Great review. I only got a C in differential equations and this help. I studied a lot for this class but went blank on the final. I was happy just to survive it.
You've got a sub, it is 2020, this is like the most simplest explanation I have ever come across, clicked on the first video (which was yours) and I am impressed. Litereally, going to note it down the way you have explained it.
Hi Mujeeb! Thank you so much for the feedback, and I'm glad you subscribed! This really motivates me to get back to making videos. If you have more feedback in the future, I'd be happy to hear it! -Greg
You honestly have the most concise teaching style for higher math topics on youtube. I was shocked to see you only had two playlists when you are such a gifted educator. I read from an earlier comment that you wanted to or are working on software to show more complex examples. Based on the lack of videos recently obviously other things have occupied your time (such is life). However, you should honestly approach 3Blue1Brown and see if you can work with him, since he's already developed a platform (in python) to make the visualisations. He hasn't covered a lot of topics, including Laplace and Taylor series, so there's work to be done there. There is even a repository on GitHub. I assume you might know all this. Your teaching style combined with his visualisation platform would honestly be a great gift to society, and would do justice to the beauty of mathematics. If not though, many many thanks for what you've put up so far, I literally have been screaming around my room for the last week saying "but what is actually the difference between a differential and algebraic equation!!!!!" and then you literally directly addressed my question in your video. best of luck in your endeavours of choice!
Wow, thank you SO much for your encouragement and insights. I’m really happy that you found the videos helpful! Messages like these have a very positive impact on me and the work I'm doing for the channel; I have been working diligently behind the scenes, apart from a few periods when I've been occupied with work and life. Unfortunately, since the channel isn’t big enough yet to support me financially, I can only spend a relatively small amount of time on it each day. I do want to use visualizations to enhance my teaching style, so it's nice to have someone who's watched my videos see value in that! Since you were so thoughtful in your message, I’ll catch you up on what I’ve been doing with a little progress report. ITERATING Before I published the first video in the differential equations playlist, I actually went through a bunch of iterations in format and style. After each iteration, I asked my students and friends for feedback, and their feedback helped me improve. Thanks to comments like yours, I feel good about the results of that process! Once the style was established and people expressed interest in the first set of videos, I decided to go through a similar process with the software and the curriculum. SOFTWARE Thank you for the suggestion about 3Blue1Brown. Grant Sanderson has been a major inspiration to me, so it feels great to have you mention my channel alongside his! I would love to do a collaboration. Python was actually one of my first programming languages after C and MATLAB, and I think it’s a great language. So, it wasn’t an easy decision to start from scratch in a different language as I have. There is definitely an opportunity cost to that decision. Part of my motivation is that, in addition to programming visualizations myself, I want to teach others to do it. I want the coding to be as accessible as possible, and the most accessible project I’ve found is p5.js. Imagine using the differential equations you learn to code your own interactive visualization, physics simulation, or video game right in the browser! There's already at least one channel that has used p5.js to do 3Blue1Brown style videos: th-cam.com/channels/8bvGUlaTahF0lHhMsZFhfw.html However, no one has contributed a p5.js library that can do what I need, so I decided to make one. The first iteration of my library is called visualODE (it’s built on ProcessingJS, which is sort of a precursor to p5.js). I’ll be adding more features as needed, and the code and design choices are outdated, but you can click through a working demo here: www.khanacademy.org/computer-programming/visualode-demo/6127327220629504 My next step is to convert visualODE to an open-source p5.js library, which among other things will make it easier to incorporate stunning 3D graphics! Learning to convert the library is the biggest portion of the work for the channel that I'm doing right now. PROGRAMMING CURRICULUM I’m planning to have my first set of programming videos cover the irreducible minimum of core programming concepts; these are concepts that are needed for basically any modern programming language. The idea is to have an introduction to coding for math learners that gets them solving math problems as quickly as possible. I looked around and didn’t see any existing resources along the lines of what I have in mind. I’ve already planned out most of this set of videos. I’ll probably incorporate this into the differential equations course after I've finished launching the p5.js library, so that I can get feedback for the second and third parts of the programming curriculum. DIFFERENTIAL EQUATIONS CURRICULUM Another thing I’ve been doing is rethinking the differential equations curriculum. I put a lot of time into this, thinking through not only how I want to explain things and which concepts are most important, but also scouring as many relevant books, historical references, and research papers as I could. I want to make something new, and to do that, I need to know what’s already out there so I can build on the best of it! My goal has been to plan out Part 1 of the course in a fair amount of detail, since each part of that material affects the other parts. I’m mostly finished with the planning stage. This will cover a lot of material, but again, I’ll publish it before I start work in earnest on any later parts, so that I can keep improving as I go! I'm just waiting on finishing the p5.js library, since I'll need to use that soon after I start making videos again. THANKS Thank you again for your suggestions and encouragement! It really means a lot to me! -Greg
Wow, thank you! I haven't given up :) There are a lot of videos that I have planned and am eager to work on. Usually, though, my time is consumed by my online tutoring service (I work for myself full-time as an upper-level math tutor). I love working with students one-on-one, but I'd like to spend a bigger chunk of my time on these videos. The problem I've been trying to figure out is the following: how can I find the time to make these free videos while still supporting myself financially? I've been working on a solution for a while now. Devoting a serious amount of time to free videos will mean a big financial hit; so, over the last couple years, I've been working extra hours to build up some savings. Starting in 2018, I plan to reduce my tutoring load on a trial basis. Once I'm able to put out videos regularly over the course of a semester, I'll see if enough people find them helpful. If they do, then I'll try to make the project sustainable through a platform like Patreon. Thanks again for the encouragement!
