Tutor Belloᵈʸᵈˣ is live

[00:00:01]= 2.

[00:00:13]So you want to differentiate y^ and y functions of x.

[00:00:24]So the z x with respect to X. We have Y square as Y - 2 1 Y square is same as Y - 2. So this is going up. So we have something like Y - 2 Y X + 1 X. If this comes up, if this comes up, we have this one 1 - 2.

[00:00:57]Do we get that idea that = x y?

[00:01:04]So we have a function like this. Now the next thing you going to look for is that z = y^ - this is our z. So we can basically z here.

[00:01:19]Now x. So whatever is in front of this which is minus multiply this equation with y.

[00:01:29]Whatever is in front of this which is minus in this case it is minus that is in front of you multiply it by you multiply this equation through by that >> it's getting like what I just want to do is that I want to make sure this this gate is here.

[00:02:00]I want to make the Z X here.

[00:02:07]But it's only this part that I have here. It makes this minus.

[00:02:11]Do you get? It makes only this minus to make this complete.

[00:02:18]So I multiply everything through by minus. It does not change the equation at all.

[00:02:24]Now look at this. - y - 2 dy dx - 1 / x y prime y - 1 sorry = you get what I did there.

[00:02:39]So now I have the z this my equals to this.

[00:02:46]So I can now write this in terms of z prime - 1 / x = - x.

[00:03:00]So let's understand integral of - 1 x which is e power x - x - 2 - x - 2. Let me write it out so you don't confuse. This is - x -1. So I'm multip= 1 - 1.

[00:03:58]Am I correct?

[00:04:00]This x this one 1 x^ 1 - 1 0 x so I have this now x -1 z prime z prime same as dx so we said after using integrating factor this side is going to become one just pick your z remove this from That's how it is. You have to compare something like trigonometric function. You have to integrating we have - x + right. So it means = multip = - x I have z = cx - x² this same as c - x so I have z = cx - x² so I now bring in my initial z = y - 1 Bring it back and say y - = c x - x². So y = cx - x² - 1 and that's all. So we have found the solution to the first equation is only not.

[00:05:36]So the first step is just to make sure it's in the in this exact order so that you can know what your end is going to be. So after find as Z here is 2. So our Z is going to become Y^ 2. So our Z = Y 1 - 2 which is Y - 1.

[00:06:02]Then now find Z dx dz= to differentus y - 2 y prime you are differentiating.

[00:06:15]So if you compare what is in front of this one you see that minus is missing.

[00:06:19]So this minus that you use to multiply.

[00:06:22]Once you multiply by minus you are going to see your prime you going to see your z. So you just transform everything.

[00:06:30]Everything will be transformed into into this kind of equation integrating and that's I go to second and I Yeah. Yeah, I think so.

[00:07:32]You want someone?

[00:07:49]>> Okay. One more question.

[00:07:56]Can we see the board?

[00:07:58]Those who are watching online can't see the I don't think All right.

[00:08:33]Now look at this second one.

[00:08:36]It's not in ben form yet.

[00:08:39]Not in ben form yet. So the first going to do is just make sure y prime + x² y = y^ 4 is standing by - x. So we going to have - x - x - x. So this is gone. So we have yus this 1x I mean 1 x y = y^ 4 - cos x I can also bring it up as x - 4 = 4. So z = y^ 1 - x this subition we are using. So the same thing = y 1 - 4 and that is y - 3. So this is the subition we are using for z = y^ 3. Now let's differentiate dx equals to differentiating this we have - 3 y - 4 d y >> dx are >> so now let's go back to our initial equation so what do we do the first divide by y which is y 4 so we have y - 4 right I'm dividing by y^ y prime - 1 /x y y y y y y y y y y y y y y y y y y y y y y is already here so this one so if I remove from it so it's going to remain - 3 going to remain - 3 = - y so I can bring this one up and say - x - 3 x please you understand now looking at this you can see this y - 4 y prime y - 4 y prime are we >> the only thing that is missing is - 3.

[00:11:01]>> Do you get what I'm saying now? The only thing that is missing is - 3. So we multiply everything by - 3. So we're going to have - 3 y - 4 y prime + 3 / x y - 3 = + 3 x - 3 cos x. Do we get it?

[00:11:24]>> I multiply by minus 3. Now this is looking exactly like Z and we have our Z = Y - 3. I mean, so we are now going to rewrite this as Z prime + 3 / X z = 3X - 3 cos X.

[00:11:50]So what's our factor that we're going to egal dx and this gives us e^ x which is x^ of log can come to you. So you will cancel out x cub.

[00:12:21]So we multiplying by x cube. Now xq z prime + 3 xq / x = 3 x - 3 x - 3 + x. I mean x can turn this one to 2.

[00:12:46]This will cancel out. So we can cancel these two of them out. So we have p= x.

[00:12:53]So we have something that we can now work with. Thank you. So looking at this if we differentiate x get 3x 3. If you differentiate get z prime. So this is a product of whatever. So we can now integrate both sides.

[00:13:13]Integrating this side we have x = integ 3x = = Thank you.

