👉 This Differential Equation Looks Impossible 😳 | GTA Part 14 #jeeadvanced

Added: 2026-05-10

[00:00:00]Hello everyone. Outside the rain is finally arriving after days of scorching heat.

[00:00:07]The thunder is loud, the air feels lighter, and the earth is slowly cooling under evening shower. And strangely, this problem feels very similar. For many students, this differential equation question has been a nightmare for many days.

[00:00:28]complicated terms, confusing structure, and no clear direction.

[00:00:34]At first glance, it looks like the kind of question that refuses to open its door, even the windows.

[00:00:46]But every difficult problem has a hidden crack.

[00:00:52]And once this differential equation is unlocked, everything suddenly begins to flow naturally.

[00:01:01]The relief of seeing the right manipulation uh in fact is the achievement of the calmness after the struggle.

[00:01:14]It feels almost like this rain outside my window tonight.

[00:01:20]In this peaceful weather, I invite you to sit together with me and solve this differential equation because we are going to dissolve this question step by step into something unexpectedly elegant. Let's dive in. All right. So this question demands a very sound manipulation. So let's man manipulate it.

[00:01:50]So the question has a gigantic differential equation that we have to manipulate. So let's uh see how we can do it. So the first thing is that uh first I try it by just substituting 0 and 0 at the place of x and y. So if you substitute x and y as 0 and zero, you can notice that I will get only this term left out dx and dy because every term will become zero. So what I thought first that maybe I have the differential equation dx plus dy is equal to zero because already said that the function is passing through the origin. But this is not the complete solution we get from here because we only get that curve which is passing through origin as the differential equation polomial grows.

[00:02:40]Means the term grows the other differential equation solution will emerges will emerge actually. So this is not the correct way to think about any differential equation not only this one.

[00:02:55]So this failed.

[00:02:59]Now once I saw this thing I thought that it could come in the derivative of the tan inverse of x by y + 1 or tan inverse of y + 1 by x because when you differentiate it this will become 1 by x by y + 1 the whole square and you manipulate this will go down but differential equation has this kind of a format that or this kind of history or maybe you have seen this kind of questions many times that it uh this can be written anywhere but after manipulation it can be taken down and we simply use our uh idea.

[00:03:44]So for that purpose we have to collect all the terms containing y + 1 the whole square together if there are so let's first keep the left hand side as it is and I'm going to expand this right hand side so let's call it this is rhs and rhs I'm just manipulating for my purpose so I'm going to multiply this thing so before that we can see this one that x dx and this dx x but can be taken together when dx is taken out as a common factor. So what I get there is x + 1 into y + 1 and here I'm getting x + 1 into dx and similarly I will get y + 1 into dy.

[00:04:37]Now once you multiply this thing I get this as x + 1² into y + 1 dx + x + 1 into y + 1 the whole square dy and when I did this thing it clicked to my mind that I can just combine this term with the right hand side the term present with y + 1 the whole square.

[00:05:05]There is one such challenge. We'll face it and we'll crack it. So let's see.

[00:05:10]Just I'm writing this thing as y * x² y + 1 the whole square dx + x into x + 1 dy and right hand side just I have manipulated in a little bit simpler form. So I'm writing that form as it is.

[00:05:34]Now our aim is to bring this term this term there. So I also at the same time I'm going to bring this guy on the right hand side.

[00:05:47]So what I do there? Let's see. I'm taking this whole term together. It will become a negative term. So I'm going to write y * x² + y + 1 the whole square dx - x + 1 y + 1 the whole square dy and there x + 1 the whole 2 y + 1 dx - x * x + 1 dy and I'm actually just revealing the thought process what I did in this question and there no cheating I'm not just uh uh solving or just unengineering this thing means I I know the answer what it is but not actually doing the reverse engineering to get back to this differentiate equation. I thought in the same way uh in the second attempt not the first attempt. All right. So uh what I thought now after that we can do this thing that x + 1 is there and y + 1 the whole square there.

[00:06:51]But I want to take this whole terms as a factor out. It means I need again there x² + y + 1 the whole square. So for that I'm going to do is this thing x² + y + 1 the whole square as it is with dx.

[00:07:08]And there I'm deliberately writing this thing. Let's see x + 1 uh x + 1 is there. And then I need see this thing that I need x² + y + 1. So let me write in this way that I'm writing x + 1 factor and then x² + y + 1 the whole square and then dy so that I can take this thing together factored out. But in this what we did extra that I just subtracted this x + 1 into x² with this term because this was not present there. It was only this term multiplied with this. So I just made this thing for our purpose. So we did a crime mathematically. So we have to actually pay it off for this crime by just adding extra term. What is this thing that is x + 1 into x²?

