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

  Рет қаралды 20,543

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: 61
@debajitroy7300
@debajitroy7300 5 ай бұрын
one of "The" best channel for net,Gate,JAM
@divyaraju-y4k
@divyaraju-y4k 3 ай бұрын
best channel i ever seen thank u sir! pls continue more
@mathbasicbox5249
@mathbasicbox5249 3 ай бұрын
❤❤ what a approach
@shivamupd
@shivamupd 10 ай бұрын
It's very helpful video for us.
@mymathspartner6019
@mymathspartner6019 2 ай бұрын
Nice video sir, thankyou sir
@Ayoknyising
@Ayoknyising 10 ай бұрын
Sir you are alys great... ❤
@SADDAMHUSSAIN-mw3cv
@SADDAMHUSSAIN-mw3cv 10 ай бұрын
Superb...
@yashodhas1174
@yashodhas1174 4 ай бұрын
Thanks a lot sir, for this wonderful and informative class..😊
@anshikathakur7085
@anshikathakur7085 3 ай бұрын
Such a great explanation thanku so much sir means a lot God bless uh with lot of success and more❤
@alokkprajapati
@alokkprajapati 2 ай бұрын
Thanks Sir ❤
@professorragavan9783
@professorragavan9783 3 ай бұрын
Thank you sir
@simisaharan8754
@simisaharan8754 5 ай бұрын
Nice explanation sir.
@mathbasicbox5249
@mathbasicbox5249 10 ай бұрын
Thanku so much sir ur great sir
@bakulmanna7123
@bakulmanna7123 10 ай бұрын
Thank you so much sir ❤❤
@shyamnagar8917
@shyamnagar8917 10 ай бұрын
Tnkyu sir ❤
@professorragavan9783
@professorragavan9783 5 ай бұрын
Thank you sir!
@triptikumari9020
@triptikumari9020 10 ай бұрын
Thank you 👍👍 Sir
@justiceforsidhumoosewala6225
@justiceforsidhumoosewala6225 10 ай бұрын
Nice explanation sir ❤
@ritutiwari4687
@ritutiwari4687 4 ай бұрын
Nice ❤
@Manish-uq5le
@Manish-uq5le 9 ай бұрын
Thank you sir❤
@theexpress85
@theexpress85 4 ай бұрын
No words 🙏🙏🙏🙏🙏🙏🙏
@easymathematics1341
@easymathematics1341 10 ай бұрын
Ever greatest sir
@ronneysaini7763
@ronneysaini7763 10 ай бұрын
Thanks Sir❤❤❤😊
@simonrai7300
@simonrai7300 4 ай бұрын
Thank you Sir. ❤❤❤❤
@parveenkumar8050
@parveenkumar8050 10 ай бұрын
Thankyou sir
@RanjangamingYT7477
@RanjangamingYT7477 12 күн бұрын
1:3:25 check y=y0 e^x Then y0 non zero but y0 positive and negative ho sakta ha some time increase and some time decrease Therefore, not monotonic
@nishajyoti5896
@nishajyoti5896 8 ай бұрын
एक्सीलेंट 🎉
@mamtakeswani3066
@mamtakeswani3066 10 ай бұрын
sir please make videos for solutions of ring theory and group theory
@subratpradhan3581
@subratpradhan3581 6 ай бұрын
Sir has no videos on Group and Ring theory, he has mastery in applied portions + Real Analysis+ CA + Linear Algebra
@SasankaKayal-b8s
@SasankaKayal-b8s 10 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 10 ай бұрын
Nahi banayenge sir
@naveed6396
@naveed6396 2 ай бұрын
At 7:01 what about the case when A0.
@Seema-jf9nr
@Seema-jf9nr 4 ай бұрын
Answer to last ques " blows up in finite time" is option B not D, sir plz recheck
@subratpradhan3581
@subratpradhan3581 6 ай бұрын
🙏🙏🙏 Thanks sir 50:16
@memebaaz4005
@memebaaz4005 10 ай бұрын
Superhlb
@poojasingh7200
@poojasingh7200 3 ай 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
@sibashishdas8806
@sibashishdas8806 7 ай бұрын
1:06:33 I think it's z=1/1+t
@someshnogia2009
@someshnogia2009 6 ай бұрын
Yes you are right.
@Ekshivbhakat
@Ekshivbhakat 10 ай бұрын
Thank you so much sir for helping us 🙏🙏
@vindhyavijayan3826
@vindhyavijayan3826 4 ай бұрын
Sir in 1st question Since every polynomial is continuous atleast one solution will exists. How is possible for no solution?
@Kawaionnanoko8777
@Kawaionnanoko8777 2 ай бұрын
Maybe the roots lie in the complex plane
@anshukumartiwari1361
@anshukumartiwari1361 4 ай бұрын
Result @7:03 is incorrect for A
@kshamashreek6126
@kshamashreek6126 3 ай бұрын
Last question... i think the right answer is option b
@meghavar
@meghavar 10 ай бұрын
Sir WhatsApp group join ni ho pata iss link se
@shefalithakur4580
@shefalithakur4580 4 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 3 ай бұрын
Option D is correct😊..!!
@poojasingh7200
@poojasingh7200 3 ай бұрын
​@@Zeelkhokhariyayeah D is also correct bcz It's taking about A solution That is one so
@sibasankarbarada4169
@sibasankarbarada4169 7 ай бұрын
But sir e^z is unbounded because it is entire
@naveenpondara4903
@naveenpondara4903 7 ай бұрын
No bro e^z is bounded
@VlogswithJay
@VlogswithJay 5 ай бұрын
My dear brother who told you that e^z is bounded its unbounded
@naveenpondara4903
@naveenpondara4903 5 ай бұрын
@@VlogswithJay Haa yes...
@VlogswithJay
@VlogswithJay 5 ай бұрын
Bounded or unbounded ?
@naveenpondara4903
@naveenpondara4903 5 ай бұрын
@@VlogswithJay if there is no condition for real and imaginary parts then it is unbounded
@aweirdhimalayanbiker
@aweirdhimalayanbiker 5 ай бұрын
1st ka d hoga
@spp626
@spp626 10 ай бұрын
Thank you sir
@Learnmathematics-k5o
@Learnmathematics-k5o 5 ай бұрын
Thank you sir ❤
@positivethoughts6419
@positivethoughts6419 4 ай бұрын
Thank you so much sir 😊
@shrawankumarpatel2148
@shrawankumarpatel2148 10 ай бұрын
Thank you sir
@subho7125
@subho7125 5 ай бұрын
Thank u sir❤
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