[00:13:41]We now have Z = 3 X - 3 sin X + C X right.

[00:13:54]So this our z= this z = y = 3 x - 3 sin x + c x - 3 y = 3x - 3 sin x + c x - 3 and that's it.

[00:14:53]Is it >> when you come to the other side?

[00:15:11]>> Sorry.

[00:15:29]So this is one sorry this will be one I'm coming to this one if you don't I know that's Hello.

[00:16:27]>> We are starting second order. Are we still there?

[00:16:46]So I think those >> all right all right please share the link with others on the group so that they can see I don't want to leave the page to avoid um transmission.

[00:17:28]You write in just small zoom.

[00:17:36]I share the link. We are starting second now.

[00:18:07]We have this form. The first thing that comes to your mind is characteristic solution.

[00:18:16]I'm going to write down the equation for this one.

[00:18:21]Equation is that you are going to use um the substitution e^x.

[00:18:43]So this is the substitution that we use = So if y= y prime different rx right y prime second r² e rx now let's solve the first one using this number one we have y prime - 3 y prime + 2 y = our y prime is this. So just write it r² e rx - 3 cos y prime e r x + 2 cos y e r x = 0.

[00:19:32]We cannot divide by e^x.

[00:19:37]So you can say this is r² - 3 r + 2.

[00:19:51]So this is what we call our characteristic solution. Our characteristic equation. So we equation solutions. No, now this is - >> r² - 2 r - r + 2 = it = 0. So this means r = 1 or r = 2. So this are the roots of our so we can say y = e^x is either one or two. So let's plug it in. We can say y = e^ 1x or e^ 2x. So y1 = v^ x and our y2 = v^ 2x.

[00:20:58]Since both of them are solutions to this, since both of them are solutions, we can add them together and see our y = e^x + e^ 2x but most c.

[00:21:15]So this solution. So this is solution to this to this number equation.

[00:21:23]Just use this y = it and The val of this house matter a lot.

[00:21:53]Matter a lot.

[00:21:56]The best the best scenario is for us to have two different two different = this first.

[00:22:20]So the second will now become Y2. This is Y1. Y2 will not become C X^ 4 X if they are three this is one if it is repeated you know this is a quadratic you get r= 4= 4 so if it is a cubic and we have four in three places so you going to do y3 = c3 x² square x^ 4. If equation y = c4 x 3 x^ 4x2 y 5 c5 x^ 4 just like that just continue.

[00:23:18]So if it is just one single we have complex if we have complex we should be able to we should be able to encounter one of that.

[00:23:36]Can I change this please?

[00:23:53]questions.

[00:23:55]Maybe contact Can we start?

[00:24:22]>> Yes. Okay. So the second one we have y prime - 4 y prime + y = 0. So I can write this as r² r - 4 r + 5 = 0. I don't want to jump to the you can easily write this second r² this is the first so we have r² - 4 r + 5 = satisfy is going to give us let's just quadratic straight away. So formula says roo<unk> b 2 b square is 16 - 4 a c - 4 a c is 5 that's - 20 4 * 5 is 20 all 2 a is 1. to this two. So it means = 4 + 2 = 4 plus or minus 2 - 1.

[00:25:54]So it means our y = e^ 2 + i x.

[00:26:03]We basically don't need this plus or minus. All we need is our plus. We can use minus but still.

[00:26:15]So let's multiply by.

[00:26:18]So this e^ 2x + i right?

[00:26:24]E^ 2x + i. So what are we going to do?

[00:26:29]This is indices.

[00:26:31]We add indices. I can separate this and say e^ 2x e^ i.

[00:26:40]Now according to you said power i= what sin x i cos x.

[00:26:53]>> Okay sin x cos x i sin x. So this is the constant.

[00:27:01]So we take it as e^ i in what was it called in o we take it as a x = a cos x + b sin x. So this is the first solution. It means y1 = c1 cos x and y2 = c2 sin x both of them satisfy this equation.

[00:27:27]So we now go back to this and say this= = e^ 2x into bracket e^ c1 cos x + c2 sin x and That's the solution for that.

[00:28:17]It was factorial.

[00:28:28]So this one is something I can use. Look at what we going to get.

[00:28:38]= b² a 4 a c. So 4 a 2 a = 3 + 8 - 1 is 1 8 - 9 - 8 is 1 square root of this is 1 / 2. So this become 3 + 1 4 - 1 2 4 / 2 or 1. So I can easily use factor method.

[00:29:15]So it gives r= 2 or = 1 which is the same thing the same thing quadratic you can solve quadratic using >> are we following any issue?

[00:29:34]We always divided It is for >> Yes. Yes.

[00:30:15]>> Why? Why prank? Why prank? Why?

[00:30:24]Sorry about that. Y= Y which is the first order.

[00:30:36]>> So why do we put back?

[00:30:44]There's no >> order = r= r y= r.

[00:31:16]So the third one, can we go for the third one?

[00:31:26]Can I just >> We have slept.

[00:31:52]Where's my data?

