Exceellent lecture. I learnt all this stuff in engineering long ago, and even went through the stuff after seeing your video, which is not so easy to understand! You made this look like pure magic! You just need to carry a wand and wear a hat in your next video, and whoa, there you are- Math Magic! Thanks so much Mr. Bazet (I am from India, and 60+). You make math fun. God Bless you
In 9:15 why did we divide y1+y2 by 2 and didn't just let the general solution be y1+y2? I, also, have one more question. Are the general solutions we found the only solutions to tis ODE, or there are other solutions too?
Dr., since all solutions can be expressed as linear combinations of the two linearly independent ones, are the complex solutions linear combinations of the 2 real solutions? In other words, when we talk about a multiplicative "constant" to make linear combinations, "i" is included as a possibility too? I always thought these multiplicative constants as reals, but apparently they can also be complex?
Exactly. If modelling real world phenomena it is easier to keep everything real which is why we do this shift. But yes we can have complex functions and complex coefficients and it works either way.
Yes, and that's precisely how you reconcile imaginary solutions, by setting up your two constants to be a complex conjugate pair. Eventually, you can show that it is a linear combination of sine and cosine, given an imaginary solution to the characteristic equation.
Hi Trefor, I was in your calculus class last summer. Just wanted to say big thanks for putting out content like this. Thanks to you and other math/physics channels on youtube I went from "lets just get though the class and get the grade" to starting to see the beauty in math. Your work is very much appreciated, thank you!
You may be wondering why the case with repeated roots behaves so differently from the case when every root is distinct. This video presents having two real roots and two complex roots are being different cases, but the video acknowledges that in both cases, you can just simply write the general solution as A·exp(r·t) + B·exp(s·t) if you really want to, where r and s are the roots to the characteristic polynomial. This is definitely not so in the case of repeated roots, where one of the solutions has a factor of t, and this happens only specifically in this case. Why? In the previous video, in the comments section, I explained that, in fact, you can solve second-order linear equations with constant coefficients without having to "guess" the solutions, and the method I presented for solving the example equation relies on the fact that differential equations can always be rewritten so that they look like linear algebra equations, and that is because the nth order detivative, for every natural n, is always a linear operator, and so it behaves like a matrix. Specifically, the example equation was y''(t) - y'(t) - 6·y(t) = 0, which I said can be written as (D^2 - D - 6·I)[y(t)] = 0, where D is the derivative operator and I the identity "matrix." Since D^2 - D - 6·I is a polynomial in D, this can be factored as (D + 2)·(D - 3), and this was the key to solving the equation. The same concept actually does apply to an arbitrary second-order linear differential equation. In particular, a·y''(t) + b·y'(t) + c·y(t) = 0 can always be written as (a·D^2 + b·D + c·I)[y(t)] = 0, and since a·D^2 + b·D + c·I is a polynomial of degree 2, it can always be factored as a·(D - s·I)·(D - r·I), where s and r are the roots of the polynomial. I already gave the details in that one comment I wrote in the previous video in this channel, so for the rest of this explanation, I am going to directly work with this factorization instead. So when solving the equation [a·(D - s·I)·(D - r·I)][y(t)] = 0, what to do? Let (D - r·I)[y(t)] = z(t), so the equation to solve is simply a·(D - s·I)[z(t)] = 0. This is just a first-order linear equation, which after dividing by a, is simply (D - s·I)[z(t)] = 0, which can be rewritten as z'(t) - s·z(t) = 0. This has solutions z(t) = A·exp(s·t), where A is just a constant of integration. So (D - r·I)[y(t)] = A·exp(s·t), which can just be written as y'(t) - r·y(t) = A·exp(s·t). The integration factor is exp(-r·t), so multiplying by it results in exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A·exp[(s - r)·t]. This is the key moment. This is the moment where having repeated roots, as opposed to distinct roots, makes an important difference. Why? Because if s = r, which is the case with repeated roots, then s - r = 0, so A·exp[(s - r)·t] = A. Therefore, when you antidifferentiate both sides of exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A, you simply get exp(-r·t)·y(t) = A·t + B, so y(t) = A·t·exp(r·t) + B·exp(r·t), and this gives you the same result as in the video. However, if s and r are distinct, then s - r is nonzero, so A·exp[(s - r)·t] is simply an exponential function. Therefore, when antidifferentiating exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A·exp[(s - r)·t], you get exp(-r·t)·y(t) = A/(s - r)·exp[(s - r)·t] + B, hence y(t) = A/(s - r)·exp(s·t) + B·exp(r·t), which matches the result in the video if you simply acknowkedge that, since s - r is nonzero, A/(s - r) is just another arbitrary constant. The differences and similarities between both cases are more clear when you just leave A/(s - r)·exp(s·t) + B·exp(r·t) written as {A/(s - r)·exp[(s - r)·t] + B}·exp(r·t). Both solution forms {A/(s - r)·exp[(s - r)·t] + B}·exp(r·t) and (A·t + B)·exp(r·t) have the factor exp(r·t) in them, and the repeated roots case replaces exp[(s - r)·t]/(s - r) with t. This makes sense if seen as a way of avoiding division by 0, but ultimately, it results from the fact that, in the repeated roots case, a constant was being antidifferentiated, instead of an exponential. Another way to interpret this is via a strange limit argument. If we expand exp[(s - r)·t] as its Maclaurin series definition, then exp[(s - r)·t]/(s - r) = 1/(s - r) + t + (s - r)·t^2·f(t), with f(0) = 1/2, and f(t) is the Maclaurin series (s - r)^(n + 1)·t^n/(n + 2)! for every natural n. 1/(s - r) does not exist if s - r = 0 or even if s - r -> 0, but if one could somehow "regularize" this summand, so that it actually becomes 0 when s - r = 0, then the result would just be t, as expected, since then, (s - r)·t^2·f(t) = 0. In fact, the integral of exp[(s - r)·t'] on the interval [0, t] with respect to t' is (exp[(s - r)·t] - 1)/(s - r), which does have limit t as s -> r.
Thank you so much for going further to prove exactly why the form of the particular integral is that . Really appreciate what you do and you deserve so much more recognition . Thank you so much !!!!
Greetings from Brazil. Your explanations are awesome. Thank you very much for your kindness in helping so many people around the world learn/review so important stuff.
ooooh, I do want to do this and would fit nicely with my series on vector calculus and line integrals. Not sure exactly when though, not for a few months at least:)
Hmmm, I'm wondering what's the motivation for arbitrarily having te^rt as the second solution in the case of a repeated root? is there perhaps a more direct approach Edit: Nevermind, just saw the brilliant comment explaining this all via a direct approach with the derivative as a linear operator, and the te^rt guess beautifully turns out to be very much analogous to the first order case when your inhomogeneous part solves the homogeneous equation. Loving this series by the way - it really makes the course seem so much more approachable than it might look from afar.
9:45 hold up, these are BOTH general solutions according to the principle of superposition . I thought any differential equation could only have a single general solution? Or do we just _choose_ which solution is our general solution, or are we allowed to how multiple general solutions? Here, you just call them “solutions”
@@pepehimovic3135Yes, you can make a linear combination of the complex root case. And that is precisely how you reconcile it to have a real solution. Let the two constants be a complex conjugate pair, with equal and opposite imaginary parts. You'll eventually cancel out the imaginary constants in front of your terms, and show that imaginary solutions to the characteristic equation produce a linear combination of sine and cosine. For 1st order diffEQ's there is only one solution. For 2nd order diffEQ's there are two fundamental solutions that are related to each other, and a linear combination of the two. In general, nth order diffEQ's will have n fundamental solutions, and n arbitrary constants to combine them with each other. In a process that produces other arbitrary coefficients, like variation of parameters and method of integrating factor that has you integrate, you'll produce more arbitrary constants, that ultimately can be combined with the original ones. Such that only a number of arbitrary constants equal to the order of the DiffEQ remain.
Professor Bazett, thank you for explaining the different cases that's involved in the Constant Coefficient Ordinary Differential Equations. The three cases are Real, Repeated and Complex Roots, which comes from solving the characteristic equation.
Let them equal a complex conjugate pair, which are C1 = a + b*i and C2 = a - b*i. Then make a linear combination of (a + b*i)*e^(r*t) and (a - b*i)*e^(r*t), with the solutions you got for r. After applying Euler's formula, you'll see that the solution is a linear combination of sine and cosine, with the magnitude of the imaginary solutions, as the frequency. You don't need to prove this from first principles every time. Once you know this is the result, you just set up A*cos(w*t) and B*sin(w*t) as your general solution, when you get r = +/- w*i. Note that capital A & B do relate to lowercase a and b in this method, but aren't strictly equal. I'll leave it as an exercise to you, to determine how they do.
I’m actually breaking up differential equations into a few different playlists. This one has a half dozen more. But Laplace, Fourier, systems etc each get their own miniseries.
