Reading my math textbook I became able to solve second order ODE questions, but I didn't understand or have any intuition why the result and characteristic equation arises. The textbook literally just gave me the general solution form and said this is how it is. Thank you Dr. Trefor for actually helping me understand WHY this happens.
You can actually solve the equation y''(x) - y'(x) - 6·y(x) = 0 without having to guess the form of the solutions. In the comments section to a previous video, I discussed a principle, by which you can think of differential equations as being analogous to equations in linear algebra, where y is the unknown vector, and D^2 - D - 6·I is a linear operator, a "matrix" of some sort, which, when applied to y, outputs the function y'' - y' - 6·y. This is motivated by the idea that the derivative operator D is a linear operator, so you can rewrite y'(x) as D[y(x)], and y''(x) as (D^2)[y(x)]. If you also define I as the identity operator, so that I[y(x)] = y(x) is true for any y, then as such, y''(x) - y'(x) - 6·y(x) = 0 is equivalent to (D^2)[y(x)] - D[y(x)] - 6·I[y(x)] = 0. Furthermore, if you think of the distributivity property, you can rewrite this as (D^2 - D - 6·I)[y(x)] = 0. You can think of D^2 - D - 6·I as a single linear operator L, as a single "matrix," and write the equation as L[y(x)] = 0. Why is this useful? Because what this tells you is that solving the differential equation is just equivalent to finding the null space of L! This is nice and all, but how does this actually help in solving the equation? All I have done so far is present a different way of thinking about the equation, not an intuitive solution method. But now that I have presented the idea on how to conceptually recast the equation as a problem about linear operators or "matrices," I can actually present the solution method. For starts, notice that L = D^2 - D - 6·I is a second-order polynomial in D, noting that D^0 = I. What is something that you can typically do with second-order polynomials? You can write them as a product of two first-order polynomials. In general, s^2 - s - 6 = (s + 2)·(s - 3), so perhaps you may think that, similarly, D^2 - D - 6·I = (D + 2·I)·(D - 3·I). Can this be justified rigorously? Yes. Keeping in mind that the multiplication T·S of two linear operators T and S is defined as T{S[y(x)]}, which is also how matrix multiplication is defined, you can use the distributive property to prove the above equality, and the order of the factors is not relevant, because D commutes with I, as do all linear operators. What does this imply? It implies I can write (D^2 - D - 6·I)[y(x)] = 0 as [(D + 2·I)·(D - 3·I)][y(x)] = 0, which you can also rewrite as (D + 2·I){(D - 3·I)[y(x)]} = 0. This is the key to solving the equation. Why? Because now, the structure of the equation calls for a substitution, one which is definitely far more obvious than any other possible substitution you can make here, and that substitution would be (D - 3·I)[y(x)] = z(x), resulting in the equation (D + 2·I)[z(x)] = 0. This is masterful, because this turns a second-order problem into simply two first-order problems! In particular, we already familiar with solving any first-order linear equation. This is more clear once we actually rewrite the equations. For example, (D + 2·I)[z(x)] = 0 can be written as z'(x) + 2·z(x) = 0, and now, surely, this looks like a familiar equation that people who are watching this video should know how to solve. I actually provided additional insight into how solving first-order equations is related to linear algebra in my comments to a previous video, the same ones I mentioned in my first paragraph. Getting into that is obnoxious here, though. For now, all you need to know is how to solve first-order linear equations. z'(x) + 2·z(x) = 0 is a separable equation, and it simply has the solutions z(x) = A·exp(-2·x). Once you know this, you can back-substitute to obtain y(x), so (D - 3·I)[y(x)] = A·exp(-2·x). If you would like, you can definitely just rewrite this as y'(x) - 3·y(x) = A·exp(-2·x). Then, you would just solve this as usual: multiply by the integrating factor, which in this case, is exp(-3·x), to obtain exp(-3·x)·y'(x) - 3·exp(-3·x)·y(x) = A·exp(-5·x). exp(-3·x)·y'(x) - 3·exp(-3·x)·y(x) is the derivative of exp(-3·x)·y(x), so by antidifferentiating, you get exp(-3·x)·y(x) = B - A/5·exp(-5·x). Therefore, y(x) = B·exp(3·x) - A/5·exp(-2·x). Now simply let B = C1, -A/5 = C2, and you obtain the exact same result in the video. Instead of simply guessing the solution is of the form exp(k·x), though, you can intuitively think of this substitution using the concept eigenvalues. The functions of the form A·exp(k·x) are the solutions