[00:31:55]>> Okay.

[00:32:06]>> This is my You can finish it now.

[00:32:30]>> Wait. Okay.

[00:32:45]>> Is What is >> I know I know they all right let's do the third y prime y prime 16 y= something like r² - 8 r + 16 >> so this um characteristic equation so - 4 now can you look at this now can you see this -4 - 4 - 8 - 4 * -4 16 so this is -= so it means = something like this now = 4 or four you get = 4 so we can = 4 x so my y1 will not be c1 e^ 4 X and my Y2 will not be C2 X^ 4 X.

[00:34:23]So my total c my final my final y my final solution c1 e^ 4x + c2 x we can usually we usually factoriize this and say y = e^ 4x c1 + c2 x so that's number to four Yes.

[00:34:53]R= four twice at the back then we can now write our How is going to do the He's after There are only two parts difference equations equations vectors part of probably want to record.

[00:36:44]>> We send it before tomorrow.

[00:36:48]>> So, after this one, we do those that are not in Jesus.

[00:37:02]So that's me run for varation of parameters they call varation of parameters but I I don't take them as single because there are like three methods you can use to solve three different methods to solve it's not just Okay.

[00:37:43]Yeah. So, we have 2 + 10 y = 2 - 2 + 10.

[00:38:05]= 2 plus or minus I mean square root of b square - 4 aus 2 + 36 - 36 / 2 so this will give us 2 + 6 I over 2. So we have = 1 + 3 I. So y = e^ 1 + 3 i x.

[00:38:42]So this is same thing as e^ x + 3 i x.

[00:38:47]So it means y = e^ x into bracket. Now 3 i is the same thing as what?

[00:38:54]a cos 3x + v sin 3 x and that's final answer 3x that's what we use for you understand so this the final one for homogeneous second order then there is one there is one whenever you have an o that y is not inside The transformation we use let's say does not have y is missing. So we going to use a transformation b = dy dx are y is missing. So this b is a function of x.

[00:39:44]If x is missing you still use v= dx but this time around is a function of y.

[00:39:56]If you have any question or so. Now look at this one. If is a function of y and v = to dy dx. What is what y >> dx?

[00:40:15]We want to differentiate v. But v is a function of y. But we are differentiating with respect to x.

[00:40:24]So since v is a function of y, our square v square will now become v y.

[00:40:35]Please can I use somebody?

[00:40:46]Yes. Exactly. Exactly.

[00:40:52]I can clean this now.

[00:40:56]>> All right, let me clean this sh.

[00:41:07]I cannot come. I follow on YouTube. I don't see you now.

[00:41:19]Tik Tok.

[00:41:30]>> I've already started.

[00:41:37]>> How much does live on.

[00:42:11]I also have what we call super rather 1,000.

[00:42:33]Yes.

[00:42:46]We got all these ones quite well.

[00:42:49]>> They don't go far.

[00:43:04]So looking at it now X is Y is not yes.

[00:43:07]So Y is missing. This one both of them are even missing. So this one just take it as X is missing. This one Y is missing. X is missing. So for that Y is missing. So our v = y dx. But since y is the one missing, we are going to use v is a function of x.

[00:43:30]V is a function of x.

[00:43:34]So it means b = y prime= what? Y.

[00:43:42]I mean just differentiate both of them.

[00:43:51]y is missing in this equation.

[00:43:54]>> So the substitution we use is v = dy dx for both of them v= to y dx but the difference is that anytime y is missing your v is a function of x anytime x is missing your v is a function of y do you get?

[00:44:11]So since v is a function of x for this one dx dv dx equals to second order you get. So this is what we are now going to throw inside this equation.

[00:44:31]Do you get? So now let's do this. So we have x² is dx dx is just like you are reducing the reducing equals to 2 dy dx what v² minus dy dx is vow >> so you have the other >> I'm still going to write.

[00:45:09]So this is everything here is a function of dx.

[00:45:16]>> So I can write this as x dx = 2 v ^ 2 - v / You can transpose or whatever dx = d² minus v. Are we?

[00:45:37]So what else we have here? We integrate both sides.

[00:45:47]x.

[00:45:49]>> So x let me take this to dx.

[00:45:54]Now let's integrate this one.

[00:45:58]So for this one we have something like g2 I can bring my 1. This is V into bracket V minus oneh.

[00:46:17]>> So I can separate this using partial fraction equals >> let me just write this out.

[00:46:29]>> Now let's use partial fraction to separate this. So we have a b minus one b into bracket v a v - a + dv must be equal to 1 / v into bracket vus.

[00:46:50]So everything I've written up must be equal to 1. So this means a + = a and b there's no function of b on topus a = 1us a = 1 here= basically bals we know how to use we know how to use partial function we can We can't do this.

[00:47:35]Should I please a function that is like f let's say this one now v into bracket v minus one which is a product I can sepate it into something like a pick one of product v over the B minus one.

[00:48:07]Now let's join this together. Bracket B.

[00:48:11]We have goten what is here. So what is on top must confform to what is on top here.

