PDE Part - 5 CSIR NET July 2026 Memory-based Question

[00:00:00]Hello everyone, welcome to the next lecture on the memory based question.

[00:00:05]Today I will upload the another video related to the partial differential equations of the CSN at July 2026.

[00:00:13]Myself Dr. Harishkar. You can follow and subscribe my YouTube channel. Now the question is related to the constant coefficient. So whenever you have seen the constant coefficient I can convert the given partial differential equation into the terms d² - 2 into d into d prime + d prime² 2d - 2d prime + 1 into z is equal to e^x + 2 y. Now clearly say from the first three terms I can write as a d minus d prime square plus I can taken two as a common d minus d prime + 1 into z is equal to e^x + 2 y if you look about the left hand side it is a a - a + b² is it okay? Now once you have simplified the expression then your target is to find the complimentary functions and the particular inagle. Fine or partial differential equation playlist theory lecture playlist just before your one month examination detail course explain. So how you can write the complimentary functions? I can write because my homogeneous it's a non-homogeneous part fine so it will be e power min - x into 51 of y + x because it's a repetition I can write as a x of 52 y + x now how you can find the particular integral so I can write for the particular integral is 1 / d minus d prime + 1 square e power x + 2 y clearly say if I substitute d = 1 d prime is equal to 2 it's a case of failure I can multiply with the x and take that derivative again you can see it's a case of failure fine so instead of x I can twice the differentiate and it will be 2 e to power x + 2 y so your complete solution will be this number. Fine. Now your target is to find the value of zed of 2 0 with the initial condition.

[00:02:46]Z of 2 I it is e power minus of 2 51 of 2 + 52 of 2 + 4 / 2 is a 2 e to power 2. Now I can apply the initial condition zed of 1 comma 0 sorry zed of 1 y is my zero. So which implies e power -1 [snorts] 51 1 + y 1 1 into 52 1 + y + x = 1 e + 2 y it will be my zero or second condition z of 0 y is equal to e power y. So if I substitute the left hand side will be e power0 is a 1. Fine. So that will be one x will be zero. x will be zero. It is e power 1. Now because my target is to find the 51 and 52, I can substitute the value is e right side. So it is e power 1 + y + 52 of 1 + yus or e power divide. So that will be 2 + 2 y / 2. Fine. So you 52 easily 52 will be 1 + y is - e power 2 + 2 y / 2 - 1 + y now I can substitute all the values in the given requirement e² e power minus of 2 51 of 2 I can substitute y = 2 plus 52 of y= 1 and then so that number will be minus y value 1 f that's a four * 2 it's a minus y = 1 f it's e² + e² now you can open the bracket it's a 1 - e² - 2 + e² the answer will E² -1 is the correct answer of the problem is the answer. Yes, the B is my correct answers. So, let me know question.

[00:05:19]If you are interested to watch more videos, let me know in the comment box and subscribe my YouTube channel. I will upload for you always in a front matter.

[00:05:27]Thank you very much students. Desktop.