to the eigenvalue equation D[y(x)] = k·y(x). Since f(D) = D^2 - D - 6·I is a polynomial in D, and from linear algebra, you would know that D[y(x)] = k·y(x) implies [f(D)][y(x)] = f(k)·y(x) for polynomials f, you may realize that, therefore, the equation [f(D)][y(x)] = f(k)·y(x) is solved by A·exp(k·x) as well. However, as the actual equation says [f(D)][y(x)] = 0, you need f(k) = 0, which is just the famous characteristic equation that you always get in these problems, and in this case, that is simply the equation k^2 - k - 6 = 0, which has solutions k = -2, k = 3, which tells you that A·exp(-2·x) solves the equation, based on the eigenvalue principle, and so does B·exp(3·x), and thus, so does their sum, which is the general linear combination.
Took a course on differential equations recently, never understood why the characteristic equation works but your explanation is crystal clear! Thank you very much!
Dr Bazett i'm an engineering student from italy. I just want to thank you for creating helpful videos and making maths visually more interesting, your videos like the SIR models, renews my love for math. So please keep doing helpful videos as you keep doing interesting and more advance matters too. THANK YOU!
finally you had made it public, thanks to you ......i had tried to join your channel but my card didnt work . Can you use other payment method like UPI which is more prevalent in india for online transaction .
Yup! I usually only have the up for members for a week or so before they go public, thanks for trying but I actually have no control over how youtube does the payment system!
Hello, always loved your video. Do not miss a single one. This is the real blessing of internet. I am thousands mile away from you and learning by your video like in your class room. I have a question and i asked it before also. Do you have any plan to make video on mathematical analysis?
Higher order diffEQ's with functions of t as the coefficients, are a lot harder to solve. A common strategy is to use the Laplace transform, which allows you to reduce it to a constant coefficient diffEQ or a lower order diffEQ, but this only works for polynomial functions of t, or exponential functions of t (or linear combinations thereof) as the coefficient.
Very pretty me like. Question to self: how do I guess the solution? Using e^something is a good start, as it's constant. But from experience, the primary way is to just do many problems to get a large reference pool
Hi Dr Trefor Bazett. I have a few questions. Shouldn't we check if the solution is correct? Since we have guessed the form? Are there any other forms of solutions to ccde? I understand that we should also give y=0 as part of the solution? Or is it not necessary? I am currently on my 3rd week into the subject (at the university), hence all the questions. Thank you for your hard work on these videos, they are a lot of help. Kind Regards.
Great questions! We could check, but note we have already substituted y=e^rt into the equation and so know it works, unless we made an algebra mistake. And y=0 definitely is a solution, that would happen when both constants are zero. Could there be more? Potentially. But our theory of second order linear equations (see previous video in the playlist) says that if you have two linearly independent ones, everything can be written in terms of them so we don't have to search for more.
@@DrTrefor Right. I have forgot that C might equal 0. YT proposed only this video, I didn't know about the other one. I have watched it and now I do understand how are all the solutions taken into consideration. Thank you.
You would know your guess is good or bad based on what happens to the characteristic equation. Some things would lead to unsolvable or very difficult to solve equations: For example: Guess y=t² Substitute y=t² into y′′−y′−6y=0 The equation becomes: 2−2t−6t² This is clearly not zero (as you can figure out from the equation of roots of polynomials). I think one good way to think about is with physics. Perhaps you are trying to guess the equation that can be used to solve a real world harmonic oscillator, you of course aren't going to choose something silly like y = t⁴ because how does that even closely resemble the motion? So it's about picking something that seems like it will make sense, which would seem like a high level explanation that usually isn't useful for maths and physics but here it sort of applies :)
It's an easy guess, e^x is the only function that stays the same after differentiation and integration. So, if you can rearrange your equation into a function and its derivative (or second derivative) being the same times a constant, well it's most likely an exponential.