Wow you replied.Since, you have replied. All I have is a question to ask, What is the point of slope field ? I can't seem to get my head around it. Edit: Hopefully you can start making new videos soon,maybe start off by visualising higher order derivatives, My brain fails to understand more than third order derivatives.
generic name: Hey there! I realized I missed a few comments, so I’m just replying now. I'm really glad you asked about the point of a slope field! Right now, I’m working on some software so that I can program custom visualizations that would be difficult to do by hand (including more complicated slope fields). Once I’m finished, though, I was thinking of making a video addressing your exact question. I wasn’t sure how many students would wonder about the point of a slope field, but now I’m thinking it’s a good idea to discuss it! Regarding higher-order derivatives, I will probably eventually do a review video on Taylor polynomials once they are needed. These are functions that make use of higher-order derivatives to better describe the behavior of a graph near a point, so that may answer your question at least partially. I hope you’ll let me know once I get that far!
My whole problem with this class has been that it's been hard for me to really visualize the concepts and understand what they mean, unlike say when I took Calc 1. Thanks so much for this video
Thanks! I appreciate the encouragement and kind words. I'm still here on the ground :) I've been developing computer visualizations and a new course plan. More videos are on their way!
@@HigherMathNotes o really, soory i thought you were gone, because i saw " last video 2 years ago" , 😉 you know that was that sarcasm, well, good luck for new work (we love mathematics) 💥 💥 💥
Why is it we add the homogeneous general solution to the inhomogeneous particular solution when working with the solution to second order differentials?
The term “homogeneous” gets used in many contexts. Even when it’s used to describe a differential equation, it can mean completely different things! My guess is that you are asking about equations like y’’ + 5y’ + 6y = 0 or y’’ + 5y’ + 6y = sin(x). Why are particular solutions and general solutions so important in this context? I’ll talk about this in later videos, but I’m glad you asked. EXAMPLE: To help you understand, I’ll start with two simpler, algebraic equations. (1) 2x + y = 0 (2) 2x + y = 6 Equation (1) is homogeneous (the constant term is zero) and Equation (2) is nonhomogeneous (the constant term 6 is nonzero). How are the solutions to these two equations related? ANSWER: First, consider 2x + y = 0. Can you find a solution? Here’s one: (1, -2). Plugging in the coordinates produces the true equation 2(1) + -2 = 0, so this is one particular solution. In this context, the “general solution” is the set of all possible solutions. How can we find this “general solution”? Solving 2x + y = 0 for y, we get y = -2x. For every x, we multiply by -2 to get the y. So, the “general solution” consists of all the points that look like (x, -2x), and x can be any number we want. (The particular solution (1, -2), from before, comes from setting x=1). Second, consider 2x + y = 6. Can you find a solution? Here’s one: (0, 6). Now we get to the interesting part: we already found all solutions to 2x + y = 0, so could we just modify those slightly to get all the solutions to 2x + y = 6? Hint: try graphing 2x + y = 0 and 2x + y = 6 on the same pair of axes. They look like two copies of the same line but one crosses the y-axis at 0 and the other crosses at 6. Now remember that the graphs are pictures of the solutions: every point on the graph of 2x + y = 0 represents a solution to that equation, and the same is true for 2x + y = 6. So, if we could shift the solutions along the line 2x + y = 0 up by 6, we would get the solutions to 2x+y=6. We can do this by adding 6 to the y-coordinate, and since we’re shifting straight up, we’re adding 0 to the x-coordinate. Specifically, the solutions to 2x + y = 6 are all the points of the form (x, -2x) + (0, 6) = (x + 0, -2x + 6) = (x, -2x + 6), where x can be any number. [If you haven’t ever added points like this, don’t worry; you’ll learn about that whenever you learn about vectors for the first time.] ANALYSIS: What do we learn? The point (0, 6) that we added isn’t arbitrary. It’s a particular solution to 2x + y = 6. So, we found all solutions to 2x + y = 6 by (a) finding the “general solution” of the homogeneous equation 2x + y = 0, and (b) adding a particular solution of the NONhomogeneous equation 2x + y = 6. This also works for differential equations like y’’ + 5y’ + 6 = 0 and y’’ + 5y’ + 6 = sin(x). The visual representation is different, but the overall process is the same. To get the general solution to a nonhomogeneous equation like y’’ + 5y’ + 6 = sin(x), you just have to find one particular solution! Then you add that to the general solution of the homogeneous equation y’’ + 5y’ + 6 = 0, which is easier to find. I hope that helps!