[00:48:16]So this is now a into bracket. If this comes out this comes out it must be equal to our is one d must be equal to 1 / brackal.

[00:48:38]So it means open = Okay must be equal to 1.

[00:48:49]Do we get so we function that V every function of V this same as writing plus V there's no V here so it's 0 V. So this I can say a + a plus b a + b= mean a + b = it means a= which means a - 1 + b = a is - 1. So I just put in - 1 b= 0. It means b = what? 1 p. Do we get?

[00:49:44]So b = 1 = 1. So I can now rewrite this integral as 1 instead of writing 1 - 1 / v plus what my b 1 / b - 1 everything is in d integrating this one I can bring my minus I can't bring 1 / 2 I can't bring it out because this one belongs to both of them you get 1 belongs to both of Now let's integrate.

[00:50:18]So this is minus v.

[00:50:25]>> Okay. Yeah.

[00:50:27]>> Yeah.

[00:50:27]>> V - one. So multiply we have - 1 / 2 v + 1 / 2 v= x + I can write this one as >> yeah yeah >> I can write as so as to join them together. A = CX let me multiply by - 2 whatever. Yeahus 2. So I'm going to have if I multiply by minus 2. No let me join it together first.

[00:51:03]>> Yeah let me join it together please. If I join this together I have 1 / 2 into bracket l - one minus l v. Please do do anybody get how it is 1 / 2us. So I factored 1 out.

[00:51:29]>> Yeah. Yeah.

[00:51:32]>> Do we get the same thing?

[00:51:36]>> L C X.

[00:51:39]>> Yeah. Yeah.

[00:51:40]>> Now difference of what was it called?

[00:51:43]difference of >> loging function log when it was return.

[00:51:50]So I have 1 / 2 v - 1 over = c x.

[00:52:06]So this one will not leave here and come here.

[00:52:12]These two can see here and come here.

[00:52:16]>> Yeah.

[00:52:18]>> Do you get what I'm doing? Please >> please do you get what I'm doing?

[00:52:22]>> So if everything here is equal link that means what is inside is equal to what is inside? So it means v = cx.

[00:52:34]So do this 1 - 1 v = what? CX² I'm bringing this to the other side and this to the other side = 1 - CX² I mean what was my V what bx you're not done I'm shifting this >> please do you I go to this one. So it means 1 / b is the n y = 1 - you don't understand. Let me know.

[00:53:22]>> It's not the answer. We looking for one >> - 1 - we integate this side we have dy = x 1 - cx² integral of this but there is a catch here this minus this c is a constant This is a constant. There's two things.

[00:53:54]There are only two things that happen.

[00:53:56]Either this is negative or positive.

[00:53:59]>> If it is negative and we have y = dx + c x² >> or let me just say y is also equal to both of them are valid. If you can both of them attemp this one 1 + >> Yeah. Yeah. Yeah.

[00:54:28]We have to square.

[00:54:53]>> I think alone should be there.

[00:54:59]Yes, you can factor out C from the denominator.

[00:55:05]>> 1 C.

[00:55:15]Like that. Like this.

[00:55:18]Like this.

[00:55:28]1 + one over there's a way to do that.

[00:56:03]So it end with you.

[00:56:23]This one. This one is not without taking I will soon call the live stream now.

[00:57:28]>> You guys are see people are watching now. I'm on my way to >> one person.

[00:57:34]>> Don't tell me >> two people.

[00:57:53]Who is Gibbart?

[00:58:02]That guy said be like one mass lecturer.

[00:58:12]>> Can we finish?

[00:58:12]>> It's not clear again.

[00:58:19]Yeah.

[00:58:34]>> X Yes.

[00:58:50]>> Okay. The video. I'll be sorry.

[00:58:58]>> Is it clear now?

[00:59:00]Oh my god.

[00:59:06]>> But wait, why did they keep on >> badaming?

[00:59:31]exam.

[00:59:41]Okay.

[00:59:55]Depends. It depends. The man's questions are not >> so knowledge.

[01:00:05]>> Yeah. Yeah.

[01:00:23]>> Physics is shaking me like this.

[01:00:29]I don't like I also have 103 level I love vector electromagnetic vector 102 those are my best course inside should have pou some spirit inside my I'm very close my very very close >> or let me say it was my very very close come out.

[01:01:43]>> One person watching.

[01:01:44]>> Okay.

[01:02:02]Zero person watching for I will do live stream I see like 500 people watching >> so please >> the integration of this one is easier so let's do this one first so y = dx we going to this looks like it looks like so what we going to do is that we write this as the x= x². So we say let u = what? Let u = roo<unk> c x which means d u = roo<unk> c dx. Sorry and dx = d u.

[01:03:13]Do you get what I'm doing?

[01:03:16]Please don't root C is a constant 1 + 1 integrating this right standard integral. So I have 1 / roo<unk> act isn't you get what I think as 1 / roo<unk> c.

[01:03:51]Okay.

[01:03:53]So this is the final answer.

[01:03:58]This y1 this the first y that I can get. There are still other y that I can get especially from this one. Let's integrate the second one.

