PDE Part - 3 CSIR NET July 2026 Memory-based Question

[00:00:00]Hello students, welcome to the part 12 on the memory based questions of the CSR net July 2026. Today's lecture is again on the partial differential equation.

[00:00:13]Myself Dr. Harishkar. You can follow and subscribe my YouTube channel. The question that I had received from the student is given in your screen. I don't know whether this statement is a correct or wrong because it's a memory based questions. I can assume that this statement is correct and I can provide you the answer of this statement.

[00:00:34]Statement statement misprint. You can send the correct statement in the comment box. I will upload the again the solution for you. Now the question are asking about the singular solution. Fine. So first of all whenever you are talking about the singular solution definition singular solution key the first of all it must be the solution.

[00:01:00]First of all it must be the solution and second is singular means you are talking about the inverine check option solution.

[00:01:12]So first option z = 1 + y. If I substitute the value, what is the partial derivative of the zed with respect to x? 0ero. Partial derivative of zed with respect to y is 1. So that number will be zero.

[00:01:28]Second quantity is also zero. Now look at the only the first quant second quantity zed is 1 + y square into 1 sin² x = 0. 1 + y can never be zero. Because if 1 + y will be zero then z=0 is the trivial solution then you can get as a sin square x will be zero that means sin x will be zero again which is not possible because the domain is my 0 to by 2 and we know sin of x is zero only when x is my zero but zero is not the domain is z = x + y is not A solution solution singular solution fine. Now look at the second one. Z = 3 that is my constant function. So clearly say Z X is 0ero Z Y is also zero. So that case is also 0 is =0 satisfied. So therefore Z = 3 is my solution.

[00:02:36]Singular.

[00:02:39]Look at the third option. Z = Y - 3. So clearly say Z X is my zero. Z Y = 1.

[00:02:48]Again once the Z Y is 1 expression again it implies sin square X is my zero because Z can never be zero to previous solution but that can never be zero on the given domain. So therefore z = yus 3 is not a solution solution singular.

[00:03:17]Now look at that z = 3 is a solution to fine z= 3 is a solution is fine but is it a singular? Now look at that z= 4 solution z= 5 solution. It means every constant is my solution. Fine.

[00:03:40]Z= 3 is one of the constant. It is not an invel.

[00:03:46]Fine. Inpution.

[00:03:54]This option is also cancel. Only the D singular solution does not exist. Is the right answer of the problem. Fine.

[00:04:04]substitution satisfied every constant satisfied by given partial differential equation. So the right answer is D is the correct answer.

[00:04:14]Examination let me know in the comment box. Second question that I had received from the student is given in your screen. Clearly say the given partial differential equation is of the second order partial differential equation. Fine. And your target is to find the values at the point 2 0 second order differential to boundary condition.

[00:04:47]Fine comment whether this statement is a correct statement or something is missing in the given statement or correct statement. You can send me in the comment box statement correct comment. Yes, it's a correct answer, correct statement. Then I will upload the solution of these questions in the next video. I'm waiting for your comments about the initial conditions whether zed of 1 comma y is zero and one more condition must be there because it's a second order partial differential equations. So I will wait for your comments. Till then you can subscribe my YouTube channels, like and comment on the videos and mark interested to watch the videos. Best of luck students.