homogenous/non homogeneous i know this homogenous the right side = 0 non homogeneous the right side in not 0 and that's it. You should make a vdo about this.
Thanks for the feedback! Very helpful. I'm working on a lesson for one or more upcoming videos that will explain linearity, including homogeneous vs. nonhomogeneous linear equations. I'll include the relationship of these concepts to the idea of general and particular solutions. Currently, I'm reorganizing my original course plan, but most likely, I will be making a few other videos before I get to the video that answers your question. This will allow me to incorporate all the important topics in an organized way. Thanks again! Your feedback is always welcome.
How many solutions can a graph of ODE have.? Like if we put y(0)=0 in our equation and we get 0=0 how can we say that it doesn't have unique solution but solution exists.? Please explain?
But what is complementary functions? Why do we care about doing that the general solution of any ODE is the sum of complementary functions and particular integrals. How does it makes sense ? Please help me to get over from this question
Hi Meet! Thanks for your question. It looks like you're studying (1) linear independence in the context of solutions to (2) second-order linear ordinary differential equations. I'll cover those topics in later videos. I'd direct you to those videos now, but I haven't made them yet (as of the time this comment is posted).
Hi Mukesh! It's a question of notation. The inverse of the tangent function can be written as arctan(x) or as tan^(-1)(x). These two notations mean the same thing. Some authors prefer arctan(x), since tan^(-1)(x) can cause confusion. Specifically, tan^(-1)(x) is sometimes confused by students to mean tan(x) to the negative first power, i.e. 1/tan(x). To be clear, 1/tan(x) is what we call the multiplicative inverse of tan(x), which is a different function with its own name (cotangent). On the other hand, arctan(x) and tan^(-1)(x) both refer to the inverse function (the compositional inverse). This confusion is avoided by always writing (tan(x))^(-1) when the multiplicative inverse is intended and writing tan^(-1)(x) when the compositional inverse is intended. This is fine, as long as the conventions are made clear. I'm glad you mentioned this, since now I get to share a little bit about the etymology of the term arctangent. The prefix "arc-" comes from Latin "arcus," meaning "bow" in English, as in an archer's bow. Being curved, the term was eventually applied to a curved piece of a circle. In the unit circle (the circle with radius one), the measure of a central angle in radians is by definition equal to the length of the corresponding arc. So, you can remember what arctan(x) means by thinking that it's the arc (i.e. angle) whose tangent equals x. (cf. The Words of Mathematics by Schwartzman, or the Wikipedia article on inverse trig functions). Thanks for the comment!
SORRY BROTHER BUT YOU ARE WRONG!!!!!!!!!!!!!!!!!! IF YOU WANT TO MOVE THE CURVE ""e"" TO THE POWER OF ""x"" TO THE RIGHT OR TO THE LEFT. YOU HAVE TO RAISE ""e"" TO ""(x-a)"" OR TO (x+a). BY CHANGING ""C"" WHAT YOU ARE DOING IS MAKING THE CURVE THINNER OR SQUATTER!!!!!!!!!!!!