[01:04:09]>> Let's integate the second one.

[01:04:12]>> All right.

[01:04:26]She go chop your rice now.

[01:04:38]>> Should we do it?

[01:04:47]I don't write nonsense for this any >> okay but don't what if you don't write the answer Yes, that's the reason you are sick.

[01:05:15]>> Like you are sick.

[01:05:17]>> Wow.

[01:05:18]>> There is this I borrow >> I thought a second course.

[01:05:24]>> No, no, no.

[01:05:27]There's one structure.

[01:05:36]What is the name of Adam?

[01:06:00]second semester because I thought the second I go instruction test.

[01:06:32]There is another man too.

[01:06:54]Monday.

[01:07:19]>> Looking at this.

[01:07:20]It's just like this. I want the shape of x. If we differentiate is 1 / 1 + x square.

[01:07:39]So what I wanted to do here is make this thing a perfect square. The square of something. So this one is square already make this also in form of a square. that C squ I mean you can see if I open this bracket the same thing as this do you get what I'm saying so I I let my u= to this let= to this thing u = cx bracket so if I differentiate both sides u= this is a constant = dx you get what I'm saying >> d u = dx it's just like saying this is u = to roo<unk> cx dx X if I differentiate this with respect to X is this X will disappear. So C.

[01:08:26]So it means my du = roo<unk> c dx. What I need is to make this x formula. So I can transform this thing from dx to du.

[01:08:36]So my dx = to d u / roo<unk> c. I divide it. So after I do this the next thing I will do is to write everything here in form of in terms of so I have something like integral of dx instead of writing dx dx= root coox= roo<unk> instead of writing dx I write u root c you get what I'm 1 + 1 + this is my u I want to rewrite everything in 1 + u² but since this one is a constant I can bring it out 1 /<unk> c integral of d u 1 + u ^ 2 now this is in form of if we integrate 1 + u square or 1 + anything square when both of them are in the same what's it called This is U.

[01:09:42]This is U. This is actu.

[01:09:44]So I now have 1 / roo<unk> C U plus C. I can now with U is roo<unk> C 1 / roo<unk> C at roo<unk> C X plus that's what I did.

[01:10:07]Now let's do the second one.

[01:10:14]I can clean everything.

[01:10:21]>> Now for the second one this time around we are operating with minus not plus. So this means our um c. So we integrating dx over 1us cx².

[01:10:37]So what do we do? Make it perfect square integral of dx over 1us roo<unk> c x².

[01:10:46]Do you get what I mean? Please do you understand why I'm >> to make this a perfect square?

[01:11:00]To make it a perfect square.

[01:11:04]So what I do after this one? This is one. One is always a perfect square.

[01:11:09]Difference of squares. So I now have integral of dx over 1 -<unk> cx 1 + roo<unk> cial fraction. Thank you.

[01:11:28]have a 1 - roo<unk> c x + b 1 + c x. If I do this, I have 1 -<unk> cx 1 + roo<unk> c.

[01:11:47]>> Let me * everything here. A + c plus b * everything here. B - B roo<unk> C X everything here is equal to what is equal to one everything here is equal to one I don't comparing this is a constant this is constant it is equal to one so a + b= 1 a roo<unk> c - b a c x - b c x = what >> so it means A roo<unk> c x = b<unk>t c x. I can cancel out. So what does that mean? It means a= to a + b= to 1. So I can now say 2 a = 1.

[01:12:42]>> Do you get what I So a + a= 1. 2 a= So it means a= 1.

[01:12:51]>> A same as which means b is also equal to >> 1. Do you get what I'm saying please?

[01:12:56]>> So now let's plug it back please. You all understand?

[01:13:00]>> Yes.

[01:13:01]>> Do you understand this? Can I >> So a is 1 / 2 is also 1.

[01:13:11]>> So I can now put it here. Since we are integrating >> yes >> since we integrating a 1, I can bring it outside. 1 / 2 integral of dx over 1 -<unk> c x plus 1 / 2 integral of b of 1 + roo<unk> c x. Now we can finish this one. Can we finish it?

[01:13:39]>> Yes.

[01:13:40]>> C roo<unk> c. What do we notice?

[01:13:44]>> L.

[01:13:45]>> It is a link. But if you differentiate this one, what are we going to get?

[01:13:49]root.

[01:13:52]So what do we do?

[01:13:53]>> Multiply.

[01:13:54]>> We introduce itus roo<unk> cus roo<unk> c. This one too. C plus<unk> c.

[01:14:05]So this is going to come out and stand here - 2<unk> 6. So this one will stay up. So that if I differentiate everything that is down, it will become what?

[01:14:18]it will become what it is. So I have 1 - 2<unk> c 1 -<unk> c x plus this one. This one will come out here 2<unk> c + 1 / 2<unk> c l 1 + roo<unk> c x if I differentiate this I will get this.