Is there a way to find complex roots of higher degree polynomials (say degree 4) that has no real roots? For example, consider the characteristic equation: r^4 + 8r^3 + 26r^2 - 40r +25 = 0. In the book that I am using for self study, this is given as example of repeated complex roots (the roots of this polynomial are r(1,2) = 2 +/- i and the other two roots are repeated complex roots r(3,4) = 2+/- i. The book gives you the roots but it does not tell me how they found the complex roots given that the degree 4 polynomial has no real roots. I already know how to find complex roots if I can manage to reduce the higher degree polynomial to a quadratic (by dividing the higher degree polynomial by the real roots until I get a degree 2 polynomial) and then I can find the remaining complex conjugate with quadratic equation but I do not know how to find complex roots of higher degree polynomials if there are no real roots.
There is a general cubic formula and a general quartic formula, but they are complicated and seldom taught outside a math degree. I can refer you to this video that gives a great explanation: ruclips.net/video/N-KXStupwsc/видео.html For quintics and anything beyond, there is no such formula. Sometimes you are lucky enough to have one that has a real/rational root to get you started. Or you might have a special kind of quartic that is really a biquadratic, where you can simply let w = x^2, and solve it in terms of w. As an example: x^4 + 2*x^2 + 5 = 0. You can even do this with 6th order polynomials, and solve them as bicubic equations. Like x^6 - 9*x^2 + 28 = 0. This has no real solutions for x, but it does have 1 negative real solution for w, when you let w=x^2, which you can easily find with the cubic equation or rational roots theorem. With polynomial division or the cubic formula, you can also find the two complex conjugate solutions for w. Knowing all 3 solutions for w, you can then find the corresponding pairs of solutions for x.
It has to do with the fact that exponential functions differentiate to always form a scalar multiple of themselves. You can assume 2^(r*t) instead if you prefer, but you'll accumulate ln(2) factors that make it more complicated. So you keep it simple and stick to the most elegant form of the exponential function that differentiates as r*e^(r*t), without additional constants. Sines and cosines can also be the solution, in the event that the only values of r that work, are imaginary numbers. Since sines and cosines ultimately are connected to exponentials.
Because i to an even power is real, and i to an odd power is imaginary. Since the Taylor series of cosine has all even exponents, cosine gets to be the real part of Euler's formula. Sine has a Taylor series of all odd exponents, so we have to multiply by i to cancel out the imaginary part of the coefficient we accumulate.
![Yailin La Mas Viral, @kreizyk.official - Dale 2 (Remix) [Video Oficial]](http://i.ytimg.com/vi/JaYHBzgqz94/mqdefault.jpg)
For the second time, thank you for saving my upcoming engineering math exam next week!
Exceellent lecture. I learnt all this stuff in engineering long ago, and even went through the stuff after seeing your video, which is not so easy to understand! You made this look like pure magic! You just need to carry a wand and wear a hat in your next video, and whoa, there you are- Math Magic! Thanks so much Mr. Bazet (I am from India, and 60+). You make math fun. God Bless you
ruclips.net/video/wLoTF_DRF-E/видео.html
bro looked so hyped explaining the quadratic formula
One of the best contents on mathematics 💯
Great Explanation sir!
In 9:15 why did we divide y1+y2 by 2 and didn't just let the general solution be y1+y2? I, also, have one more question. Are the general solutions we found the only solutions to tis ODE, or there are other solutions too?
Is there a mathematical demonstration supporting
y=e^(rx) ? I'm of the conviction that mathematics doesn't rely on guesswork or chance.
It is just a handy trick because e^(rx) can never be zero so you are just reducing the problem to find polynomial roots.
thank you soooo muchhh
Dr., since all solutions can be expressed as linear combinations of the two linearly independent ones, are the complex solutions linear combinations of the 2 real solutions?
In other words, when we talk about a multiplicative "constant" to make linear combinations, "i" is included as a possibility too?
I always thought these multiplicative constants as reals, but apparently they can also be complex?
Exactly. If modelling real world phenomena it is easier to keep everything real which is why we do this shift. But yes we can have complex functions and complex coefficients and it works either way.
can the constants be imaginary?
Yes the method still works
Yes, and that's precisely how you reconcile imaginary solutions, by setting up your two constants to be a complex conjugate pair. Eventually, you can show that it is a linear combination of sine and cosine, given an imaginary solution to the characteristic equation.
