Existence and Uniqueness Theorem Examples

[@filipeferoninho6738]

Great explanation! Clear concise. Really helped me a lot. Thank you

[@kevinebedi5257]

honestly this video was exactly what i needed, clear concise , wonderful had to come back to subscribe .gracias

[@lucarioind2203]

THANKS A LOT MADAM RHIS VIDEO WAS PERFECTLY PRECISE FOR ME

[@vijaysinghchauhan7079]

She is a great Teacher.

[@amalap3815]

Great lecture Mam. Thank you.

[@dioutoroo]

Thank you for your lecture!

[@mdtaifurhassan4911]

Nice explanation... Thanks mam

[@simonprice8638]

I thought she was writing everything backwards till I realised shes writing with her left hand

[@sadiqurrahman2]

Thanks a lot. Would you please explain how do we get (-∞,-2), (-2,2) and (2,∞). With regards.

[@aaronward4388]

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]

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]

​@@prathamsood244 Can I write it as (-∞,-2) U (-2,2) U (2,∞) ?

[@anshusharma2255]

Ty mam

[@JuanPerez-m3g]

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]

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]

@@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]

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.

[@CommodoreKat]

are you writing backward?

[@simonprice8638]

nah the footage is flipped

[@raizenzoldyck7563]

does that mean everything about her is flipped too?