[01:14:44]>> Yes. property of l I can bring I can factor this one and say 1 2<unk> into bracket this oneus this one so that it will become easier l since it is this one minus l this one I can write it as l 1 +<unk> c x 1 x do we get it the final This is the final answer or the second one or this one that has minus. So this is the integral fin y y y y y y y y y y y y y y y y y y y y y = to this for number one that's the answer we can get the two possible answers you can get this and the first one that we did >> I didn't >> which are sir >> number one after exactly >> so that's Please do get what I just do.

[01:16:00]>> It's just long.

[01:16:09]>> All right. So that's so that's number one. So if there is no y is using dy dx y is a function of x Yeah. Yeah. Even when he taught us a new course that I want to experiment on our head.

[01:17:21]Good night. Yo, >> I know they follow up again.

[01:17:41]First one for guy is I need for this. I hope you all know that.

[01:18:25]Okay. For the second one, the square y dx² = d y dx 2 + 2 d y dx. So for this one we said so for this one we say let x be the one that is missing. So it means b is a function of y and it is equal to dy dx is a function of y and= y.

[01:19:01]So which means differentiate both sides it will mean square y dx square y square= what? B D V D Y = V D V Y.

[01:19:18]So we are not going to plug it inside D Y = Y is V² + 2 V I can divide by.

[01:19:31]So which means Y dy = what? V + 2 the V / V + 2 = to Y integrate both sides. So we have something like Y = V + 2.

[01:19:54]So y + 2 = e^ y + c = c - 2. I can turn this one to c I can bring it c^ y. So v is dy dx dy dx dy dx = to c^ y - 2. So we can separate again dy over c e y = this one.

[01:20:54]That was >> I'm looking at what I can do with this 120 g integration.

[01:21:18]>> It's all about integration.

[01:21:35]See guys, this is the score.

[01:21:56]You know what?

[01:22:18]>> No.

[01:22:27]So I can clean this one.

[01:22:37]>> So that's why we're going to enter parameters and we going to use the scan scan method but there's one clarification I'd like to make about this variation of parameter method but the way both of them is drastically different but I'm going follow exactly the way they did that but this run method I don't think they did I still All right.

[01:24:14]>> So, we have only questions.

[01:24:50]Don't worry.

[01:24:59]Or you know what?

[01:25:03]We are going to use only these three questions like we solve three of them using one question. Look for this only works for x^ and sin x.

[01:25:19]So we going to use the three of them to the three of them. So we've also the three of them using this one. Three of them using this one. Three of them using this.

[01:25:31]We start. Shall we start?

[01:25:41]>> So whenever we have a nonomogeneous whenever we have a nonomogeneous the first thing is what? find the particular solution of the compliment solution sorry or the characteristic or whatever of the homogeneous part. So let's do that.

[01:25:59]So this one we say y prime - 5 y prime + 6 y = 0 r 2 - 5 + 6 - not equivalent to zero they are not just part - 2 r - 3 r + 6 = - 2 - 3 - 2 - 2 = 0 = 3 or = 2. So y = 2x or u 3 x isn't it? You get so it means y = the compliment. Don't forget please y our final y final y = y + y please don't forget that so y= to the addition of c1^ 2 x + c2 x y comp >> yes we want to use that first we have not started we not started this one is the first you have to do always Find the compliment solution first. The solution of homogeneous part is it first. You know what?

[01:27:48]>> Yes.

[01:27:54]>> Maybe that's another slide.

[01:28:11]y= to this. Can I >> now undetermined coefficient is that undetermined coicient I prefer to call it initial guess method initial guess method.

[01:28:30]So it's just a guess.

[01:28:39]Now this is x² if we ask if it is equal to x you are going to add a x + turn it into a poltal if it's a pol to pol a x + b x + cic.

[01:29:03]So what we now need to do is find you get this one.

[01:29:16]>> So now let's find the value of a B and C. Ah please let me write this for number one y = what? C12.

[01:29:36]>> So continue with number one.

[01:29:40]Number one, our y our particular y using initial method is ax plus bx + z because this is x². So you start with a x square. If it's zq, you start with a xq.

[01:29:57]x^ 4 plus b x^ 3 you will increase it until you get to the final constant. Now there are two the second this x² so it's going to be a x² + b x + c if it is xq it will be a x + b x + c x + 3 you get. So we are now going to be differentiating until we get so now this method of variation of we are differentiating twice why because this is second order we need a white particular which is a second order so y let's ax right >> plus bx plus b sorry >> y prime what 2 a so we going to substitute into the original equation. Y prime is 2 - 5. What was y >> 2 a x + 6 y? What was y? a x + b x + c = = x².

[01:31:24]Now let's open the brackets. 2 a - 10 a x I mean - 5 b + 6 a x² + 6 b x + 6= to s the first thing you do is look at x square compare compare so it mean 6 a square= to s >> you get what I'm saying >> the ones that have s on this side have square you compare them so it means 6 a A² = A². So I can cancel out that 6 A = A = 1. I got it. Now for P.

[01:32:09]For B, let's compare the ones that have X. The ones that have X. Let's write them out.

[01:32:16]- 10 X.

[01:32:20]Okay. Plus 60X.

