PDE CSIR NET July 2026 Memory-based Question

[00:00:00]Hello everyone. Welcome to the part 10 on the memory based questions of the CS and net July 2026. In this video, I will solve the questions related to the partial differential equation. Myself Dr. Harish Kar, you can follow and subscribe my YouTube channel.

[00:00:18]The question I had received from the student is related to the Lagrange equation. How you can solve the Lagrange equation? It's a very very simple. You can also solve with the help of the Cauchy equation. You can write the Cauchy equations or the Lagrange equation. DX over and so on.

[00:00:38]Then, you can clearly say if I start from that because the coefficient of the XYZ are separate, I can start with the DX over DY.

[00:00:46]Fine. So, you get a negative 2Z.

[00:00:49]Plus Y, so it's a minus of Y. So, if I simply add DZ over Z, so Y and Y cancel.

[00:00:57]2Z, it's will be the minus of Z. So, I can add 2 into DY over Y. So, clearly say the required denominator will be zero. Therefore, I can integrate that.

[00:01:09]It will be log X.

[00:01:12]Z and it's a Y squared is my constant.

[00:01:17]Fine. So, I can say X Z Y squared is my e to the power c1 is another constant.

[00:01:26]Now, it involves all those X, Y, and Z.

[00:01:29]I can substitute direct value. You will get X. X and Y are same. So, that will be X cubed is equal to c1. But, c1 is always a constant number. But, you have a constant number. Yeah, there is no need to find c1.

[00:01:45]Then, I have a second expression that you have c2 in the name say. Fine. So, we can do that expression to make it all. Now, you can see it's a 2Z.

[00:01:55]Here is a minus Z minus Z. So, I can write as a Uh you can see it's a X 2 ZX is a 4x ^ 2 Z. If I multiply by first quantity by 2x, you can see it's a 2x DX. The belly expression can be 2 2 = 4 x ^ 2 Z plus 2x^2 y. The 4x ^ 2 Z minus can be clearly up to a plus of Z and here minus of yz or minus of yz. It's missing minus of me dy can be a. So, again you can see the denominator will be zero.

[00:02:35]Now, you can integrate them. It's a x ^ 2 by 2. 2 and 2 cancel plus Z minus y is my another constant. Fine. Now, you have the two arbitrary constants. You can write x ^ 2 plus Z minus y is a function of x y z ^ 2. Now, I can substitute the value of the initial condition. Z of x {comma} x x ^ 2 Z will be 1. Y will be x, which is equal to phi of x ^ 3. Fine.

[00:03:10]x y z ^ 2 x y z ^ 2. So, I can taken A is my x ^ 3. The value can be ready up to here. A raised to power 2 by 3 plus 1 minus A raised to power 1 over 3, which is equal to phi of A. I can substitute the value in this equation. The after equation can be x ^ 2 plus Z minus y, which is equal to phi of x z y ^ 2. I can substitute x y z ^ 2 at this point. x y z ^ power 2 by 3 plus 1 minus x z y squared power 1 1/3. Now you can see use the initial condition and get the value of alpha. X Z is my alpha cube. Y is equal to root 8. Fine? It will be x z y squared. X z y squared is root 8 y squared into alpha cube. That will be 8 alpha cube raised to power 2 by is 4 alpha squared. Plus 1 minus is a It will be 2 alpha cube. So Q will be cancelled out. It's a 2 into alpha. One and one will be cancelled. Your target is to find alpha cube. Minus 4 alpha squared plus 2 alpha which is equal to root 8.

[00:04:40]Root 8 will be 2 root 2 is the right answer of the So you can see that a very simple by using the Lagrange equation you can find the two values and get the simplest way to solve this problem.

[00:05:00]Fine? We will see the next lecture very soon. Till then, you can subscribe my YouTube channels, like and comment on this video as well. Best of luck students. Happy learning.