Problem

For each of the differential equations in Exercis…

14:16

Problem 11 Easy Difficulty

For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition:
$(x+y) d y+(x-y) d x=0 ; y=1$ when $x=1$

Answer

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Calculus 2 / BC

NCERT Class 12 Part 2 - Math

Chapter 9

Differential Equations

Section 5

Methods of Solving First order, First Degree Differential Equations

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 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Video Transcript

we will not find a particular solution of this given differential equation. So here I have about playing the steps that we follow to find the particular situation. First we determine if they're given differential equation is homogeneous. And then we find the general solution. We find that generation is in the middle of solving homogeneous differential equations. And finally we find the particular solution by substituting the initial conditions into the general solution. So first let's do the first part which is held with gentlemen if this is uh more genius differential equation. So for that I need to write this in the form of dvds. So first let me write down the first part as it is that is expressed white times of the way. And I'm going to transfer this term to the right side which means it will take its negative form. So we can put this as negative of minus it's minus Y times objects and this is equal to when I distribute. Then I do this is uh y minus X times of the X. So I can rewrite this us y minus x times of dx. Now I can divide both sides by X plus Y as plus by dx. So that I'll get the real media. So let me do that for the D V. D x is equal to y minus X. They were a boy express white to prove that it is a homogeneous uh We can read on this in the form of export and and then function of labor. It's if we can write on this right side part in this form. Then we say that this is a homogeneous differential patient. So let's try to write down let me factor the X from both numerator and denominator. And when I do that I'll be getting ex uh ex minus one. To buy my factor X from the denominator. Russell. We're getting one plus Y by X. So basically this is export zero since we can cancel the X as one times one I'm writing as expo zero. And this is no formal for a four X. Because every term has webex as opposed the constant. Um So this proves that this is a homogeneous differential equation. So let's find the general solution of this given differential equation for that. Let me write down this differential commission. That is, I'm going to take this form. The wide body X is equal to y minus X, divided by X plus Y. Let me write down that first you wait by dx is equal to y minus X over X plus Y. It's called the situational situation number one. Uh To find a general solution of homogeneous differential equation Will be placing this institution, Y Equal two weeks. And then we've differentiate both sides with respect to X. That is so we get everybody Mexico too. We'll be getting we plus extremes of dvds. I have used the product rule of differentiation to differentiate why call two weeks. So let's call this equation as uh equation to what he said, this is uh alternative for the very body X. Which means I can substitute this uh expression into. Do you have the dx? Let me do that. So on the left side I'll be having B plus X. Times of devi by dx. And on the right side I have y minus X over X plus y. Remember that we are making this substitution. Why call two weeks? So we have to replace all white with this which means are beginning we x minus X over express we expect. So I can factor the expression in the numerator and denominator. I'll be getting X times of v minus one, divided by X times soft one plus sleep, which means nowhere cancel these extremes. And this gets reduced to we minus 1/1 plus B. So let me I don't. The next step. Now I'm subtracting this way from bosses so that I am writing X T V by D X. This is equal to we minus one over me plus one minus we. So let's get this right side part simplified. So basically we have to multiply it this way With the denominator factor we plus one. So when we do that we'll be getting we minus one minus b -3 times a day is minus we square and minus three times of oneness minus B. I'll divided by me plus one. So you can notice that this we and this be negative. We'll get canceled. And finally we are left with I can factor the negative one minus or one plus we square Over We Press one. So let me write on this in the next step. So we have this X D V by dx The sequel to negative of one. Press we square that would have broke it divided by we press one. Now we have to separate the variables that is all the victims has to come on to the website and extends has to go to the right side. Which means if I transpose this to this side, I just have to multiply by its reciprocal. So let me do that. That is we plus one divided by one plus we square times of levi. I keep the devious it is and then I just have to multiply by dx and do it by X. What is that? We already have an active. So I put the neck to over here now we can absorb that. We have successfully separated the variables that is all the way. And V terms on the website as well as all the X and and the dx on the right side. So we can go ahead and integrated. So we put this integration symbol like this and uh to integrate the left side, we're going to write down in two parts. That is we over one plus we squared times of T V. Press