PYQ on Uniqueness and Existence in ODE | Short Cut Tricks | CSIR NET 2011 to 2023

This lecture explains the PYQ on Uniqueness & Existence in ODE Short Cut Tricks CSIR NET 2011 to 2023

[@debajitroy7300]

one of "The" best channel for net,Gate,JAM

[@yashodhas1174]

Thanks a lot sir, for this wonderful and informative class..😊

[@theexpress85]

No words 🙏🙏🙏🙏🙏🙏🙏

[@SADDAMHUSSAIN-mw3cv]

Superb...

[@positivethoughts6419]

Thank you so much sir 😊

[@justiceforsidhumoosewala6225]

Nice explanation sir ❤

[@simisaharan8754]

Nice explanation sir.

[@ritutiwari4687]

Nice ❤

[@Ayoknyising]

Sir you are alys great... ❤

[@easymathematics1341]

Ever greatest sir

[@Seema-jf9nr]

Answer to last ques " blows up in finite time" is option B not D, sir plz recheck

[@bakulmanna7123]

Thank you so much sir ❤❤

[@triptikumari9020]

Thank you 👍👍 Sir

[@parveenkumar8050]

Thankyou sir

[@ronneysaini7763]

Thanks Sir❤❤❤😊

[@shrawankumarpatel2148]

Thank you sir

[@shivamupd]

It's very helpful video for us.

[@professorragavan9783]

Thank you sir!

[@shyamnagar8917]

Tnkyu sir ❤

[@Manish-uq5le]

Thank you sir❤

[@mathbasicbox5249]

Thanku so much sir ur great sir

[@simonrai7300]

Thank you Sir.
❤❤❤❤

[@memebaaz4005]

Superhlb

[@vindhyavijayan3826]

Sir in 1st question
Since every polynomial is continuous atleast one solution will exists. How is possible for no solution?

[@nishajyoti5896]

एक्सीलेंट 🎉

[@shefalithakur4580]

In 1st question :- a polynomial and its derivative is always continuous and bdd so it satisfies Lipschitz condition
Hence there is a unique solution in an interval containing 0 (d) is true

[@subratpradhan3581]

🙏🙏🙏 Thanks sir
50:16

[@user-yt6gx4xb6e]

Sir please making vidios for super tricks solution of algebra group theory and ring theory .....other vidio are very helpful ...so thank you so much sir ❤

[@ronneysaini7763]

Nahi banayenge sir

[@mamtakeswani3066]

sir please make videos for solutions of ring theory and group theory

[@subratpradhan3581]

Sir has no videos on Group and Ring theory, he has mastery in applied portions + Real Analysis+ CA + Linear Algebra

[@Ekshivbhakat]

Thank you so much sir for helping us 🙏🙏

[@meghavar]

Sir WhatsApp group join ni ho pata iss link se

[@anshukumartiwari1361]

Result @7:03 is incorrect for A

[@sibashishdas8806]

1:06:33 I think it's z=1/1+t

[@someshnogia2009]

Yes you are right.

[@aweirdhimalayanbiker]

1st ka d hoga

[@sibasankarbarada4169]

But sir e^z is unbounded because it is entire

[@naveenpondara4903]

No bro e^z is bounded

[@VlogswithJay]

My dear brother who told you that e^z is bounded its unbounded

[@naveenpondara4903]

@@VlogswithJay Haa yes...

[@VlogswithJay]

Bounded or unbounded ?

[@naveenpondara4903]

@@VlogswithJay if there is no condition for real and imaginary parts then it is unbounded

[@spp626]

Thank you sir

[@subho7125]

Thank u sir❤

[@reemuchauhan219]

Thank you sir ❤