Boundary Value Problem (Boundary value problems for differential equations)

[@lilianadelgado6961]

Thank you so much for taking some of your precious time explaining this in detail. Really appreciated.

[@BriTheMathGuy]

You're welcome :)

[@mattklinger8644]

I am wondering if you'd be able to explain the following question: "Determine whether a member of the family can be found that satisfies the boundary conditions." My textbook gives examples of boundary cases where you don't have a solution, but I don't really understand the logic behind it. I understand how to work them at least!

[@rubyexotic]

hey! im struggling with this type of exercise when the boundaries are given in y and y'. e.g. y(o)=0 , y'(pi)=0. how can I solve this type of questions?

[@WinterFlare]

Write out your general solution. See if you can solve for y(0) = 0 for at least one constant. If not, take the derivative of your general solution, and then try to solve for y’(pi) = 0, then go back and solve the other. Hope this helps!

[@funwatchtv763]

can you explain why wrote r^2 + 2 = 0, what happened to the y

[@kosmoch]

Thank you, this helped me a lot.

[@BriTheMathGuy]

You're welcome!

[@TheTrollingKoala]

Thank you 🙏🙏🙏❤️ Helped me so much, you're a star

[@BriTheMathGuy]

So glad!

[@NTHStudio]

what is separated boundary conditions. please explain.

[@danielmoss7133]

isn't root 2*I the same as positive or negative 2i?

[@crashcoptr]

No. We can show this by testing if they're equal: case 1: root2*i=-2i root2=-2 case 2: root2*i=2i root2=2 In both cases, a contradiction (always-false statement) is found.

[@BriTheMathGuy]

3 Integral Tricks Teachers Don't Tell You! kzbin.info/www/bejne/eYXZaWptjNRqnJY

[@michelle_mensimah03]

Thank you

[@algorithmo134]

Can you do videos on real analysis I and real analysis II basically the full book of ruin

[@anirbanpandit445]

Why u picked y as r ?

[@abro7827]

You lost me at r^2 ? what are you doing there? 1 derivative? where did the 2 come from?

[@hans3331000]

so he broke up -4AC into perfect roots. it's something you learned way back so it's not obvious right away. so he broke up -4(1)(2)=-8 (notice how -8 is NOT a perfect root), so he broke up -8 into (sqrt(-4x2)), now he knows sqrt of 4 is 2, so you can rewrite that as [2xsqrt(-2)]but again, you cannot have a negative under the root, BUT we do know a special number "i" is equal to (sqrt(-1), so now[2sqrt(-2)] turns into[2sqrt(2)xi]. And the whole time this equation is actually being divided by 2A (we've only been dealing with -4AC) so he divided the '2' outside the sqrt by 2A which is 2/2(1)=1. so your final statement is +-(sqrt2i)

[@vishvajitsinhkosamiya7154]

If a differencial equation is homogeneous you can use this trick somehow when you replace dy/dt with r and d^2(y)/dt^2 by r^2 Or am i missing something?😅

[@vishvajitsinhkosamiya7154]

m.kzbin.info/www/bejne/qaC4Xq2ImdWmpaM This link might help. I had the same trouble. Cheers.😅

[@WinterFlare]

Mid engineering student here, but this is the way that I view it: It’s a little trick with differentials. We know that in order to solve this, y’ must be c*y. Aka, a constant multiplied by Y. This is so everything cancels out and we’re left with 0. For example, to help visualize this. Whenever differentiating y, we get y’ = c*y, so y’’ = c^2*y. Let’s apply this to y’’ + 2y = 0. If we use the two terms above and factor out y, we get y(c^2+2)= 0. We know y isn’t 0, because then y is always 0 so, y(0) ≠ 1. Because we know we can’t set y=0 as a solution, instead want to solve for c^2+2 = 0, specifically the c part, since that’s sort of like a ‘scaling factor’ (probably not what the actual term is) that helps make the second derivative equal to the original in order cancel out. This is why they write as r^2 + 2 = 0. I’m sure you see the pattern of how y = 1, and y’’ becomes c^2. It’s just replacing c, the constant, as r.

[@ruhisinghsidar8506]

Thank you sir

[@anuaj3767]

Thanks sir

[@vishvajitsinhkosamiya7154]

Anyone noticed final solution has y=0 irresepctive of value of t? Y= cos(sqrt(2)t) -Cot(sqrt(2)t)*sin(sqrt(2)t) Y= cos(sqrt(2)t) - cos(sqrt(2)t) Y=0 Or am i missing something?

[@BriTheMathGuy]

Vishvajitsinh Kosamiya the cot function is evaluated at pi, not at t, so they don’t cancel. Thanks for watching and have a great day!

[@engahmedou2024]

This video is totally terrible... anlamıyorum!!!

[@BriTheMathGuy]

Very sorry you thought so, but I hope you have a nice day.

[@vegbetle]

mad cause you didn't learn complex roots + cope