one divided by one plus we square times of devi this is equal to minus of uh D X By X. It's just integration of one by X. And let me integrate this part first. So this is basically using the formula that is uh uh you could remember this formula D X by one plus X squared integration of this is equal to turning us off X. Press the integration constant. So this means I can put this earth integration of one by one plus B squared is equal to turn inwards of week. And uh so let me also integrate the said this will be negative of logics plus the constant of integration that is sick And to do this integration part I'm going to use the substitution that is let me make the substitution one plus we squared equal to t take differentials on both sides which means the two V. D V is equal to D. T. We can solve for we tv many times D V. Is equal to D. T. Over to because I can see that I can replace this entire part in terms softy after the substitution which means that this will be VDV is basically deep. Deep divided by two. I put this in front of the integral like this that is a half DT there will be one plus B squared. I can replace it as tea. So we have like this So the next step we could we could integrate the spot. We have one way to Decoration of one Brady's logo. Absolutely lofty. So in place of T I'm going to replace one plus B squared. Let's turn inwards of we the call to minus of. I forgot to put the small X. No go Absolute value of X-plus six. Now let me add this term to the side. Alright transports this to the safe. So then it will become uh this factor this half. I'm going to raise it to the power of this one plus the script so that I can combine the algorithm. So this will be one plus we square square plus no golf, absolute value of eggs. Yes. Turn inwards of we this is equal to see and we had to do the back substitution for we. So first let me combine these two logarithms in the log properties log a plus lobbies, log tv. So therefore this will become one plus B squared times of square times of X. Let's turn in rules of we this is equal to see. Yeah. So therefore now we'll do the back situation for we we made the substation Y equal to be X. This implies we equal to Y by X. So you can do the back separation. So therefore this will become longer one place Y squared by X squared. Always good times of X. Everything under the algorithm. Yes. This will become turn inwards off. Why buy X? He called to see Now let's get this part simplified. So this is uh equal into this will be if I simplify this inside part, I'll be getting X squared plus Y squared by X square. We already have this X multiplied in the numerator. So you can reduce this one of the excesses will get reduced to one power. And finally we have this stop. Let's turn inwards off. Why buy X And called to see? So let me clean this up so I can write down this as log off X squared. Press y squared by X less. Turn in Warsaw, why buy X a call to see? So this is basically the general solution of the given differential equation. I'm sorry, it looks like I have made a mistake. We were here. Let's get this corrected as you can see here when you raise it to the power of veto this in fact will become square root. So let me clean this up and we do this from this spot. So this is not correct. So this implies the next step, this will become log of one plus we square when you raise it to the power of on by two, it is equal to the square root. So I put a square root symbol, Press certain inverse of we plus log see I'm sorry log x is equal to see. So let me do the back substitution for Robbie when I do the back separation. I'm going to simplify this to uh so we know that these two logarithms we can come back. So therefore this will become log up one plus we squared times of X. Okay, press turn inwards off Y by X is equal to see now I do the back substitution for we So let let me simplify this over here. This is one plus Y squared by X square. Everything under the square root multiplied by X. So this is equal to X squared plus Y squared under the root divided by square root of X squared, which is basically X. So therefore these to access will get cancelled. So we we just have a log off X squared plus y square under the root then plus turn inwards off. Why buy X. The call to see? So this is basically the general solution of the Cuban differential equation. So now let's mourn to the last step that is. We find a particular solution. The single given conditions. Let's go back and review the conditions. That is why I called one when x equal to one able to substitute these values into this equation so that it will help us to find the value of C. So we have y equal to one And x equal to one, which means if I do the substitution into that equation, this will be logged off. One squared plus one squared is two. So that this basically route to plus turn inwards off. Why is one and X six months? So therefore one way one is one. So it is certain in words of one To an inverse of one is basically buy baby food. So can replace this has they before equal to see which means we now found the value of C. This is the value of C. So all that we have to do is we substitute the value of C into this general equation so they're full. I can relate this. This is we have a logo X squared plus Y squared the route plus turn inwards off, turning words of labor X. Yes. The call to Log off route two place by before. So this is the particular solution of the given differential equation.

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