Problem

For each of the differential equations in Exercis…

20:33

Problem 10 Easy Difficulty

For each of the differential equations in Exercises 1 to 10 , find the general solution:
$e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0$

Answer

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Calculus 2 / BC

NCERT Class 12 Part 2 - Math

Chapter 9

Differential Equations

Section 4

Formation of a Differential Equation whose General Solution is given

Discussion

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Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Video Transcript

we will not find the general solution of this given differential equation using the variable separable matter. So in this method we basically separate all the white terms onto the website and all the X term onto the right side. And then we integrate both sides with respect to X. And then we solve for way in some cases where it involves trigonometry expressions like this, it's quite challenging to solve for way. So in those cases we just retain those expressions as such. So let's go ahead and solve this differential equation 1st I'm going to transfer this term to this site. So when I do that I'll be having one minus of epo x times of seconds. Quite why the way I'll be having this term as such, this will be and when you transfer this term to the other side it will take it it will take the negative form. That is we will be having minus orb forex turn why? Dx uh huh And to separate the variables. That is to separate uh the white terms to this side and all the X terms to the site. Uh If you could observe me how uh this uh seconds quite. Why do you Why? And we also have this term one minus or B products. So we don't need this term. If we could remove it from this term, this will be imperfect of white terms. So we need to divide by 1- of epochs and also notice that this side we should not have this certain way area. That term is an ex So we need to get rid of this town way. So basically we just how to divide by this whole expression that is we're going to divide both sides by this expression that is one minus F or X. Times of can wait. So let me write down here. So in the numerator we have one minus of epo x times of seconds. Quite why do you weigh? I'm going to do it by the stuff that is divided by 1- of People X. Time. Soft dan way. And on this side I have this ah epo X dan white dx Well what the term that I had to do it 1- New Politics time soft and white. We can simplify this now uh this will get cancelled one minus E products will get cancelled with this one minus E products and similarly on the right side this stand way will get cancelled and in the next step we could simplify this a little bit further. So this side we have only seconds quite why The same place. 2nd squared Times of Dy over 10 Y. This is basically second squared why do I? So this is equal to we have this negative sign minus of epo X. Times of dx divided by one minus e. Perec's. So that's what we have After we have simplified we can absorb that, we have successfully separated the white terms to the website, this left side we have all the items and now on the right side we have older X terms. So we can go hurt and integrate this now. So let's do that. I'm going to put this integration symbol over here which means I'm integrating both sides with respect to the X. No let's do this in to the for two parts. I'm going to first we can consider this as equation one. Since we need to apply the method of substitution. Let's let's consider this part one as integration of this. You can squared white. Do you weigh devoted boy dan white. So for this I'm going to apply the matter of substitution So let me substitute dan Y. He called to be. Now I take differentials on both sides when I do that I'll be getting sick and squared Y. The way because the differentiation of tanga is seconds quite Y. And then we have to put this freeway. So this side differentiation of T. Yes did he? So this is what we have and if you could observe we already have this term on the numerator that is seconds. Quite quite the way. So you can replace that as did the so let's do that. So therefore this integration will become bP they would have by tan why we could replace it. Dusty. So this is what we have and similarly let's integrate this so we will be getting integration of D. To liberty. This is equal to log off. T. Because we know from the fact that integration of D. X over X. This is equal to log x. Using the standard integration formula. So we just applied this form. So this is what we have let's go back and re substitute or We'll do the back substitution. 40. So this we will rewrite this as a log off tan white Which means we have successfully integrated the left side of the equation number one Question # one. Now it's in the same way I'm going to integrate this pot. So let me take this pot that is negative four X divided by one minus E pro X. Yes We will consider is as part two. What do we have? Negative eh poor x dx over one minus E. Poor X. We have to integrate this. So here also we are going to play this substitution. So for this I'm going to put 1 -30x. Let me put some different variables. That is you. Now I take the differentials on both sides. So when I do that differentiation of oneness, a constant and differentiation of negative people access we'll have this negative differentiation of products, E products times dx and this side will be getting to you. So as you can notice we have this negativity Parex times dx we already have it in the numerator which could be replaced us deal. So let's do that. So therefore this integration will become Bu over one minus E. Products, we can replace this as you. So therefore this would become like this. And uh we're going to play the formula which we applied for part one. So therefore this will become log off. You we do the back substitution for you. That is uh Lago we can replace this us one minutes of free products 1- or the products. So we have successfully uh integrated both sides. You can go back and substitute in the original equation at least in the equation number one. And went to substitute here we know the integration of this side. We know the integration of this side. So let's substitute in this one. We know that we know the integration of the third is equal to this is a log off done way. So on the left side we have log often way and on the right side we have log of one minus E products. Which means if you substitute into the equation number one will have log off and way on the left side. And on the right side we'll have this log off 1- E. Power X. And then we have to put this constant integration constant that we see ah this. See I'm going to relate this as a log off. See for some convenience. So la galaxy because luxi is also another constant. Now let's ah simplify this. The side. It will be logged off the runway is equal to we can play the property of logarithms that is that these two will get combined as lago one minus F or X. Time softly because log a plus log B is equal to log A. B. I have used that property. Now you can see that both sides we have algorithm. This is actually part of this algorithm. So we can remove this liberal thumb from bosses. And when we do that we'll be getting tan. Y is equal to see times of one minus F. O. X. So this, in fact, is still general solution of the given differential equation.

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