot to mention that this also doesn't work for partial derivatives. If your function has two or more variables, like f(y, x), then the derivative with respect to each variable is a partial derivative. This means we cannot integrate ∂f/∂x ∂x to obtain F(x, y).
However, the differential form can be expressed as a multiplication of variables using the multivariable chain rule:
∂f = (∂f/∂x ∂x + ∂f/∂y ∂y)
And for single-variable functions, it is:
df = (df/dx) dx
Divide both sides for “dt” and you get the good old chain rule
df/dt = df/dx dx/dt
@@Mysoi123 If df = (df/dx) dx, doesn't that just prove that we can divide out the dx in the single variable case?
@@jeremykievit897 The equation "df = (df/dx) dx" only works when you relate two differential variables with a function in between, where that function is the derivative of "f." For example, a tiny change in distance (ds) equals velocity as a function of time multiplied by a tiny change in time (dt), expressed as ds = v(t) dt = (ds/dt) dt. In this context, an integral is akin to an infinite continuous sum of tiny distances, providing a helpful perspective when studying motion using calculus.
Great video 👍
Thanks 👍
@8:00 why is +C written on the RHS with the x, and not on the LHS with the y?
It's arbitrary which side you write the +C on. Ultimately, there are two constants, that you then consolidate into just one constant.
Why are we allowed to change (dy/dt)*dt to dy? Or how come that you can treat dy/dt as a fraction in this specific case.
There are two related perspectives on this. One is that we are DEFINING dy to be dy/dt*dt. The second is thinking of this as a change of variables (i.e a u-sub with u=y) and then applying chain rule. Basically differentials are always defined to be obeying chain rule, and chain rule is what makes "cancelling the dt on top and bottom" turn from a convenient fiction to something that actually works.
But why is this the case? That's the question people are having difficulties in comprehending.
It's the shift from one dimension to another, hense the shift in domain and range.
@@Infinitesap Actually, this has nothing to do with "shifting dimensions." The dimensionality of the problem is the exact same. Also, I already replied to your objection elsewhere.
@@angelmendez-rivera351 Please help me here then, because if you shift from (dy/dt)*dt to dy then there has been a tour across the dy/dt and as I see it this is indeed a change in range (of dy) to another domain (of dy/dt).
@@Infinitesap Yes, a change in range, and in domain, but not a change in dimensionality. A change in dimensionality would be like if the domain went from R to R^2.
This is amazing
I genuinely love u
but integrating each side means we add more information! because now the area under the curve of the LHS must be equals to the area under the curve on the RHS.
When dealing with x^2 = x it becomes quite obvious, that this is not true. The area on the LHS would be 1/3 * x^3. whereas the area on the RHS would be 1/2 * x^2.
Why is this allowed?
In this example, x^2=x is ONLY true at 0 and 1, so integrating it and considering other points doesn't work. But if f(x) =g(x) for all the x values in a domain, you CAN integrate both sides and it will be equal.
Hi Trefor I like you but I already know this;)
Can someone explain dy/dx=ky? How does a derivative of y equal y times a constant? Where does this equation even come from?
It doesn't necessarily do that in all cases. That's simply a given, for this particular family of problems.
It would come from the application of the differential equation. One such application is Newton's law of cooling, where the rate at which a body cools is proportional to the difference between its temperature and its environment. Consider a body at a positive temperature T, put in a 0 Celsius environment. Its temperature would be governed by:
dT/dt = -k*T
The solution would be:
T = T0*e^(-k*t)
The constant in front of the variable that changes is positive for exponential growth, and negative for either exponential approach or exponential decay.
Side note: if its environment were any temperature (Tbg) other than zero, then T would be replaced with (T - Tbg), and the solution would be: T = (T0 - Tbg)*e^(-k*t) + Tbg.
thank you !
Pretty sure I solved the DE but can someone verify? I multiird everything by 2 and put the 2 in the exponent of the ln. Then just raise to e and w lambert and sqrt to get y=+/-sqrt(w(e^(2x+2c))). Again, not sure if this works.
"dropped absolute sign because exponential is always positive"
BIG blunder here, the absolute value of y is equal to the exponential hence y itself could be equal to the negative of an exponential
for example if y = -e^x then |y|=e^x
but |y|=e^x DOES NOT imply y = e^x
instead: |y|=e^x implies y=plus or minus e^x
HOWEVER since you wrote C tilde infront of the exponential you somewhat saved it because C tilde can also represent a negative constant
It's a case of two wrongs make a right, when they are the right two wrongs.
Why no plus C on left side with y ?
Bc if u add a constant, let’s say C1 to the right and a constant C2 to the left, just subtract C1 from both sides. On the left you end up with C2 - C1. This is a constant minus a constant, which is just another constant. After all this, you end up with just one constant on the right, so they only add it on the right in the first place.
X'(t) = X^2(t) AND X(0) = a.
Hi Trefor please help me and let me know what you think about it. I know and my family knows I am very smart, as a 7th grader I do calculus 2 right now and understand everything really quick. My parent is strict about grades and keep pressuring me about bad grades in middle school will not get you to good high school. And yes my grade is very bad and I don’t know why I’m not doing so well in as online school, for some reasons my motivations towards listening well to teachers, doing work in time and ect.., I don’t want to do work hard as I used to. For first quarter grades, I got 5 A’s and 2 B’s, second quarter I got 4 A’s and 3 B’s. Now in third quarter I now have 2 A’s 2 D’s and 5 C’s. Mostly the B, C, grade is form science, health, art, social studies. I want to be an inventor, most my interests is about teleportation. Is it still possible for me to do well now from on and actually see my self graduating one of the top college and succeed in life? Or just I just give up at this point. Please let me know.
David K You can do everything.. you are capable of everything, don't be disappointed 🙁 maybe the online school isn't the best for you.. Try learning new things or take a rest every once in a while .. don’t care about grade coz if you did u wouldn’t get good grades.. that’s what I believe in .. hope I could help u 💖
support,really nice videos
i love you i love you i love youuuuuuu
awesome
Another person to add to the wall of fame: Eddie Woo, Blackpenredpen, …
Organic Chemistry Tutor
@@omikhe_ no I’m sorry if you like him but he’s quite bad
@@omikhe_ he doesn’t explain properly just gives too many examples
thank you sir :)
Nice video :)
Great video as always! Thank you!
Glad you enjoyed it!
He looks like Jack Dorsey.
How can we write dy/dx times dx as dy?
Can someone explain?
DX gets cancelled by numerator in the DY/DX leaving DY
@@joshuasflowers3203 but it is a notation
We can not cancel right?
I see it being treated as a fraction when you separate variable such as y dy = x dx.
It also has something to do with the chain rule of implicit differentiation though
2:21 so basically, you can cancel the dt's, just don't call it that or you'll lose friends
🙏🙏🙏
1:00
4:15
j
I don't like it. dy is what? It's notation. You're manipulating notation as if were a real object with a real definition. I think the dy is just notation used in conjunction with the integral sign that means, integrate with respect to y. It's not a thing.
You are half right. Integral in itself is not something completely abstract, it rather represents an infinite sum, hence an integral sign is a deformed letter S. dx on the other hand stands for infinitely small change of x, so integrating is actually adding up small pieces of (f(x) times dx), written down as [Integral sign]f(x) dx. When we write derivative this way: dy/dx, notation here also represents infinitely small values. In some way, the definition of limit is built inside the d(something) so we can actually work with it.
6:36
Hi Trefor I like you but I already know this;)
Come Erika badu me