Wow, you're very emphatic haha. I was not wrong, but it might look like I was. I'll explain. The graph of y = Ce^(2t) consists of the ordered pairs (t, Ce^(2t)) for all values of t. So, when we change the value of C from 1 to 2, we are doubling the y-coordinate of each point on the graph. In other words, we are stretching the graph vertically. However, it may appear to you that I translated the graph of y = e^(2t) to the left to obtain the graph of y = 2e^(2t). This is actually also true, due to a key property of the exponential function: e^x e^y = e^(x+y). To see how this applies when we multiply by 2, notice that 2 = e^(ln 2). Therefore, y = 2e^(2t) = e^(ln 2)e^(2t) = e^(ln 2 + 2t) = e^(2(0.5 ln 2 + t)). So, scaling the graph vertically by 2 and shifting the graph to the left by 0.5 ln 2 (which is approximately 0.35) have the same effect. To be extra clear, this is due to that key property of the exponential function. If we try this with, say y = x^2, then scaling the graph vertically and shifting the graph to the left will have different effects. SIDE NOTE: Be careful about "thinner" and "squatter." This is a related mistake that's easy to make. In general, multiplying y = f(x) by C to get y = C*f(x) makes the graph taller or squatter, not thinner or squatter. I'll show you what I mean. For a function like y = x^2, multiplying by 2 to obtain y = 2x^2 can again be interpreted in two ways, this time due to a property (sometimes referred to as scale invariance) of the power function y = x^2. Since y = 2x^2 = (sqrt(2) x)^2, stretching vertically by 2 and compressing horizontally by sqrt(2) have the same effect. Due to this, it's easy to think that multiplying a function by 2 makes the graph thinner (compresses it horizontally). That is true for y = x^2, but it is not true for all functions. An example I like is y = sin(x). We know that y = 2*sin(x) has twice the amplitude of y = sin x, so it's stretched vertically, but it's not compressed horizontally, since the x-intercepts don't move at all. Even y = x^2 - 1 will be stretched vertically but not compressed horizontally when multiplied by 2. So, multiplying y = f(x) by C to get y = C*f(x) only sometimes has the same effect as a horizontal stretch or compression, but it always stretches or compresses the graph VERTICALLY. That's because the graph of y = f(x) consists of the ordered pairs (x, f(x)) for all x in the domain of f, whereas the graph of y = C*f(x) consists of the ordered pairs (x, C*f(x)). So, the y-coordinate (i.e. the vertical coordinate) is certainly scaled by C.
Well, I didn`t really intend to be mean or disrespectful to you. Quite the contrary I respect your effort and your work is very laudable. However, this time my friend I´m sorry to tell you that you are wrong. I think that you owe to your viewers a truthful explanation. I just want to remind to you (and to many out there) that Mathematics is not ""words"". I know the properties of logarithms. What we have at the "end" is a "C" multiplied by "e" to the power of "2t", and that`s when you call the equation "general" (which indeed is general, but not in the way you show it) and you change C to 2 and move the curve to the left. That is when I say that ""that is incorrect"". The correct equation for the curve that "you" show is "e" to the power of (t+2). In the equation you show (which is rightfully the result of the derivative of the fuction you showed at the beginning) the explanation of "C" is incorrect. I guess you will agree that "e" to the power of any "t" or to the power of any "at" ("a" being any number) is rooted to "1" when "t" equals zero. The "C" as I said before only elongates or shortens the curve. But the curve no matter what the value of "C" is, stays """as it were""" in the same """""""place"""""""Trust me I know what I am talking about. For god`s sake, just go to any graph plotter and graph your equations and the ones that I suggest and you`ll see!!!!! Keep the good work!!!!!
If it's hard to understand my original reply without visuals, you might try plotting y = e^(2t) and y = 2e^(2t) yourself, as you've suggested. A link to a plot on WolframAlpha is below. Notice that the graph of y = 2e^(2t) is above (and to the left) of the graph of y = e^(2t), as I've drawn them in the video. www.wolframalpha.com/input/?i=y+%3D+e%5E%282t%29%2C+y+%3D+2e%5E%282t%29 In case others have the same question as you do, I'll leave this discussion up. If you reread my original reply and have any specific questions, I'd be happy to answer them. I hope you're genuinely interested, but if you're trolling, then this will be my last reply!
Great review. I only got a C in differential equations and this help. I studied a lot for this class but went blank on the final. I was happy just to survive it.
I'm so glad this helped, and thank you for the feedback! I really appreciate it.
You've got a sub, it is 2020, this is like the most simplest explanation I have ever come across, clicked on the first video (which was yours) and I am impressed. Litereally, going to note it down the way you have explained it.
Hi Mujeeb! Thank you so much for the feedback, and I'm glad you subscribed! This really motivates me to get back to making videos. If you have more feedback in the future, I'd be happy to hear it! -Greg
You honestly have the most concise teaching style for higher math topics on youtube. I was shocked to see you only had two playlists when you are such a gifted educator.
I read from an earlier comment that you wanted to or are working on software to show more complex examples.
Based on the lack of videos recently obviously other things have occupied your time (such is life). However, you should honestly approach 3Blue1Brown and see if you can work with him, since he's already developed a platform (in python) to make the visualisations. He hasn't covered a lot of topics, including Laplace and Taylor series, so there's work to be done there. There is even a repository on GitHub. I assume you might know all this.
Your teaching style combined with his visualisation platform would honestly be a great gift to society, and would do justice to the beauty of mathematics.