I don't know you would react to this anyway I have to say that this is the day that I knew maths sometimes can be illogical 😂
Hi Trefor, I was in your calculus class last summer. Just wanted to say big thanks for putting out content like this. Thanks to you and other math/physics channels on youtube I went from "lets just get though the class and get the grade" to starting to see the beauty in math. Your work is very much appreciated, thank you!
Hey cool, thanks for your kind words:)
Nice to see some people are passionate about what they teach. Cheers.
You may be wondering why the case with repeated roots behaves so differently from the case when every root is distinct. This video presents having two real roots and two complex roots are being different cases, but the video acknowledges that in both cases, you can just simply write the general solution as A·exp(r·t) + B·exp(s·t) if you really want to, where r and s are the roots to the characteristic polynomial. This is definitely not so in the case of repeated roots, where one of the solutions has a factor of t, and this happens only specifically in this case. Why?
In the previous video, in the comments section, I explained that, in fact, you can solve second-order linear equations with constant coefficients without having to "guess" the solutions, and the method I presented for solving the example equation relies on the fact that differential equations can always be rewritten so that they look like linear algebra equations, and that is because the nth order detivative, for every natural n, is always a linear operator, and so it behaves like a matrix. Specifically, the example equation was y''(t) - y'(t) - 6·y(t) = 0, which I said can be written as (D^2 - D - 6·I)[y(t)] = 0, where D is the derivative operator and I the identity "matrix." Since D^2 - D - 6·I is a polynomial in D, this can be factored as (D + 2)·(D - 3), and this was the key to solving the equation. The same concept actually does apply to an arbitrary second-order linear differential equation. In particular, a·y''(t) + b·y'(t) + c·y(t) = 0 can always be written as (a·D^2 + b·D + c·I)[y(t)] = 0, and since a·D^2 + b·D + c·I is a polynomial of degree 2, it can always be factored as a·(D - s·I)·(D - r·I), where s and r are the roots of the polynomial. I already gave the details in that one comment I wrote in the previous video in this channel, so for the rest of this explanation, I am going to directly work with this factorization instead.
So when solving the equation [a·(D - s·I)·(D - r·I)][y(t)] = 0, what to do? Let (D - r·I)[y(t)] = z(t), so the equation to solve is simply a·(D - s·I)[z(t)] = 0. This is just a first-order linear equation, which after dividing by a, is simply (D - s·I)[z(t)] = 0, which can be rewritten as z'(t) - s·z(t) = 0. This has solutions z(t) = A·exp(s·t), where A is just a constant of integration. So (D - r·I)[y(t)] = A·exp(s·t), which can just be written as y'(t) - r·y(t) = A·exp(s·t). The integration factor is exp(-r·t), so multiplying by it results in exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A·exp[(s - r)·t].
This is the key moment. This is the moment where having repeated roots, as opposed to distinct roots, makes an important difference. Why? Because if s = r, which is the case with repeated roots, then s - r = 0, so A·exp[(s - r)·t] = A. Therefore, when you antidifferentiate both sides of exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A, you simply get exp(-r·t)·y(t) = A·t + B, so y(t) = A·t·exp(r·t) + B·exp(r·t), and this gives you the same result as in the video. However, if s and r are distinct, then s - r is nonzero, so A·exp[(s - r)·t] is simply an exponential function. Therefore, when antidifferentiating exp(-r·t)·y'(t) - r·exp(-r·t)·y(t) = A·exp[(s - r)·t], you get exp(-r·t)·y(t) = A/(s - r)·exp[(s - r)·t] + B, hence y(t) = A/(s - r)·exp(s·t) + B·exp(r·t), which matches the result in the video if you simply acknowkedge that, since s - r is nonzero, A/(s - r) is just another arbitrary constant.
The differences and similarities between both cases are more clear when you just leave A/(s - r)·exp(s·t) + B·exp(r·t) written as {A/(s - r)·exp[(s - r)·t] + B}·exp(r·t). Both solution forms {A/(s - r)·exp[(s - r)·t] + B}·exp(r·t) and (A·t + B)·exp(r·t) have the factor exp(r·t) in them, and the repeated roots case replaces exp[(s - r)·t]/(s - r) with t. This makes sense if seen as a way of avoiding division by 0, but ultimately, it results from the fact that, in the repeated roots case, a constant was being antidifferentiated, instead of an exponential. Another way to interpret this is via a strange limit argument. If we expand exp[(s - r)·t] as its Maclaurin series definition, then exp[(s - r)·t]/(s - r) = 1/(s - r) + t + (s - r)·t^2·f(t), with f(0) = 1/2, and f(t) is the Maclaurin series (s - r)^(n + 1)·t^n/(n + 2)! for every natural n. 1/(s - r) does not exist if s - r = 0 or even if s - r -> 0, but if one could somehow "regularize" this summand, so that it actually becomes 0 when s - r = 0, then the result would just be t, as expected, since then, (s - r)·t^2·f(t) = 0. In fact, the integral of exp[(s - r)·t'] on the interval [0, t] with respect to t' is (exp[(s - r)·t] - 1)/(s - r), which does have limit t as s -> r.