[00:08:17]Now the right hand side I will keep it as it is x + 1 the whole square y + 1 dx - x into x + 1 dy.

[00:08:30]Now notice these two terms uh you can see they're together that were the whole purpose of this whole manipulation drama. We can take this thing common. So x² + y + 1 the whole square common and we get y dx minus x + 1 dy by the way there dy and again I'm sending this guy what we have uh received from the previous step to the right hand side. So we have got x + 1 the whole square into y + 1 into dx - x into x + 1 - x + 1 into x² and since this has also dy that I forward that right because this was coming from the dy I will take these two things together with a plus sign in between with dy.

[00:09:30]I hope it's clear.

[00:09:33]And by the way, this is not there. Let me just take this thing because there I have a space here. All right.

[00:09:41]So now we have this whole situation. So let's let's deal with this. So what we do? You can realize this not actually very quickly but you can realize this thing. This is inviting us to take the quotient rule of y by x + 1 or because I need x + 1 whole square cancel. I can also write x + 1 by y but maybe the minus sign we will uh manipulate it. So what I'm going to do I'm writing this minus in front and this whole is coming with all this things. So let minus is front. So this you can write x + 1 dy - y dx and I'm dividing the entire equation by x + 1 the whole square where it is this is here.

[00:10:34]So let's divide it x + 1 the whole square and why I'm dividing because I saw this structure that this is going to be x + uh yeah that's correct y by x + one derivative.

[00:10:52]Now once you divide this x + 1 the whole square throughout that will give us y + 1 dx and minus here we can multi divide by x + 1 the whole square. So I get it here x by x + 1 + x² by x + 1 and dy.

[00:11:14]We are almost there. So let me write this thing because we have done this for the same purpose.

[00:11:23]So I can write this the derivative of y / x + 1.

[00:11:32]And on the right hand side I have these two terms it actually manipulated.

[00:11:39]How you manipulate? You see that if you manipulate I can take this x + because the denominator is same x + x² there x common. So x + 1 cancel. This is nothing but just x. So I can write there y + 1 dx minus x dy.

[00:11:58]And now we can just divide this entire equation by this quantity.

[00:12:03]So this quantity is sitting over here.

[00:12:06]So this is d by dx y by x + 1. And this is going to be y + 1 dx - x dy and divided by x² y + 1 the whole square. Now there's a chance and mathematics invite actually invites it's a subject in discipline like this that we can take a chance. So because our guess was that x² + y + x² + y + 1 the whole square will come in the denominator of derivative of tan inverse x y + 1. So let's just hope it. Why it is hope it? Because you notice that this thing uh suppose you have d by dx of tan inverse of x by y + 1 then we can differentiate it.

[00:12:56]What we get it here is y + 1 the whole square and this is y + 1 dx - x dy uh rather I have done a mistake there.

[00:13:10]It's not like that we differentiate it.

[00:13:13]So what we get 1 by 1 + x² by y + 1 the whole square and into this quotient rule there which is present this uh present there by the way. So this is going to be what uh y + 1 the whole square then y + 1 dx and minus x dy. So this is actually the exact derivative because these two terms will get cancel. You notice this will come there and after LM and all this two terms will get cancel. So this will become Y + 1 the whole square and after that we get it. We can also think in think like this. What we can think that since I know this fact here that this term is coming in the derivative of this what this x by y + 1. So I can multiply and divide by it. So it'll give you y + 1 the whole square there. And again after this I can write y not y rather this is going to be 1 by 1 + x by y + 1 the whole square into the derivative of uh x by y + 1.

[00:14:39]And now you notice that this is the derivative of tan inverse and you're further differentiating x by y + 1. It means there is giving the sense that this is exact derivative of tan inverse of x by y + 1. So by the way there was a minus sign I'm just leaving there otherwise the okay it means what I got is uh got it is what dy dx of y by x + 1 is equal to tan uh dx of tan inverse of x by y + 1 and so we got it y by x + 1 is equal to either minus sin tan inverse of x by y + 1. And since the graph is passing through origin, it means the constant term is not required. And so we got tan inverse x by y + 1 + y by x + 1 is equal to zero is the required curve we got. Now after this any student can do it but because they're asking that the word is passing through a comma b. So we substitute this and even all the students have uh solve this portion after getting the answer by looking at the solution. The solution the official solution in chhattanyaki has gi like given so much difficult manipulation which is not easy to understand. It's very it require like a very sophisticated skill to understand.

[00:16:18]I tried my way to make it a little bit simple.

[00:16:24]So that is the solution for this question. So that's all for today. See you in the next video with some other interesting question and solution. Till then stay math active.

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