Try out Walter Lewin. He doesn't give you the same kind of lectures as BPRP does, but he gives out weekly highschool problems that are fairly hard, and then at the end of the week he tells us the answer and explains why it is like that.
you all prolly dont care at all but does anyone know of a tool to log back into an instagram account?? I somehow forgot the password. I love any assistance you can offer me
Thank you for the video! I noticed that when we substituted v' = w and divided, we obtained an equation of the form y' + f(x)y=g(x) which would usually require the integrating factor method but since g(x) = 0, the equation is separable
MY friend, this equation y' + f(x)y=g(x) couldn't be separated. But if g(x) multiplied with y then it could be separated, else it's need an integrating factor !
"so, why is my son in tears while solving a math equation?" "ma, I'm not gonna win tomorrow's exam..." aaaaaand I'll have to brace for impending emotional damage tomorrow... thanks for the tutorial, but I'm definitely suffering from the time attack tomorrow
There are also techniques like Reduction to Riccati Change of an independent variable I didnt see a video in which you present these techniques (Reduction to Riccati works only for second order linear but not always equation after reduction is easier to solve)
I thought of a question, and as far as I can tell this can't be done, but can you give it a go? Find a nontrivial function (aka not f(x)=0 because this is the only solution I could find) that satisfies the fact that: f ∘ f(x) = f'(x) I don't think it's possible to solve, but you might be able to do it. If it isn't possible, how about: f ∘ f(x) = (f'(x))^2
Austin Blake Hey that's interesting! I will have to think about it. Have you seen th-cam.com/video/ZH5rfWcR7l4/w-d-xo.html I called this the fake product rule. So ur problems might be doable but I just have to work on it later.
Austin Blake Question is whether you'd allow functions of complex variables and complex-valued functions to be used. For example, if f(x) = Cx^(i+1), f( f(x) ) = C(Cx^(i+1))^(i+1) = (C^(i+2))(x^2i) and f'(x) = C(i+1)x^i, (f'(x))^2 = (C^2)2i(x^2i). So the resulting equation that you have to solve is (C^i)*(C^2)*(x^2i) = (2i)*(C^2)*(x^2i). You can cancel out some terms and get C^i = 2i, so C = exp(π/2 + 2nπ)*2^(-i), for n ϵ ℤ. That's kind of ugly though.
Can anyone tell me can we put the general soultion like first solution + second soultion as first solution is given put this eq into diff eq and find second solution of this eq if not then why
It is because of Linear Algebra! The differential equation is linear meaning L(a*y1 + b*y2) = a*L(y1) + b*L(y2). So if L(y1) = 0 and L(y2) = 0, L(a*y1 + b*y2) =0. That's also why y1 and y2 are supposed to be linear independant :). Also, it's not that bad to show that the differential equation is linear.
i wish i could find someone for physics also just like him
Try out Walter Lewin. He doesn't give you the same kind of lectures as BPRP does, but he gives out weekly highschool problems that are fairly hard, and then at the end of the week he tells us the answer and explains why it is like that.
you all prolly dont care at all but does anyone know of a tool to log back into an instagram account??
I somehow forgot the password. I love any assistance you can offer me
Looking at you gives a positive vibe which keeps me engaged with the problem. God bless!
Thank you for the video! I noticed that when we substituted v' = w and divided, we obtained an equation of the form y' + f(x)y=g(x) which would usually require the integrating factor method but since g(x) = 0, the equation is separable
Albert Silvans ahhh it is huh! I was doing the problem on the spot and just chose I.F. right away lol
MY friend, this equation y' + f(x)y=g(x) couldn't be separated.
But if g(x) multiplied with y then it could be separated, else it's need an integrating factor !
"so, why is my son in tears while solving a math equation?"
"ma, I'm not gonna win tomorrow's exam..."
aaaaaand I'll have to brace for impending emotional damage tomorrow... thanks for the tutorial, but I'm definitely suffering from the time attack tomorrow
There are also techniques like
Reduction to Riccati
Change of an independent variable
I didnt see a video in which you present these techniques
(Reduction to Riccati works only for second order linear but not always equation after reduction is easier to solve)
Fantastic video! Very clear and easy to follow, thank you so much!
nice information
10:56 - Wouldn't d/dx of (xw) just be w? That part doesn't make sense. But if I check your final answer, it does equal zero?
w is a function of x
thanks a lot for existing ...........
Integrating factor may not be needed. Those 2 terms (with w) are separable.
Thank you :)
how do we know that the second solution is the product of the first and v. Anyone?
Nice proof
I thought of a question, and as far as I can tell this can't be done, but can you give it a go?
Find a nontrivial function (aka not f(x)=0 because this is the only solution I could find) that satisfies the fact that:
f ∘ f(x) = f'(x)
I don't think it's possible to solve, but you might be able to do it. If it isn't possible, how about:
f ∘ f(x) = (f'(x))^2
Austin Blake Hey that's interesting! I will have to think about it. Have you seen th-cam.com/video/ZH5rfWcR7l4/w-d-xo.html I called this the fake product rule. So ur problems might be doable but I just have to work on it later.
Austin Blake Question is whether you'd allow functions of complex variables and complex-valued functions to be used. For example, if f(x) = Cx^(i+1), f( f(x) ) = C(Cx^(i+1))^(i+1) = (C^(i+2))(x^2i) and f'(x) = C(i+1)x^i, (f'(x))^2 = (C^2)2i(x^2i). So the resulting equation that you have to solve is (C^i)*(C^2)*(x^2i) = (2i)*(C^2)*(x^2i). You can cancel out some terms and get C^i = 2i, so C = exp(π/2 + 2nπ)*2^(-i), for n ϵ ℤ. That's kind of ugly though.
Fully sent
Please do that reduction of order with symmetry of equation
Can anyone tell me can we put the general soultion like first solution + second soultion as first solution is given put this eq into diff eq and find second solution of this eq if not then why
It is because of Linear Algebra! The differential equation is linear meaning L(a*y1 + b*y2) = a*L(y1) + b*L(y2). So if L(y1) = 0 and L(y2) = 0, L(a*y1 + b*y2) =0. That's also why y1 and y2 are supposed to be linear independant :). Also, it's not that bad to show that the differential equation is linear.
this is euler differential equation , right?
Yes Cauchy Euler is applicable here since the variable coefficients have the same power as the derivatives
Wonderful and keep it up Mr
Nice haircut my dude!
Like is 216=6^3 and dislike 6-1
dablu