If not though, many many thanks for what you've put up so far, I literally have been screaming around my room for the last week saying "but what is actually the difference between a differential and algebraic equation!!!!!" and then you literally directly addressed my question in your video.
best of luck in your endeavours of choice!
Wow, thank you SO much for your encouragement and insights. I’m really happy that you found the videos helpful!
Messages like these have a very positive impact on me and the work I'm doing for the channel; I have been working diligently behind the scenes, apart from a few periods when I've been occupied with work and life. Unfortunately, since the channel isn’t big enough yet to support me financially, I can only spend a relatively small amount of time on it each day.
I do want to use visualizations to enhance my teaching style, so it's nice to have someone who's watched my videos see value in that! Since you were so thoughtful in your message, I’ll catch you up on what I’ve been doing with a little progress report.
ITERATING
Before I published the first video in the differential equations playlist, I actually went through a bunch of iterations in format and style. After each iteration, I asked my students and friends for feedback, and their feedback helped me improve. Thanks to comments like yours, I feel good about the results of that process!
Once the style was established and people expressed interest in the first set of videos, I decided to go through a similar process with the software and the curriculum.
SOFTWARE
Thank you for the suggestion about 3Blue1Brown. Grant Sanderson has been a major inspiration to me, so it feels great to have you mention my channel alongside his! I would love to do a collaboration. Python was actually one of my first programming languages after C and MATLAB, and I think it’s a great language. So, it wasn’t an easy decision to start from scratch in a different language as I have.
There is definitely an opportunity cost to that decision. Part of my motivation is that, in addition to programming visualizations myself, I want to teach others to do it. I want the coding to be as accessible as possible, and the most accessible project I’ve found is p5.js. Imagine using the differential equations you learn to code your own interactive visualization, physics simulation, or video game right in the browser! There's already at least one channel that has used p5.js to do 3Blue1Brown style videos:
th-cam.com/channels/8bvGUlaTahF0lHhMsZFhfw.html
However, no one has contributed a p5.js library that can do what I need, so I decided to make one.
The first iteration of my library is called visualODE (it’s built on ProcessingJS, which is sort of a precursor to p5.js). I’ll be adding more features as needed, and the code and design choices are outdated, but you can click through a working demo here:
www.khanacademy.org/computer-programming/visualode-demo/6127327220629504
My next step is to convert visualODE to an open-source p5.js library, which among other things will make it easier to incorporate stunning 3D graphics! Learning to convert the library is the biggest portion of the work for the channel that I'm doing right now.
PROGRAMMING CURRICULUM
I’m planning to have my first set of programming videos cover the irreducible minimum of core programming concepts; these are concepts that are needed for basically any modern programming language. The idea is to have an introduction to coding for math learners that gets them solving math problems as quickly as possible. I looked around and didn’t see any existing resources along the lines of what I have in mind.
I’ve already planned out most of this set of videos. I’ll probably incorporate this into the differential equations course after I've finished launching the p5.js library, so that I can get feedback for the second and third parts of the programming curriculum.
DIFFERENTIAL EQUATIONS CURRICULUM
Another thing I’ve been doing is rethinking the differential equations curriculum. I put a lot of time into this, thinking through not only how I want to explain things and which concepts are most important, but also scouring as many relevant books, historical references, and research papers as I could. I want to make something new, and to do that, I need to know what’s already out there so I can build on the best of it!
My goal has been to plan out Part 1 of the course in a fair amount of detail, since each part of that material affects the other parts. I’m mostly finished with the planning stage. This will cover a lot of material, but again, I’ll publish it before I start work in earnest on any later parts, so that I can keep improving as I go! I'm just waiting on finishing the p5.js library, since I'll need to use that soon after I start making videos again.
THANKS
Thank you again for your suggestions and encouragement! It really means a lot to me!
-Greg
Thank you very much for this, I don't know what I would've done without this.
You're very welcome! I'm so glad to have helped!
Why did you stop making videos?Honestly,you are a great teacher.
Wow, thank you! I haven't given up :) There are a lot of videos that I have planned and am eager to work on. Usually, though, my time is consumed by my online tutoring service (I work for myself full-time as an upper-level math tutor).
I love working with students one-on-one, but I'd like to spend a bigger chunk of my time on these videos. The problem I've been trying to figure out is the following: how can I find the time to make these free videos while still supporting myself financially?
I've been working on a solution for a while now. Devoting a serious amount of time to free videos will mean a big financial hit; so, over the last couple years, I've been working extra hours to build up some savings. Starting in 2018, I plan to reduce my tutoring load on a trial basis. Once I'm able to put out videos regularly over the course of a semester, I'll see if enough people find them helpful. If they do, then I'll try to make the project sustainable through a platform like Patreon.
Thanks again for the encouragement!