Another easy guess is involves another kind of "cycling" function. There are a family of functions that return to themselves after 2 or 4 operations, the trigonometric ones depending on whether they're standard or hyperbolic. If you see y'' = -y that's a sinusoid (and unsurprisingly a solution to the harmonic oscillator).
Reading my math textbook I became able to solve second order ODE questions, but I didn't understand or have any intuition why the result and characteristic equation arises. The textbook literally just gave me the general solution form and said this is how it is. Thank you Dr. Trefor for actually helping me understand WHY this happens.
How can I get the textbook?
You can actually solve the equation y''(x) - y'(x) - 6·y(x) = 0 without having to guess the form of the solutions. In the comments section to a previous video, I discussed a principle, by which you can think of differential equations as being analogous to equations in linear algebra, where y is the unknown vector, and D^2 - D - 6·I is a linear operator, a "matrix" of some sort, which, when applied to y, outputs the function y'' - y' - 6·y. This is motivated by the idea that the derivative operator D is a linear operator, so you can rewrite y'(x) as D[y(x)], and y''(x) as (D^2)[y(x)]. If you also define I as the identity operator, so that I[y(x)] = y(x) is true for any y, then as such, y''(x) - y'(x) - 6·y(x) = 0 is equivalent to (D^2)[y(x)] - D[y(x)] - 6·I[y(x)] = 0. Furthermore, if you think of the distributivity property, you can rewrite this as (D^2 - D - 6·I)[y(x)] = 0. You can think of D^2 - D - 6·I as a single linear operator L, as a single "matrix," and write the equation as L[y(x)] = 0. Why is this useful? Because what this tells you is that solving the differential equation is just equivalent to finding the null space of L!
This is nice and all, but how does this actually help in solving the equation? All I have done so far is present a different way of thinking about the equation, not an intuitive solution method. But now that I have presented the idea on how to conceptually recast the equation as a problem about linear operators or "matrices," I can actually present the solution method. For starts, notice that L = D^2 - D - 6·I is a second-order polynomial in D, noting that D^0 = I. What is something that you can typically do with second-order polynomials? You can write them as a product of two first-order polynomials. In general, s^2 - s - 6 = (s + 2)·(s - 3), so perhaps you may think that, similarly, D^2 - D - 6·I = (D + 2·I)·(D - 3·I). Can this be justified rigorously? Yes. Keeping in mind that the multiplication T·S of two linear operators T and S is defined as T{S[y(x)]}, which is also how matrix multiplication is defined, you can use the distributive property to prove the above equality, and the order of the factors is not relevant, because D commutes with I, as do all linear operators. What does this imply? It implies I can write (D^2 - D - 6·I)[y(x)] = 0 as [(D + 2·I)·(D - 3·I)][y(x)] = 0, which you can also rewrite as (D + 2·I){(D - 3·I)[y(x)]} = 0. This is the key to solving the equation. Why? Because now, the structure of the equation calls for a substitution, one which is definitely far more obvious than any other possible substitution you can make here, and that substitution would be (D - 3·I)[y(x)] = z(x), resulting in the equation (D + 2·I)[z(x)] = 0. This is masterful, because this turns a second-order problem into simply two first-order problems! In particular, we already familiar with solving any first-order linear equation. This is more clear once we actually rewrite the equations. For example, (D + 2·I)[z(x)] = 0 can be written as z'(x) + 2·z(x) = 0, and now, surely, this looks like a familiar equation that people who are watching this video should know how to solve. I actually provided additional insight into how solving first-order equations is related to linear algebra in my comments to a previous video, the same ones I mentioned in my first paragraph. Getting into that is obnoxious here, though. For now, all you need to know is how to solve first-order linear equations. z'(x) + 2·z(x) = 0 is a separable equation, and it simply has the solutions z(x) = A·exp(-2·x). Once you know this, you can back-substitute to obtain y(x), so (D - 3·I)[y(x)] = A·exp(-2·x). If you would like, you can definitely just rewrite this as y'(x) - 3·y(x) = A·exp(-2·x). Then, you would just solve this as usual: multiply by the integrating factor, which in this case, is exp(-3·x), to obtain exp(-3·x)·y'(x) - 3·exp(-3·x)·y(x) = A·exp(-5·x). exp(-3·x)·y'(x) - 3·exp(-3·x)·y(x) is the derivative of exp(-3·x)·y(x), so by antidifferentiating, you get exp(-3·x)·y(x) = B - A/5·exp(-5·x). Therefore, y(x) = B·exp(3·x) - A/5·exp(-2·x). Now simply let B = C1, -A/5 = C2, and you obtain the exact same result in the video.