[01:32:24]That's all. Let's compare the one that have the constant terms 2 a - 5 a + 16. Now you notice that there's no function of x= there's no conant functional x + 6 b x = 0.

[01:32:52]I can write this one as 2 which is 1 3 - 5 + 6 = 0. Looking at this one, I can write x.

[01:33:04]Do you understand what I'm doing?

[01:33:06]>> So it mean = - 10 = 36.

[01:33:16]>> I've got a binus.

[01:33:21]Oh, it's coming to the other side. Thank you. It's coming to the other side. 106.

[01:33:27]Thank you so much. No, for this one. 1 / 3 - 5 10 36 + 6.

[01:33:37]We don't know = 1 - 50.

[01:33:47]So I can turn this 126.

[01:33:55]So I will join both of them together + 60 = 0.

[01:34:03]So this will give me something like - 38.

[01:34:07]If I join 12 - 60 - 30 + 60 = 0. So I can say 60= bringing this to the other side 36. I can reduce this 2 year 19 2 18. So C = what - 19?

[01:34:33]Oh 19 / 18 * 6. What's 18 * 60 is 60 + 48?

[01:34:39]108. So that's 19 / 108. So that's you have gotten a you have gotten botten c. So our yp = what? y particular is now equal to a x² 1 6 plus bx. B is 306. I can reduce this to 5 and 2 5 8 5 8 plus C. What is C? C is 198 19 * 9.

[01:35:24]>> No 19 19 is a prime number. 19 plus 198 and these are y particular I mean this are y particularly so y for number one the answer is equal to y c + yp so y = what c1 e^ 2x + c2 e^ 3x + 1 6 >> oh my go 18 X plus 198.

[01:36:01]So this is initial method initial method.

[01:36:08]So this is the method of undetermined.

[01:36:22]>> So I start with this After differentiating everything like this, I will now take everything back into my initial equation.

[01:36:34]Initial equation this one. Y prime. Y is now functioning as Y. So everywhere I see YP, I take as Y.

[01:36:43]Do you get? Now starting Y prime, I go to Y prime 2 A. I write instead of this 2 A - Y So you= to this equal to that. So we are not going to compare. Open the bracket. After open the bracket I have 2 a - 10 - 5 + 6 + 6 b + 6 = x². So everything you are left to do now is compare. Which ones have square? Which ones don't have? This is 6 a square. This is square. So it means 6 a = 1. The coefficient of this square is 1. The coefficient of square is 6. So it mean a = 1. Do you get? So here means a = 1 / 6. I've got a Now you are going to establish another equation for the ones in terms of x. This one is for the ones in terms of x. For the ones in term of x, you write them out. - 10 a x + 6 bx there's no time of x here so it is equal to z do you get = 0 so I cannot divide by x and say - 10 = 0 but not I found a 6 >> so I cannot put it inside that's what I did - 10 a 6 you get so it means - 10 + 6 = you I can now bring this to the other side and say 6 = 10.

[01:38:28]Do you get? So I can now isolate and say dividing both sides by 10.

[01:38:36]Do you get >> you got a you got b I can simplify this 5.

[01:38:48]This one is from this one from the third equation. The ones that are linear in constant the ones that are just constant constant. So they are now going to become 2. A is 1 / 6 2 / 6. That's why I got 1.

[01:39:05]You get 1 - 5 - 5 18 + 60 = to zero. you get. So it means 1 - 25 + 60 = Now these are two constant this 18 3 is a is a factor of 18. So I can write this as 1 3 6 18.

[01:39:39]So I can now do 6 - 25 - 9 19 yes - 19 is 19 +als.

[01:39:58]Do you get what I'm doing?

[01:40:02]So it means 60= to come to that side 18= this function by 6. So that's 19 / 18 * 60 + 48. So it means what? 198.

[01:40:21]So write you have no you have no you have no C.

[01:40:26]So, but note please don't Okay.

[01:40:46]So much.

[01:40:55]>> So should we use number one?

[01:41:03]We are now going to look at which one is the best method.

[01:41:08]Which one is the best method?

[01:41:34]Is this thing showing?

[01:41:38]>> Is that showing?

[01:41:55]Yeah, I think the way you arrange it >> everything So this is what I basically did.

[01:42:29]Something like this. First aftering everything first write out the correlate with x². This is the one part that is there. So write it out as 6 a= x. Now let's go down the x - 10 a x on that x + 6 b xal.

[01:42:56]There's no function of x= 0.

[01:43:00]>> So I can divide this by x and say - 10 a + 6 b = I say 6 b = - 10 aent of x.

[01:43:22]God Jesus Christ.

[01:43:32]>> So that's 10.

[01:43:36]>> Now the ones in terms of um the ones that are just constant. Let me just So we have 2 a conant - 5 b + 6 a. There's no conant here. So it is equal to also equal to z.

[01:43:56]So those are the systems of equations system.

[01:44:21]+ y = 6. And they give you another one like x y + y.

[01:44:47]So can you finish it?

