CSIR NET July 2026 Memory based Question | Differential Equation| Part 7

[00:00:00]Hello students, welcome to the part seven of the memory based question.

[00:00:05]Today I will explain you in a very very simple manner how you can solve the CSNet July 2026 mathematics. Myself Dr. Harishar you can follow and subscribe my YouTube channel. the question that I had received from the students related to the second order differential equations and because four options examination but I had received the several different options from the students I put all of them in into the single platform. So let's see what is the right answer of the problem in a very very simple manner. Can you write the auxiliary equation of the second order differential equation? m² + alpha m + beta is equal to zero. What does the meaning of y = x e to power x that implies the auxilary equation has double root or double root be the cofficient of the e power x. Fine. Why? Because we all knows whenever x e powerx appears that mean there is a double root. So once m is equal to 1 and 1 are the roots of the auxilary equation. I can substitute into this equation. You will get 1 + alpha + beta is equal to zero. And clearly say the first option is cancel out because alpha + beta must be minus1. If alpha plus beta is a minus one d option is also cancel out examination plus one minus one then it's the correct option.

[00:01:42]Fine. Similarly the third option is also cancel because alpha plus beta must be minus one. So it can never be two. How you can find the alpha square minus beta square. For that case you must know the exact value of the alpha and beta. Fine.

[00:01:59]Now how you can find the alpha and beta since m= 1 is my double root. So double root definition.

[00:02:07]Firstly I can substitute m in the auxilary equation and second in the derivative. 2m + alpha is zero. I can substitute m= 1. So you will get alpha is my minus of 2. Once alpha is my minus2 I can substitute the first equation. You will get beta is one. So the B option is also cancel option like alpha plus alpha beta minus of three then you can substitute the value and get the answer because this a totally memory based questions so I never know what is the exact option given to you. These are first method you can find you can find the value of the alpha and beta. There is one more method you can find that that's twice on u because m is equal to 1 and 1 are my double root of the auxiliary equation or auxilary equation second order that means it's auxilary equation that will be m² - 2 m + 1 =0 is my auxilary equation. Now you can compare both the equations you will get alpha is my minus2 beta is equal to 1 to I compare and again you can discard or correct options A b C D clear part now look at the options P is a solution of the second equation the left hand side is a constant coefficient which is a similar to the previous one so it's P of X C1 + C2X E to power X fine then the how you can find the particular integral that will be 1 / D -1² E R power X / X why because it's a particular integral now I can take an e power X is outside so I can replace capital D to be the D + 1 1 is the cofficient of the X so that's a D square of 1 /x. So what is the meaning of the d square? That is a double integration of 1 /x. So can you find the double integration? The first integration is my ln x. Second integration is I can taken as a one. So that will be ln x into x - 1 /x into x. So that is again one. One is my x. x is outside ln x - 1. So your pi will be I can write pi will be e power x into x ln x minus 1. Is it okay?

[00:04:54]Now what is your target? Your target is to find the value of the p of e. So if I look at the what is the value of the p of e? What is the ln of e? ln of e is my one. 1 minus 1 is zero. So your answer is totally dependent on C1 and C2. Fine.

[00:05:16]By using the initial condition P of 1 is E X value 1 f C1 plus C2 of E X key value 1 X key value 1. What is the ln of 1? ln of 1 is my zero. So that will be minus1. I can take an e on the left hand side and cancel out e that will be c1 + c2 is equal to 2. Fine. Second is derivative at x= 1 product. Firstly I can take c2 x value 1 f plus I can take c1 plus c2 x value 1 e power 1 plus again I can use the product rule e power x dative e power x substitute x = 1. So is e 1 ln of 1 is zero is a minus one plus is x dative 1.

[00:06:17]So the remaining term is e power 1 ln 1 is zero is a minus one plus is ln x dative 1 /x. So 1 /x and x will be cancel out only e will be left behind.

[00:06:30]So that quantity will be cancel minus of e that is a twice or e cancel twice c1 plus of c2. Now clearly say from these two equations what is the value of the c1? c1 is 2 c2 will be zero. You can subtract them and get the value of c1 and c2. Therefore your answer will be p [clears throat] of e. P of E is my C1 is 2, C2 is zero, E ra to power, E is the right answer of the problem. So E is the correct answer. F is the wrong answer.

[00:07:20]Send me the comment box. I will upload the solution for you very soon. Till then you can subscribe my YouTube channel and mark in the comment box industry to watch my new video as well.

[00:07:32]Till then best of luck students.

[00:07:34]Happiness.