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

  Рет қаралды 18,467

Dr. Harish Garg

Dr. Harish Garg

Күн бұрын

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

Пікірлер: 60
@debajitroy7300
@debajitroy7300 3 ай бұрын
one of "The" best channel for net,Gate,JAM
@mathbasicbox5249
@mathbasicbox5249 Ай бұрын
❤❤ what a approach
@mymathspartner6019
@mymathspartner6019 25 күн бұрын
Nice video sir, thankyou sir
@alokkprajapati
@alokkprajapati 21 күн бұрын
Thanks Sir ❤
@professorragavan9783
@professorragavan9783 28 күн бұрын
Thank you sir
@divyaraju-y4k
@divyaraju-y4k Ай бұрын
best channel i ever seen thank u sir! pls continue more
@Ayoknyising
@Ayoknyising 7 ай бұрын
Sir you are alys great... ❤
@yashodhas1174
@yashodhas1174 2 ай бұрын
Thanks a lot sir, for this wonderful and informative class..😊
@SADDAMHUSSAIN-mw3cv
@SADDAMHUSSAIN-mw3cv 7 ай бұрын
Superb...
@shivamupd
@shivamupd 7 ай бұрын
It's very helpful video for us.
@simisaharan8754
@simisaharan8754 2 ай бұрын
Nice explanation sir.
@anshikathakur7085
@anshikathakur7085 Ай бұрын
Such a great explanation thanku so much sir means a lot God bless uh with lot of success and more❤
@mathbasicbox5249
@mathbasicbox5249 7 ай бұрын
Thanku so much sir ur great sir
@ritutiwari4687
@ritutiwari4687 2 ай бұрын
Nice ❤
@bakulmanna7123
@bakulmanna7123 7 ай бұрын
Thank you so much sir ❤❤
@professorragavan9783
@professorragavan9783 2 ай бұрын
Thank you sir!
@shyamnagar8917
@shyamnagar8917 7 ай бұрын
Tnkyu sir ❤
@simonrai7300
@simonrai7300 2 ай бұрын
Thank you Sir. ❤❤❤❤
@theexpress85
@theexpress85 2 ай бұрын
No words 🙏🙏🙏🙏🙏🙏🙏
@justiceforsidhumoosewala6225
@justiceforsidhumoosewala6225 7 ай бұрын
Nice explanation sir ❤
@positivethoughts6419
@positivethoughts6419 2 ай бұрын
Thank you so much sir 😊
@easymathematics1341
@easymathematics1341 7 ай бұрын
Ever greatest sir
@Learnmathematics-k5o
@Learnmathematics-k5o 3 ай бұрын
Thank you sir ❤
@ronneysaini7763
@ronneysaini7763 7 ай бұрын
Thanks Sir❤❤❤😊
@triptikumari9020
@triptikumari9020 7 ай бұрын
Thank you 👍👍 Sir
@vindhyavijayan3826
@vindhyavijayan3826 2 ай бұрын
Sir in 1st question Since every polynomial is continuous atleast one solution will exists. How is possible for no solution?
@sahebakhatun3806
@sahebakhatun3806 19 күн бұрын
Maybe the roots lie in the complex plane
@parveenkumar8050
@parveenkumar8050 7 ай бұрын
Thankyou sir
@Seema-jf9nr
@Seema-jf9nr 2 ай бұрын
Answer to last ques " blows up in finite time" is option B not D, sir plz recheck
@subratpradhan3581
@subratpradhan3581 4 ай бұрын
🙏🙏🙏 Thanks sir 50:16
@mamtakeswani3066
@mamtakeswani3066 7 ай бұрын
sir please make videos for solutions of ring theory and group theory
@subratpradhan3581
@subratpradhan3581 4 ай бұрын
Sir has no videos on Group and Ring theory, he has mastery in applied portions + Real Analysis+ CA + Linear Algebra
@naveed6396
@naveed6396 6 күн бұрын
At 7:01 what about the case when A0.
@nishajyoti5896
@nishajyoti5896 5 ай бұрын
एक्सीलेंट 🎉
@memebaaz4005
@memebaaz4005 7 ай бұрын
Superhlb
@user-yt6gx4xb6e
@user-yt6gx4xb6e 7 ай бұрын
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
@ronneysaini7763 7 ай бұрын
Nahi banayenge sir
@poojasingh7200
@poojasingh7200 27 күн бұрын
As you told in this vdo e^x is unbounded function this is correct but e^z is bounded function ....how can you say We know that non constant entire function is unbounded So e^z is also unbounded function
@kshamashreek6126
@kshamashreek6126 27 күн бұрын
Last question... i think the right answer is option b
@anshukumartiwari1361
@anshukumartiwari1361 2 ай бұрын
Result @7:03 is incorrect for A
@sibashishdas8806
@sibashishdas8806 5 ай бұрын
1:06:33 I think it's z=1/1+t
@someshnogia2009
@someshnogia2009 4 ай бұрын
Yes you are right.
@Ekshivbhakat
@Ekshivbhakat 7 ай бұрын
Thank you so much sir for helping us 🙏🙏
@meghavar
@meghavar 7 ай бұрын
Sir WhatsApp group join ni ho pata iss link se
@aweirdhimalayanbiker
@aweirdhimalayanbiker 3 ай бұрын
1st ka d hoga
@sibasankarbarada4169
@sibasankarbarada4169 5 ай бұрын
But sir e^z is unbounded because it is entire
@naveenpondara4903
@naveenpondara4903 5 ай бұрын
No bro e^z is bounded
@VlogswithJay
@VlogswithJay 3 ай бұрын
My dear brother who told you that e^z is bounded its unbounded
@naveenpondara4903
@naveenpondara4903 3 ай бұрын
@@VlogswithJay Haa yes...
@VlogswithJay
@VlogswithJay 3 ай бұрын
Bounded or unbounded ?
@naveenpondara4903
@naveenpondara4903 3 ай бұрын
@@VlogswithJay if there is no condition for real and imaginary parts then it is unbounded
@shefalithakur4580
@shefalithakur4580 2 ай бұрын
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
@Zeelkhokhariya
@Zeelkhokhariya Ай бұрын
Option D is correct😊..!!
@poojasingh7200
@poojasingh7200 27 күн бұрын
​@@Zeelkhokhariyayeah D is also correct bcz It's taking about A solution That is one so
@spp626
@spp626 7 ай бұрын
Thank you sir
@subho7125
@subho7125 3 ай бұрын
Thank u sir❤
@shrawankumarpatel2148
@shrawankumarpatel2148 7 ай бұрын
Thank you sir
@Manish-uq5le
@Manish-uq5le 7 ай бұрын
Thank you sir❤
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