A couple of questions. Why could we assume y was an exponential function? What if that assumption had been incorrect? Also, why was the general solution the sum of y_1 and y_2?
"But why is there no '+ C' at the end, like when you do an integral?" I'm sure that question rolls around in people's minds. Well, that "+ C" is a consequence of a problem of the form "y' = [some function]". But in this problem, it's not just y' on the left, but notably a term of y as well. Whenever it's just y' (or y'' or any combination of derivatives), then we know that any constant terms in the solution go away when we do the derivatives to get back to the original equation; but if there's a plain old y on the left in the original problem, that constant WON'T go away because the plain old y doesn't get differentiated. But consider the equation "y'' - y' = 0", without any simple "y" term. If we follow the same process as in the video, we get to: e^(rt)(r^2 - r) = 0 and r can be either 0 or 2. The 2 gives us "e^(2t)" times an arbitrary constant, but what about that 0? That gives us "e^0t" times an arbitrary constant, but "e^0t" is itself a constant, so that whole section reduces to ... an arbitrary constant. Lo and behold, we get that "+ C" because there is no simple "y" term in the initial problem! And we didn't even have to remember to add it in; it naturally emerged from the math.
Well I just had to check that the final solution worked in the original equation. So, starting from y, I differentiated to find y' and y''. Plug them into the original equation and, presto!, we get 0.
Can you do y'' + y' + y = 0 at some point? The answer I'm coming up with is ugly with a capital ug. The thing is, I bet it can be made graceful and elegant, but I don't know how to do that. I bet you do though!
@@PrimeNewtons I fiddled around with it some more, and it turns out it wasn't as ugly as I feared it would be. I got frightened because some of the arbitrary constants I was coming up with looked like they would have to be complex, but it turns out they don't have to be. Honestly, they could be real, imaginary, or complex, and it still works out the same. I came up with y = e^(-x/2)*[a*cos(px) + b*sin(px)], where a and b are my arbitrary constants and p = sqrt(3)/2. Not nearly as fearsome as I thought, once I figured some things out.
I am always primed to learn math from you, sir! 😊
I like the explanation at the beginning, of how we're making an assumption (exponential solution) and testing the assumption proves it was right.
A couple of questions. Why could we assume y was an exponential function? What if that assumption had been incorrect?
Also, why was the general solution the sum of y_1 and y_2?
Good questions. It is almost impossible to generate a differential equation of this type unless the function is exponential or trigonometric.
"But why is there no '+ C' at the end, like when you do an integral?" I'm sure that question rolls around in people's minds. Well, that "+ C" is a consequence of a problem of the form "y' = [some function]". But in this problem, it's not just y' on the left, but notably a term of y as well. Whenever it's just y' (or y'' or any combination of derivatives), then we know that any constant terms in the solution go away when we do the derivatives to get back to the original equation; but if there's a plain old y on the left in the original problem, that constant WON'T go away because the plain old y doesn't get differentiated.
But consider the equation "y'' - y' = 0", without any simple "y" term. If we follow the same process as in the video, we get to:
e^(rt)(r^2 - r) = 0
and r can be either 0 or 2. The 2 gives us "e^(2t)" times an arbitrary constant, but what about that 0? That gives us "e^0t" times an arbitrary constant, but "e^0t" is itself a constant, so that whole section reduces to ... an arbitrary constant. Lo and behold, we get that "+ C" because there is no simple "y" term in the initial problem! And we didn't even have to remember to add it in; it naturally emerged from the math.
Perfect explanation!
Impressive explanation
Well I just had to check that the final solution worked in the original equation. So, starting from y, I differentiated to find y' and y''. Plug them into the original equation and, presto!, we get 0.
@@keithrobinson2941Math sure is purty some days!
U're number one ,sir
U're the best ❤🎉
Thank u so much 👍
Thank you
can you do the one on systems of equations ,variation of parameters and homogeneous ode sir @Prime newtons
Please email me a sample problem
Can you do y'' + y' + y = 0 at some point? The answer I'm coming up with is ugly with a capital ug. The thing is, I bet it can be made graceful and elegant, but I don't know how to do that. I bet you do though!
🤣 Well, I am not sure what UGly or elegant means to you. I will do something similar in the next video.
@@PrimeNewtons I fiddled around with it some more, and it turns out it wasn't as ugly as I feared it would be. I got frightened because some of the arbitrary constants I was coming up with looked like they would have to be complex, but it turns out they don't have to be. Honestly, they could be real, imaginary, or complex, and it still works out the same.
I came up with y = e^(-x/2)*[a*cos(px) + b*sin(px)], where a and b are my arbitrary constants and p = sqrt(3)/2. Not nearly as fearsome as I thought, once I figured some things out.
Find the solution for the differential equation
(2xy-sin x)dx + (x2- cos y)dy = 0