thanks
you make the comments section awsome too
WOW, that's one of the best explained comments I've ever read. Thanks so much!
Thank you so much for going further to prove exactly why the form of the particular integral is that . Really appreciate what you do and you deserve so much more recognition . Thank you so much !!!!
Love from india
Thankyou for your efforts
It's my pleasure!
Your content is so unbelievably useful. Thanks for consistently providing us with free education. It's really helped me at university.
Greetings from Brazil. Your explanations are awesome. Thank you very much for your kindness in helping so many people around the world learn/review so important stuff.
You are welcome!
Loving this series, thank you prof
Glad to hear it!
Excellent explanation! Superb work done.
sen nasıl bir kralsın yaa
Studying my first year of engineering, and the math is way to hard. But this helps so much you don't even know.
The rules are simple, you see complex numbers and you give a like to the video
Awesome videos :)
thank you so much !!! your videos are awesome!
please make videos on other topics of math too
Thank you and will do!
Sir plz upload lecture on topology?
Your teaching method is awesome!
I plan to!
Can you please make a video on Contour Integration (Complex Analysis).
ooooh, I do want to do this and would fit nicely with my series on vector calculus and line integrals. Not sure exactly when though, not for a few months at least:)
@@DrTrefor
I'll will passionately wait for it.
Please make it as soon as possible.
Thank you so much this was very helpful!!!!
how do we know there are only 2 particular solutions
ty sir for all of your efforts, plz add PDE course too
god i wish i found you sooner
ruclips.net/video/wLoTF_DRF-E/видео.html
This is SOOOOOOO satisfying. Beautiful.
Love you sir! Thanks for the lecture.
So what if a characteristic equation with more than 2 complex roots? Any chance some video coming up with that particular concept?
Yup higher order is done a little later in the playlist
thank you! could you do some videos in the future merging diff eq concepts with linear algebra?
This is going to be a big thing for us a little later in the series, first with Fourier Series, and then with Systems of Differential Equations.
Thanks a lot dr strange! Very well explained
Hmmm, I'm wondering what's the motivation for arbitrarily having te^rt as the second solution in the case of a repeated root?
is there perhaps a more direct approach
Edit: Nevermind, just saw the brilliant comment explaining this all via a direct approach with the derivative as a linear operator, and the te^rt guess beautifully turns out to be very much analogous to the first order case when your inhomogeneous part solves the homogeneous equation.
Loving this series by the way - it really makes the course seem so much more approachable than it might look from afar.
9:45 hold up, these are BOTH general solutions according to the principle of superposition . I thought any differential equation could only have a single general solution? Or do we just _choose_ which solution is our general solution, or are we allowed to how multiple general solutions?
Here, you just call them “solutions”
Can you make a linear combination of these two solutions for the complex root case?
Homogeneous* equations
@@pepehimovic3135Yes, you can make a linear combination of the complex root case. And that is precisely how you reconcile it to have a real solution. Let the two constants be a complex conjugate pair, with equal and opposite imaginary parts. You'll eventually cancel out the imaginary constants in front of your terms, and show that imaginary solutions to the characteristic equation produce a linear combination of sine and cosine.
For 1st order diffEQ's there is only one solution. For 2nd order diffEQ's there are two fundamental solutions that are related to each other, and a linear combination of the two. In general, nth order diffEQ's will have n fundamental solutions, and n arbitrary constants to combine them with each other.
In a process that produces other arbitrary coefficients, like variation of parameters and method of integrating factor that has you integrate, you'll produce more arbitrary constants, that ultimately can be combined with the original ones. Such that only a number of arbitrary constants equal to the order of the DiffEQ remain.
Is there another method aside using characteristic equation to solve a higher order ?
Yes. Laplace transform.
