Problem

For each of the differential equations in Exercis…

05:24

Problem 7 Easy Difficulty

For each of the differential equations in Exercises 1 to 10 , find the general solution:
$y \log y d x-x d y=0$

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

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Video Transcript

we will not find a general solution of this given differential equation. That is wild log Y D x minus x. Dy equal to zero. We're going to do this using this variable separable method. So in this method basically we separate all the white terms onto the website and all the extremes onto the right side. So that will be the first step. And then the next step we integrate them. That is the, integrate both sets with respect to X. And finally we solve for the white. So let's go ahead and do this. So in the first year we have to separate the variables. That is, we should have all the white terms onto the one side and X terms onto the right side. After we have separated the variables. So for that what we do is we add this term to both sides first. So when you do that I'll be getting white time. Soft log Y bx is equal to X. Times of the way. Uh If you could notice we have this wild log way with the dx term and here we have the storm expert, the waiter. So first let's divide both sides by this time. That is way long way. So when we do that this time we will not have this way long way and this I will be having X time soft dy over why log white and this side after we have done this this side we have only the extra but this side we have uh Dy over oil log way as bless this extreme. So we just have to differentiate both states by dx now. So let's do that when we do that. I'm sorry. We have to divide by X. Term. So when we do that we'll be getting D. X over X. This time. And here will be getting Dy over white time soft log way. So now we have successfully separated the variable that is all the extremes to do one of the sides and all the way terms to the other side of the equation. So we can go ahead and integrated. This is basically this against him. So before I do that for some convenience, I'm going to transfer this all these terms to this side. That is to the left side and all this time. That is A. D. It's over X. We will transfer it to the side. We can do this in any kind of inequality. So let's do that. So I can read on this as dy over white time self log way and this is equal to D. X over X. So now I can integrate it because this left side we have all the terms in white and on the right side we have all the terms in X. So we integrated. And uh we're going to use the substitution method to integrate this uh website. So for that let me put log Y equal to P. I know differentiate both sides. So if I do that I'll be getting one by why times of dy is equal to D. D. So this is what happens when you differentiate both sides. Now observed that uh in this equation this dy over why this could be replaced as did he? Because that's what we have one by white. Times of diva is in fact a quantity. So this is part of the situation. So therefore this is going to become this left side is going to become D. P. Over. And this D way can be I'm sorry this log way can be replaced as steep. So when you do that I'll be getting DT over. Tea is equal to the side. We can write down as it is the X over X. We have not integrated. We can do that in the next few steps. So now we can integrate this easily. This is just A. D. T. Over T. Which in fact will be logged off When we integrate using the standard integration formula. Because integration of one way XDX is in fact log X. So using this logarithmic for integration formula, we integrated this. So this side this uh this all uh D X by X. If we integrated will again get log of X. And plus C. Instead of writing down this as just policy, I can write down this as log off. See because log off C will also be another constant. So we have integrated both sides. But nobody said this side is still with In terms of 40 after we have done this institution. So we go ahead and do the bad substitution. We know that we can replace the tea with log way. So let's do that. And when we do that we will we will be getting log off. No wait on the site. This is equal to we can combine these two using the standard logarithmic properties. That is a log of a low G plus lobby. This is equal to log a be using this property. I can write on this as a log off, exceed all right. Can I do this as a log of successful? So let me write it like that. This is a logo. C. X. Y. C is a constant. Now we can remove log on both sides because we have a log on both sides. We can remove this uh log this is in fact mathematically equally as taking an telegram antill algorithm on board says. So when we do that we'll be getting a log of Y is equal to six. Now we can write this logarithmic equation in why into exponential equation which means we'll be getting the Y equal to see, I'm sorry, very big. Getting why equal to He raised to the power of six. So this is the equal and exponential equation Of this equation. That is log way equal to six. When we convert this to exponential form will be getting y equal to he raised to the power of C X. And which in fact this is there solution of the given differential equation.

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