I just realized after watching the next video this was a Cauchy Euler form and explains the "just stick in an lnX if they're linearly dependent and don't ask why"-advice!
Hey John, You're absolutely right! The example shown in this video actually shows why Cauchy-Euler equations with repeated solutions for the equation for m get a log term attached to them! For everyone else, you can check out the method for Cauchy-Euler that John mentions at this link: ua-cam.com/video/_UILVgNIeTw/v-deo.html
As the video states at 0:17, this method requires the knowledge of at least one solution from the fundamental set. There is another more useful method that does not require us to already know y1. You can find videos on those here: 2nd-Order Constant Coefficients Method Intro: ua-cam.com/video/Pxc7VIgr5kc/v-deo.html Characteristic Equation Method with Distinct Roots: ua-cam.com/video/ki1laVMS3S4/v-deo.html Characteristic Equation Method with Repeated Roots: ua-cam.com/video/Qd72VS-V-_U/v-deo.html Characteristic Equation Method with Complex Roots: ua-cam.com/video/94iovfwI0tg/v-deo.html
If functions y1 and y2 are linearly independent, and we know y1, for example. Then "u" should be equal to y2/y1, yes? And we could find it by substituting y2=u*y1. This reduction of order method helps us find the function "u" so that we can then know the general solution. As we show at 5:30, the idea of this method is to reduce the order to a first-order equation, and that the substitutions shown allow us to work with either a separable or a linear first-order equation to then find that exact "u" that works in a given situation.
@@HoustonMathPrep I am sorry but I still don't understand. "Then u should be equal to y2/y1, yes?" I don't understand this exact step. Why should the two basis be related to one another by a function? In linear algebra, for example, for 2 basis to be linearly independent, say for example v1 and v2, v1 and v2 must not be related by some constant factor. [1 ; 1] and [2 ;2] are not linearly independent as [2 ; 2] = 2 * [1 ; 1] But i [1 ; 0] and j [0 ; 1] are independently but we can't write i in terms of j. The only way to get i from j is to multiply j by 90 degree rotation matrix. So I guess what I am asking is that, is multiplying by a function in case of differential equation analogous to multiplying by a matrix in linear algebra? Somehow, multiplying a base by a function transform it into another linear independent base within the solution plane and this fact just blows my mind. Thanks for replying. I find your videos high quality and the fact that you still react to every single comment. I salute your work ethic.
Thanks for this! If y1 and y2 are linearly independent (which is necessary for them to be a fundamental solution set of a 2nd-order equation), then we know that "u" is not a constant, as you reference in your mention of vectors. Just by simple properties of functions, I should be able to take y1 and multiply it by another function and get y2. For example: If y1 is 1/x and y2 is x^4, then there is some "u" that has the relationship that y2 = u*y1 (in my example here, u = x^5). And this will be true of any two non-zero functions, that we can always find a "u" that satisfies y2 = u*y1. Reduction of order is simply the process by which we find this function "u". And the reason it works out nicely is because we can use our knowledge of 1st-order equations. That's the best I can do in type chat on the internet to answer your question. I hope it helps in some small way! :)
@@HoustonMathPrep Oh it helped in a big way. Thank you so much. Your example and especially the statement "this will be true of any two non-zero functions, that we can always find a "u" that satisfies y2=u*y1" made it finally click for me. Thank you so so much!
Underrated video! Your channel is saving my life right now
Happy to help!
the best explanation I've come across , this is going straight into my differential equations playlist !!
Thanks so much for the kind words! Glad you enjoyed!
That was a solid and easy to follow video. Thanks!
You're welcome!
I just realized after watching the next video this was a Cauchy Euler form and explains the "just stick in an lnX if they're linearly dependent and don't ask why"-advice!
Hey John,
You're absolutely right! The example shown in this video actually shows why Cauchy-Euler equations with repeated solutions for the equation for m get a log term attached to them! For everyone else, you can check out the method for Cauchy-Euler that John mentions at this link: ua-cam.com/video/_UILVgNIeTw/v-deo.html
AWESOME EXPLANATION 👍👍👍👍
Glad you think so!
Wow! I never saw this method before! Thanx! 😊
You're legend.. thanks!!
You're welcome!
Thank you so much.
How r we supposed to know what y1 is if its not provided?
As the video states at 0:17, this method requires the knowledge of at least one solution from the fundamental set. There is another more useful method that does not require us to already know y1. You can find videos on those here:
2nd-Order Constant Coefficients Method Intro: ua-cam.com/video/Pxc7VIgr5kc/v-deo.html
Characteristic Equation Method with Distinct Roots: ua-cam.com/video/ki1laVMS3S4/v-deo.html
Characteristic Equation Method with Repeated Roots: ua-cam.com/video/Qd72VS-V-_U/v-deo.html
Characteristic Equation Method with Complex Roots: ua-cam.com/video/94iovfwI0tg/v-deo.html
Thank a lot buddy really appreciate it 👍
Why does this work though? Why should it be that multiplying one solution(y1) with some function(u) gives another solution(y)?
If functions y1 and y2 are linearly independent, and we know y1, for example. Then "u" should be equal to y2/y1, yes? And we could find it by substituting y2=u*y1.
This reduction of order method helps us find the function "u" so that we can then know the general solution. As we show at 5:30, the idea of this method is to reduce the order to a first-order equation, and that the substitutions shown allow us to work with either a separable or a linear first-order equation to then find that exact "u" that works in a given situation.
@@HoustonMathPrep I am sorry but I still don't understand. "Then u should be equal to y2/y1, yes?" I don't understand this exact step. Why should the two basis be related to one another by a function?
In linear algebra, for example, for 2 basis to be linearly independent, say for example v1 and v2, v1 and v2 must not be related by some constant factor. [1 ; 1] and [2 ;2] are not linearly independent as [2 ; 2] = 2 * [1 ; 1]
But i [1 ; 0] and j [0 ; 1] are independently but we can't write i in terms of j. The only way to get i from j is to multiply j by 90 degree rotation matrix.
So I guess what I am asking is that, is multiplying by a function in case of differential equation analogous to multiplying by a matrix in linear algebra?
Somehow, multiplying a base by a function transform it into another linear independent base within the solution plane and this fact just blows my mind.
Thanks for replying. I find your videos high quality and the fact that you still react to every single comment. I salute your work ethic.
Thanks for this! If y1 and y2 are linearly independent (which is necessary for them to be a fundamental solution set of a 2nd-order equation), then we know that "u" is not a constant, as you reference in your mention of vectors. Just by simple properties of functions, I should be able to take y1 and multiply it by another function and get y2.
For example: If y1 is 1/x and y2 is x^4, then there is some "u" that has the relationship that y2 = u*y1 (in my example here, u = x^5). And this will be true of any two non-zero functions, that we can always find a "u" that satisfies y2 = u*y1. Reduction of order is simply the process by which we find this function "u". And the reason it works out nicely is because we can use our knowledge of 1st-order equations. That's the best I can do in type chat on the internet to answer your question. I hope it helps in some small way! :)
@@HoustonMathPrep Oh it helped in a big way. Thank you so much. Your example and especially the statement "this will be true of any two non-zero functions, that we can always find a "u" that satisfies y2=u*y1" made it finally click for me. Thank you so so much!
ty friend
you are alegend