Problem

For each of the differential equations in Exercis…

06:19

Problem 6 Easy Difficulty

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

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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Watch More Solved Questions in Chapter 9

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 that is divided by dx is equal to one plus x squared times one plus one is quite we're going to solve this using this variable separable method. So in this matter we basically separate the variables. That's the first thing that we do that is we separate all the white terms to on off the sides, preferably on the left side of the equation and all the X terms to the other side that is on the right side of the equation. And once we have separated the variables, we then go ahead and integrate both sides with respect to the X. And then we solve for the white, which is in fact the solution to the equation in some cases where we are not able to solve for the way, we just retained the equation as it is involving their terms X. And wise. So let's go ahead and apply the steps. So first let's go ahead and separate the variables. So we have those dvds on the website which is which is in fact a combination of x and y items. So first we will multiply both sides by dx. So when we do that we'll get dy is equal to one plus x squared times one plus y squared times dx. So on the left side we just have to wait um how we're on the right side, we have a combination of on plus x squared As well as one plus for escape and DX that is fine. So we can get rid of this one plus one is good on the right side and we transfer it to the left side of the equation. Then we would have successfully separated the variables. So to do that we divide both sides of the equation by this term, that is the one plus y squared. So when we do that we'll be getting dy over one plus y squared is equal to one plus x squared times of B X. So now we could absorb that. We have successfully separated the variables on both sides, that is this side, we have all the items that is expression involving on plus y squared and then the debate. Um and this is also we have all the expressions only in terms of X, that is one plus x squared as well as the external. So we can go ahead and do the next step. That is, this is in fact we're just part of the first step. So we'll do the next step that is, we go ahead and integrate both sides with respect to X. So let me write the situation, this is what we have after we have separated the variables and we integrate both sides. So we put the integration symbol, notice that if you remember uh if you could uh integrate this side that is on this, this site, we have this uh the way over on plus y squared which in fact we can apply this formula that is uh integration of one plus x squared dx. This is in fact the formula says it is turned inwards of its plus the variable the constant of integration. So it is of the same form. So therefore integration of this left side expression will be getting than in words of Y, we don't have to put the constant of integration. This side will collectively put all the terms all the constant of integration onto the right side. So then I I don't like this. Uh this side there's right side expression, you can split it into two integral. That there's an integration of one dx plus integration of X squared bx. So let's write down this side we have successfully integrated and then we'll software. And this is the integration of the X. S. X. And integration of X squared is when you apply the power rule this will be exciting. X. Raised to the power of two plus one. That is three divided by the same power that is so three. And then we put the integration constant that is sick. So therefore this is the solution to the given differential equation

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