Instead of simply guessing the solution is of the form exp(k·x), though, you can intuitively think of this substitution using the concept eigenvalues. The functions of the form A·exp(k·x) are the solutions to the eigenvalue equation D[y(x)] = k·y(x). Since f(D) = D^2 - D - 6·I is a polynomial in D, and from linear algebra, you would know that D[y(x)] = k·y(x) implies [f(D)][y(x)] = f(k)·y(x) for polynomials f, you may realize that, therefore, the equation [f(D)][y(x)] = f(k)·y(x) is solved by A·exp(k·x) as well. However, as the actual equation says [f(D)][y(x)] = 0, you need f(k) = 0, which is just the famous characteristic equation that you always get in these problems, and in this case, that is simply the equation k^2 - k - 6 = 0, which has solutions k = -2, k = 3, which tells you that A·exp(-2·x) solves the equation, based on the eigenvalue principle, and so does B·exp(3·x), and thus, so does their sum, which is the general linear combination.
you need to register these answers on some blog post or create your youtube channel
the quality is too good to be lost in the comment section
hey angel, i watch a bunch of other channels.
Ya, Dr. Peyam made a video about this.
WOW! I'm reading through and trying to digest it
Took a course on differential equations recently, never understood why the characteristic equation works but your explanation is crystal clear! Thank you very much!
Glad it helped!
Dr. Trefor Bazett i want to tell you that i love your teaching method and i always wait for your videos. thanks alot sir.
Thank you so much for your kind words!
Praying Tanaka agrees 🌝🙏
I swear when I'm in a lecture this stuff sounds so complicated but watching this makes it seem so easy!
Dr Bazett i'm an engineering student from italy. I just want to thank you for creating helpful videos and making maths visually more interesting, your videos like the SIR models, renews my love for math. So please keep doing helpful videos as you keep doing interesting and more advance matters too. THANK YOU!
So glad they are helping!
You just saved me a weeks worth of catching up in my math course. Thank you with all of my soul
For the first time, thanks for saving my upcoming engineering mathematics finals next week
Professor Bazett, this is a solid and clear explanation of How to Solve Constant Coefficient Homogeneous Differential Equations.
such a blessing, you explain things in a more engaging tone than my professor
In tears! What an explanation. Please, make a playlist on partial differential equation.
I plan to this semester!!
Thank you so much sir. Your way of teaching is so simple yet effective.
Such a great and easy to follow video way better than reading a textbook before an exam. Thank you!
Each and every second of your video is full of knowledge.
Loved it!👍
I loved DE in Uni!!! They're so fun to work with! Thank you Dr.Bazett !
Glad you like them!
Nice work as usual Dr. , You are A W E S O M E ! ! !
OMG literally learned this yesterday! Perfect timing
Perfect!
Fantastic explanation ❤❤❤.
Dr. Bazett you literally saved my ass!
❤️ first comment from India
Nice!!
as a teacher i love one thing about india they love learning
finally you had made it public, thanks to you ......i had tried to join your channel but my card didnt work . Can you use other payment method like UPI which is more prevalent in india for online transaction .