11:37 Jordan Peterson reference
Professor Bazett, thank you for explaining the different cases that's involved in the Constant Coefficient Ordinary Differential Equations. The three cases are Real, Repeated and Complex Roots, which comes from solving the characteristic equation.
Sir you are a GEM! You really helped me get rid of many many confusions. Thank you Sir. ❤
This was beautiful. Thanks.
Thanks!
Superb! Thank you Trefor!
Thanks a lot sir 🔥🔥🔥
What are c1 and c2 when the roots are imaginary?
Let them equal a complex conjugate pair, which are C1 = a + b*i and C2 = a - b*i. Then make a linear combination of (a + b*i)*e^(r*t) and (a - b*i)*e^(r*t), with the solutions you got for r. After applying Euler's formula, you'll see that the solution is a linear combination of sine and cosine, with the magnitude of the imaginary solutions, as the frequency.
You don't need to prove this from first principles every time. Once you know this is the result, you just set up A*cos(w*t) and B*sin(w*t) as your general solution, when you get r = +/- w*i.
Note that capital A & B do relate to lowercase a and b in this method, but aren't strictly equal. I'll leave it as an exercise to you, to determine how they do.
Great work!!!!
Excellent! Excellent!
💗🌈💗🌈💗
ruclips.net/video/wLoTF_DRF-E/видео.html
I love you
Nice
Can you please tell the software you make use of for making these videos
Always waiting for your video.
nice one making the first comment!
How many videos is this series?
I’m actually breaking up differential equations into a few different playlists. This one has a half dozen more. But Laplace, Fourier, systems etc each get their own miniseries.
Is there a way to find complex roots of higher degree polynomials (say degree 4) that has no real roots? For example, consider the characteristic equation: r^4 + 8r^3 + 26r^2 - 40r +25 = 0. In the book that I am using for self study, this is given as example of repeated complex roots (the roots of this polynomial are r(1,2) = 2 +/- i and the other two roots are repeated complex roots r(3,4) = 2+/- i. The book gives you the roots but it does not tell me how they found the complex roots given that the degree 4 polynomial has no real roots.
I already know how to find complex roots if I can manage to reduce the higher degree polynomial to a quadratic (by dividing the higher degree polynomial by the real roots until I get a degree 2 polynomial) and then I can find the remaining complex conjugate with quadratic equation but I do not know how to find complex roots of higher degree polynomials if there are no real roots.
There is a general cubic formula and a general quartic formula, but they are complicated and seldom taught outside a math degree. I can refer you to this video that gives a great explanation:
ruclips.net/video/N-KXStupwsc/видео.html
For quintics and anything beyond, there is no such formula. Sometimes you are lucky enough to have one that has a real/rational root to get you started. Or you might have a special kind of quartic that is really a biquadratic, where you can simply let w = x^2, and solve it in terms of w. As an example: x^4 + 2*x^2 + 5 = 0. You can even do this with 6th order polynomials, and solve them as bicubic equations. Like x^6 - 9*x^2 + 28 = 0. This has no real solutions for x, but it does have 1 negative real solution for w, when you let w=x^2, which you can easily find with the cubic equation or rational roots theorem. With polynomial division or the cubic formula, you can also find the two complex conjugate solutions for w. Knowing all 3 solutions for w, you can then find the corresponding pairs of solutions for x.
Teachers like him instill love for maths
I am confused why you can always make the gues y=e^rt. Does it have something to do with the differentiation rule for euler's constant?
It has to do with the fact that exponential functions differentiate to always form a scalar multiple of themselves. You can assume 2^(r*t) instead if you prefer, but you'll accumulate ln(2) factors that make it more complicated. So you keep it simple and stick to the most elegant form of the exponential function that differentiates as r*e^(r*t), without additional constants.
Sines and cosines can also be the solution, in the event that the only values of r that work, are imaginary numbers. Since sines and cosines ultimately are connected to exponentials.
thank you sir. you clearly explained.
Your are awesome sir thank you!
Awesome as Always ❤️❤️❤️
Thank you!
that was fabulous, thanks!
but why not i cos why always isin
Comes from Taylor expansion
Because i to an even power is real, and i to an odd power is imaginary. Since the Taylor series of cosine has all even exponents, cosine gets to be the real part of Euler's formula. Sine has a Taylor series of all odd exponents, so we have to multiply by i to cancel out the imaginary part of the coefficient we accumulate.
thank you sir
Y did we learn abt springs
Because the mass on a spring is a common example of 2nd order diffEQ's, and is the fundamental building block of vibration analysis.