Wow you replied.Since, you have replied. All I have is a question to ask, What is the point of slope field ? I can't seem to get my head around it.
Edit: Hopefully you can start making new videos soon,maybe start off by visualising higher order derivatives, My brain fails to understand more than third order derivatives.
Yeah patron sounds like a great idea, you definitely deserve some payment for these high quality videos!
generic name: Hey there! I realized I missed a few comments, so I’m just replying now. I'm really glad you asked about the point of a slope field!
Right now, I’m working on some software so that I can program custom visualizations that would be difficult to do by hand (including more complicated slope fields). Once I’m finished, though, I was thinking of making a video addressing your exact question. I wasn’t sure how many students would wonder about the point of a slope field, but now I’m thinking it’s a good idea to discuss it!
Regarding higher-order derivatives, I will probably eventually do a review video on Taylor polynomials once they are needed. These are functions that make use of higher-order derivatives to better describe the behavior of a graph near a point, so that may answer your question at least partially. I hope you’ll let me know once I get that far!
Thanks!!
Teachers at my academy recommended your channel to us students, just wanted to let you know mr. youtuber
That makes me so happy!! Thank you for letting me know :)
Concise and wonderful. Thanks!
My whole problem with this class has been that it's been hard for me to really visualize the concepts and understand what they mean, unlike say when I took Calc 1. Thanks so much for this video
You're very welcome! Thank YOU so much for your comment :) It means a lot to me.
Brother you were doing good work, i hope you're good in the skys.
Thanks! I appreciate the encouragement and kind words. I'm still here on the ground :) I've been developing computer visualizations and a new course plan. More videos are on their way!
@@HigherMathNotes o really, soory i thought you were gone, because i saw " last video 2 years ago" , 😉 you know that was that sarcasm, well, good luck for new work
(we love mathematics)
💥 💥 💥
Thanks for understanding 😊
Wow, this vid was amazing. Keep up the good work.
Thank you so much for the encouragement!
Brilliantly explained!❤️👌👏
Hi asha!
Thanks so much for your feedback! I'm glad you liked the explanation :)
--Greg (Higher Math Notes)
You're an amazing tutor!
Wow, thank you for the encouragement!
I am very thankful for this video
I wish I would be able to attend yur class❤ as your student
Wow, I'm really glad it helped! Thank you so much for the feedback!
@@HigherMathNotes ❤❤❤
Why is it we add the homogeneous general solution to the inhomogeneous particular solution when working with the solution to second order differentials?
Great video!
I'm glad you liked it!! I'm looking forward to uploading the next video. It's currently in the works.
👍🏻
What is that homogenous/non homogeneous stuff related to this Particular and general solution concept?
Kindly help.
The term “homogeneous” gets used in many contexts. Even when it’s used to describe a differential equation, it can mean completely different things!
My guess is that you are asking about equations like y’’ + 5y’ + 6y = 0 or y’’ + 5y’ + 6y = sin(x). Why are particular solutions and general solutions so important in this context? I’ll talk about this in later videos, but I’m glad you asked.
EXAMPLE:
To help you understand, I’ll start with two simpler, algebraic equations.
(1) 2x + y = 0
(2) 2x + y = 6
Equation (1) is homogeneous (the constant term is zero) and Equation (2) is nonhomogeneous (the constant term 6 is nonzero). How are the solutions to these two equations related?
ANSWER:
First, consider 2x + y = 0. Can you find a solution? Here’s one: (1, -2). Plugging in the coordinates produces the true equation 2(1) + -2 = 0, so this is one particular solution.
In this context, the “general solution” is the set of all possible solutions. How can we find this “general solution”? Solving 2x + y = 0 for y, we get y = -2x. For every x, we multiply by -2 to get the y. So, the “general solution” consists of all the points that look like (x, -2x), and x can be any number we want. (The particular solution (1, -2), from before, comes from setting x=1).
Second, consider 2x + y = 6. Can you find a solution? Here’s one: (0, 6).
Now we get to the interesting part: we already found all solutions to 2x + y = 0, so could we just modify those slightly to get all the solutions to 2x + y = 6?
Hint: try graphing 2x + y = 0 and 2x + y = 6 on the same pair of axes. They look like two copies of the same line but one crosses the y-axis at 0 and the other crosses at 6. Now remember that the graphs are pictures of the solutions: every point on the graph of 2x + y = 0 represents a solution to that equation, and the same is true for 2x + y = 6.
So, if we could shift the solutions along the line 2x + y = 0 up by 6, we would get the solutions to 2x+y=6. We can do this by adding 6 to the y-coordinate, and since we’re shifting straight up, we’re adding 0 to the x-coordinate. Specifically, the solutions to 2x + y = 6 are all the points of the form (x, -2x) + (0, 6) = (x + 0, -2x + 6) = (x, -2x + 6), where x can be any number. [If you haven’t ever added points like this, don’t worry; you’ll learn about that whenever you learn about vectors for the first time.]