Yup! I usually only have the up for members for a week or so before they go public, thanks for trying but I actually have no control over how youtube does the payment system!
Hello, always loved your video. Do not miss a single one. This is the real blessing of internet. I am thousands mile away from you and learning by your video like in your class room. I have a question and i asked it before also. Do you have any plan to make video on mathematical analysis?
Thank you! I do hope to, but not for a little bit:)
@@DrTrefor Eagerly waiting!
Can do some videos for variable coefficient differential equations
Higher order diffEQ's with functions of t as the coefficients, are a lot harder to solve. A common strategy is to use the Laplace transform, which allows you to reduce it to a constant coefficient diffEQ or a lower order diffEQ, but this only works for polynomial functions of t, or exponential functions of t (or linear combinations thereof) as the coefficient.
Very pretty me like. Question to self: how do I guess the solution? Using e^something is a good start, as it's constant. But from experience, the primary way is to just do many problems to get a large reference pool
Thank you soo much sir 🔥🔥🔥
Hi Dr Trefor Bazett. I have a few questions. Shouldn't we check if the solution is correct? Since we have guessed the form? Are there any other forms of solutions to ccde? I understand that we should also give y=0 as part of the solution? Or is it not necessary? I am currently on my 3rd week into the subject (at the university), hence all the questions. Thank you for your hard work on these videos, they are a lot of help. Kind Regards.
Great questions! We could check, but note we have already substituted y=e^rt into the equation and so know it works, unless we made an algebra mistake. And y=0 definitely is a solution, that would happen when both constants are zero. Could there be more? Potentially. But our theory of second order linear equations (see previous video in the playlist) says that if you have two linearly independent ones, everything can be written in terms of them so we don't have to search for more.
@@DrTrefor Right. I have forgot that C might equal 0. YT proposed only this video, I didn't know about the other one. I have watched it and now I do understand how are all the solutions taken into consideration. Thank you.
what do you do if your r polynomial factors into "r^2-2r+1" in which case r=1 or r=1. what is the process then?
Sir i think you missed the Variation of Parameters Form of differential equation
what the frick are u tellıng. hell na im gonna fail diff
Did u faill
@@naida11111 yea i got F im gonna take diff again this semester
@@cagataybayrak353 sorry 4 u, i failed too shejxs but i was able to give reexam, good luckk
Geçmiş olsun benim de bu cuma.
@@anayin63Bol şanslar benim ki de haftaya pazartesi ,hangi universetde okuyorsun?
the videos are great
Thanks!
Thank you so much!
Thank you ultra very much
a lot of thanks
Thank you
good video man
How would we know if the gues doesn't work?
You would know your guess is good or bad based on what happens to the characteristic equation. Some things would lead to unsolvable or very difficult to solve equations:
For example:
Guess y=t²
Substitute y=t² into
y′′−y′−6y=0
The equation becomes:
2−2t−6t²
This is clearly not zero (as you can figure out from the equation of roots of polynomials).
I think one good way to think about is with physics. Perhaps you are trying to guess the equation that can be used to solve a real world harmonic oscillator, you of course aren't going to choose something silly like y = t⁴ because how does that even closely resemble the motion? So it's about picking something that seems like it will make sense, which would seem like a high level explanation that usually isn't useful for maths and physics but here it sort of applies :)
Please give the last video!!!!!
It actually JUST went live. I typically post videos for my members a few days earlier than releasing for the full public.
perfect
Thank you!
Why the use of exponential function as a guessed solution everyone?
It's an easy guess, e^x is the only function that stays the same after differentiation and integration. So, if you can rearrange your equation into a function and its derivative (or second derivative) being the same times a constant, well it's most likely an exponential.
Another easy guess is involves another kind of "cycling" function. There are a family of functions that return to themselves after 2 or 4 operations, the trigonometric ones depending on whether they're standard or hyperbolic. If you see y'' = -y that's a sinusoid (and unsurprisingly a solution to the harmonic oscillator).
🔥🔥🔥
🙏🙏
He talks at X2 SPEED
wtf
Thanks!