Existence and Uniqueness Theorem Examples

  Рет қаралды 23,141

Brenda Edmonds

Brenda Edmonds

Күн бұрын

Пікірлер: 20
@filipeferoninho6738
@filipeferoninho6738 Жыл бұрын
Great explanation! Clear concise. Really helped me a lot. Thank you
@kevinebedi5257
@kevinebedi5257 Жыл бұрын
honestly this video was exactly what i needed, clear concise , wonderful had to come back to subscribe .gracias
@lucarioind2203
@lucarioind2203 2 жыл бұрын
THANKS A LOT MADAM RHIS VIDEO WAS PERFECTLY PRECISE FOR ME
@vijaysinghchauhan7079
@vijaysinghchauhan7079 6 ай бұрын
She is a great Teacher.
@amalap3815
@amalap3815 5 ай бұрын
Great lecture Mam. Thank you.
@mdtaifurhassan4911
@mdtaifurhassan4911 3 жыл бұрын
Nice explanation... Thanks mam
@dioutoroo
@dioutoroo 5 ай бұрын
Thank you for your lecture!
@JuanPerez-m3g
@JuanPerez-m3g 3 ай бұрын
how is a unique solution guaranteed to exist if we have only checked if the point is continuous on f? this guarantees existence. doesnt it have to be continuous on df/dy to be unique?
@BrendaEdmonds
@BrendaEdmonds 3 ай бұрын
A function is continuous on an open region around a point (not the point being continuous on the function). But also there are 2 different theorems being used here. One theorem, for DEs of the form dy/dt=f(t,y) does require that you check continuity of f(t,y) and the continuity of the partial derivative of f with respect to y. The other theorem is specifically for LINEAR DEs, y'+P(t)y=Q(t), and has different conditions to check and different conclusions you can draw.
@フアン-f6e
@フアン-f6e 3 ай бұрын
@@BrendaEdmondsSo for linear DEs, if P(t) and Q(t) are continuous around the point given, that is enough to prove the point is a unique solution?
@visheshmangla8220
@visheshmangla8220 11 ай бұрын
Please compare the question discussed at 7:20 in kzbin.info/www/bejne/baCqeKiai9mAY9U. According to what's discussed for linear case is giving a wrong answer for this question.
@sadiqurrahman2
@sadiqurrahman2 2 жыл бұрын
Thanks a lot. Would you please explain how do we get (-∞,-2), (-2,2) and (2,∞). With regards.
@aaronward4388
@aaronward4388 2 жыл бұрын
12:22 she states that there's a discontinuity at t=-2 and t=2. So for all other values of t (the intervals above), the function is continuous
@prathamsood244
@prathamsood244 2 жыл бұрын
t^2 - 4 cannot be zero which means, t cannot be positive or negative 2. The function is continuous when t is not +2 or -2. Saying this in terms of a numberline, (-infinity, -2), (-2, 2) and (2, infinity). When we use "(" or ")" in such things, it means that the number is not included but everything between this number and other other number in the coordinate thing is included. (infinity, -2) means everything from infinity to -2 is included(including -1.9999999999.......) but not -2.
@MsSoldadoRaso
@MsSoldadoRaso 2 жыл бұрын
​@@prathamsood244 Can I write it as (-∞,-2) U (-2,2) U (2,∞) ?
@simonprice8638
@simonprice8638 2 жыл бұрын
I thought she was writing everything backwards till I realised shes writing with her left hand
@anshusharma2255
@anshusharma2255 Жыл бұрын
Ty mam
@CommodoreKat
@CommodoreKat 2 жыл бұрын
are you writing backward?
@simonprice8638
@simonprice8638 2 жыл бұрын
nah the footage is flipped
@raizenzoldyck7563
@raizenzoldyck7563 2 жыл бұрын
does that mean everything about her is flipped too?
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