ANALYSIS:
What do we learn? The point (0, 6) that we added isn’t arbitrary. It’s a particular solution to 2x + y = 6. So, we found all solutions to 2x + y = 6 by
(a) finding the “general solution” of the homogeneous equation 2x + y = 0, and
(b) adding a particular solution of the NONhomogeneous equation 2x + y = 6.
This also works for differential equations like y’’ + 5y’ + 6 = 0 and y’’ + 5y’ + 6 = sin(x). The visual representation is different, but the overall process is the same. To get the general solution to a nonhomogeneous equation like y’’ + 5y’ + 6 = sin(x), you just have to find one particular solution! Then you add that to the general solution of the homogeneous equation y’’ + 5y’ + 6 = 0, which is easier to find.
I hope that helps!
@@HigherMathNotes Thank you. Can you do a vdo about this explanation again? with more examples. please
homogenous/non homogeneous
i know this
homogenous the right side = 0
non homogeneous the right side in not 0
and that's it.
You should make a vdo about this.
Thanks for the feedback! Very helpful. I'm working on a lesson for one or more upcoming videos that will explain linearity, including homogeneous vs. nonhomogeneous linear equations. I'll include the relationship of these concepts to the idea of general and particular solutions.
Currently, I'm reorganizing my original course plan, but most likely, I will be making a few other videos before I get to the video that answers your question. This will allow me to incorporate all the important topics in an organized way.
Thanks again! Your feedback is always welcome.
i felt it in my bones but thanks to your video im sure, thanks!
Mateusz Bujnowicz I'm glad the video helped you confirm your intuition! Thanks so much for the feedback!
I'm surprised you understood what he meant. XD
Well that solved my doubt :D
Thanks
gammaking: Thank you for letting me know!
How many solutions can a graph of ODE have.? Like if we put y(0)=0 in our equation and we get 0=0 how can we say that it doesn't have unique solution but solution exists.? Please explain?
But what is complementary functions? Why do we care about doing that the general solution of any ODE is the sum of complementary functions and particular integrals. How does it makes sense ? Please help me to get over from this question
Very helpful
Glad to hear it! Thanks for the feedback :)
IF WE HAVE GENERAL SOLUTION LIKE:y=C e^(2t)+Ć e^(3t).then e^(2t),and e^(3t) are must be linearly independent. .CAN u EXPLAIN REASON BEHIND THAT?
Hi Meet! Thanks for your question. It looks like you're studying
(1) linear independence
in the context of solutions to
(2) second-order linear ordinary differential equations.
I'll cover those topics in later videos. I'd direct you to those videos now, but I haven't made them yet (as of the time this comment is posted).
There should be tan inverse t..
Because...
Derivative of tan inverse t = 1/(1+t^2)
Hi Mukesh! It's a question of notation. The inverse of the tangent function can be written as arctan(x) or as tan^(-1)(x). These two notations mean the same thing.
Some authors prefer arctan(x), since tan^(-1)(x) can cause confusion. Specifically, tan^(-1)(x) is sometimes confused by students to mean tan(x) to the negative first power, i.e. 1/tan(x). To be clear, 1/tan(x) is what we call the multiplicative inverse of tan(x), which is a different function with its own name (cotangent). On the other hand, arctan(x) and tan^(-1)(x) both refer to the inverse function (the compositional inverse). This confusion is avoided by always writing (tan(x))^(-1) when the multiplicative inverse is intended and writing tan^(-1)(x) when the compositional inverse is intended. This is fine, as long as the conventions are made clear.
I'm glad you mentioned this, since now I get to share a little bit about the etymology of the term arctangent. The prefix "arc-" comes from Latin "arcus," meaning "bow" in English, as in an archer's bow. Being curved, the term was eventually applied to a curved piece of a circle. In the unit circle (the circle with radius one), the measure of a central angle in radians is by definition equal to the length of the corresponding arc. So, you can remember what arctan(x) means by thinking that it's the arc (i.e. angle) whose tangent equals x. (cf. The Words of Mathematics by Schwartzman, or the Wikipedia article on inverse trig functions).
Thanks for the comment!
SORRY BROTHER BUT YOU ARE WRONG!!!!!!!!!!!!!!!!!! IF YOU WANT TO MOVE THE CURVE ""e"" TO THE POWER OF ""x"" TO THE RIGHT OR TO THE LEFT. YOU HAVE TO RAISE ""e"" TO ""(x-a)"" OR TO (x+a). BY CHANGING ""C"" WHAT YOU ARE DOING IS MAKING THE CURVE THINNER OR SQUATTER!!!!!!!!!!!!
Wow, you're very emphatic haha. I was not wrong, but it might look like I was. I'll explain.
The graph of y = Ce^(2t) consists of the ordered pairs (t, Ce^(2t)) for all values of t. So, when we change the value of C from 1 to 2, we are doubling the y-coordinate of each point on the graph. In other words, we are stretching the graph vertically.
However, it may appear to you that I translated the graph of y = e^(2t) to the left to obtain the graph of y = 2e^(2t). This is actually also true, due to a key property of the exponential function: e^x e^y = e^(x+y). To see how this applies when we multiply by 2, notice that 2 = e^(ln 2). Therefore,
y = 2e^(2t) = e^(ln 2)e^(2t) = e^(ln 2 + 2t) = e^(2(0.5 ln 2 + t)).
So, scaling the graph vertically by 2 and shifting the graph to the left by 0.5 ln 2 (which is approximately 0.35) have the same effect. To be extra clear, this is due to that key property of the exponential function. If we try this with, say y = x^2, then scaling the graph vertically and shifting the graph to the left will have different effects.
SIDE NOTE:
Be careful about "thinner" and "squatter." This is a related mistake that's easy to make. In general, multiplying y = f(x) by C to get y = C*f(x) makes the graph taller or squatter, not thinner or squatter. I'll show you what I mean.
For a function like y = x^2, multiplying by 2 to obtain y = 2x^2 can again be interpreted in two ways, this time due to a property (sometimes referred to as scale invariance) of the power function y = x^2. Since y = 2x^2 = (sqrt(2) x)^2, stretching vertically by 2 and compressing horizontally by sqrt(2) have the same effect. Due to this, it's easy to think that multiplying a function by 2 makes the graph thinner (compresses it horizontally). That is true for y = x^2, but it is not true for all functions.
An example I like is y = sin(x). We know that y = 2*sin(x) has twice the amplitude of y = sin x, so it's stretched vertically, but it's not compressed horizontally, since the x-intercepts don't move at all. Even y = x^2 - 1 will be stretched vertically but not compressed horizontally when multiplied by 2.
So, multiplying y = f(x) by C to get y = C*f(x) only sometimes has the same effect as a horizontal stretch or compression, but it always stretches or compresses the graph VERTICALLY. That's because the graph of y = f(x) consists of the ordered pairs (x, f(x)) for all x in the domain of f, whereas the graph of y = C*f(x) consists of the ordered pairs (x, C*f(x)). So, the y-coordinate (i.e. the vertical coordinate) is certainly scaled by C.
Well, I didn`t really intend to be mean or disrespectful to you. Quite the contrary I respect your effort and your work is very laudable. However, this time my friend I´m sorry to tell you that you are wrong. I think that you owe to your viewers a truthful explanation. I just want to remind to you (and to many out there) that Mathematics is not ""words"". I know the properties of logarithms. What we have at the "end" is a "C" multiplied by "e" to the power of "2t", and that`s when you call the equation "general" (which indeed is general, but not in the way you show it) and you change C to 2 and move the curve to the left. That is when I say that ""that is incorrect"". The correct equation for the curve that "you" show is "e" to the power of (t+2). In the equation you show (which is rightfully the result of the derivative of the fuction you showed at the beginning) the explanation of "C" is incorrect. I guess you will agree that "e" to the power of any "t" or to the power of any "at" ("a" being any number) is rooted to "1" when "t" equals zero. The "C" as I said before only elongates or shortens the curve. But the curve no matter what the value of "C" is, stays """as it were""" in the same """""""place"""""""Trust me I know what I am talking about. For god`s sake, just go to any graph plotter and graph your equations and the ones that I suggest and you`ll see!!!!! Keep the good work!!!!!
If it's hard to understand my original reply without visuals, you might try plotting y = e^(2t) and y = 2e^(2t) yourself, as you've suggested. A link to a plot on WolframAlpha is below. Notice that the graph of y = 2e^(2t) is above (and to the left) of the graph of y = e^(2t), as I've drawn them in the video.
www.wolframalpha.com/input/?i=y+%3D+e%5E%282t%29%2C+y+%3D+2e%5E%282t%29
In case others have the same question as you do, I'll leave this discussion up. If you reread my original reply and have any specific questions, I'd be happy to answer them. I hope you're genuinely interested, but if you're trolling, then this will be my last reply!
@@HigherMathNotes YOU AREN`T JUST GRACIOUS AND ACCEPT EVIDENCE!!!!!! YOU NEED TO DO A LOT OF RECKONING. BYE!!!!!!!!!!!!!!!!!!!!!