[01:44:56]>> So we areing this as 6 a² = a square cancel 6 a = 1 a = what 1. We have gotten our so using this 10 divide both sides by 6 * 1 / 6 b = 106 which I can simplify to 5.

[01:45:32]So the third one I can now 60 = 0. This 2 a1 2 a 3 1 - 25 + 60 = 0.

[01:45:54]So I can now say so I can now say this one 6 / 18 is same thing 6 - 25 - 19 / 18 + 6 = 0. So 6 = 19 / 18. So basically C = 19 / 18 * 1 / 6. 18 * 8 is 108. So we have C = 198 and that's C.

[01:46:38]Now let's go to variation of parameters using the same question. Varation of parameters using the same question.

[01:46:53]Wait, no.

[01:47:19]This one here is going to final solution.

[01:47:47]This is my wife of my particular wife.

[01:47:58]The next comes M2 is not just based on three parts. Limitation limit is there integration.

[01:49:01]But this can finish one.

[01:49:06]The worst part of this is not integration. Integration is that problem.

[01:49:45]So for does not use anything that relates to let's guess let's do this after which one is the easiest or which one is the fastest.

[01:50:04]I just hope we get the same.

[01:50:15]>> So for number one we have gotten our y1 = what 2 x and our y2 as what? C2 e^ 3x. So var say that it says that let us have a y let us just have a y which is like a compliment solution but this time around DC1 DC2 replace them with k1 e^ 2x + k 2 e^ 3x. So the difference now is that K is not a constant is also a function of X. That's the difference.

[01:51:03]Y= K1 2X + K2 X. So you have to so as not to give yourself problem. We have to establish We have to establish equation K1 prime E^ 2X plus K2 prime E^ 3X = Z must establish this in essence K1 prime Y1 + K2 prime Y2= Or this one is for second order. For second order we are doing like this. K1 prime Y1 + K2 prime Y2 = 0. Then K1 prime Y1 prime plus K2 prime Y2= to get our Y.

[01:52:09]But if it is a third order equation we are going to yes if it's a third order this will be equal to also.

[01:52:37]So can we continue?

[01:52:40]Y = K1 2X + K2 X. K is a function of X. So we have to establish this equation.

[01:52:51]Establish this equation.

[01:52:55]Now let's do for the first Y1 Y prime.

[01:53:01]What's Y prime now? K1 prime. Don't forget K is a function of X. So we differentiating K1 it will be K1 prime K1 2X + 2 K1 E^ 2X. Do you get how I differentiated this product?

[01:53:19]>> Okay.

[01:53:21]>> Yes.

[01:53:22]>> Yes. Please do you get what I did?

[01:53:25]>> Differentiation. Product room. You differentiate first.

[01:53:29]>> Yes. Thank you.

[01:53:38]>> Yes.

[01:53:39]>> Yes. True. Please. Do you get what I did? Now for this first part. Now let's do the second part. The second part will now become plus k2 prime e^ 3x + 3 k2x.

[01:53:58]Now don't forget this one that we have established right y prime 2 k1^ 2x + 3 k2 do you get what I did as just let me explain Again we are using our 2C our compliment solution that we got from this one when we made it equal to Z. So we have gotten it like this. So we are going to instead of writing C1 plus this one we are going to change C1 to K1.

[01:54:45]The difference is that C1 is a constant.

[01:54:46]It's an arbitrary constant is anything like 2 5. You constant it is equal to Z.

[01:54:52]But this K is a function of X.

[01:54:55]K is a function of X. So you notal is equal to another function of x.

[01:55:02]Do you get? So we have to establish this equation. Please don't forget just differentiate the two of them and make it equal to zero. Just write prime prime and make it equal to z please. So this one is very important. So after we have done that you're not going to differentiate y prime. What we need is y prime y prime. So we can use this equation again. Do you get? So we can use this equation again. So y prime different y prime is now what I got here. y prime differentiating I have k1 prime e^ 2x using product rule plus 2 you know differentiate 2x you get two will come down 2 e 2x.

[01:55:50]So 2 x is what give me 2 k1 e^ 2 x. If we differentiate this one k2 prime e^ 3x k2 prime e^ 3x plus this one k2 e^ 3 x do you get? So we now have a system of equation here but not forgetting the position that we made. This one is equal to Z is not compulsion you make it but to make work easier for you know it's composing method now looking for ABC.

[01:56:29]>> So we should make it so one prime this and this is equal to Z. Do we get this?

[01:56:38]Yes.

[01:56:40]>> So if it is equal to I can write my y prime = 2k1 e^ 2 k2 you get please.

[01:56:51]>> This one is zero. So I try this out please you get different again because we in the second order y prime equals to first differentiate of k 2 k1 e^ 2x prime then leave alone e^ 2x you get two come down with the one that's already there that's 4 + 4 k1 e^ 2 x do you get so I've differentiated this one again plus now let's differentiate this one 3 k2 prime e^ 3x right >> yes >> then if you differentiate 3x again three will come down with the three that is already there that's + 9 k2 e^ 3 x k prime e^ 2x plus k2 prime e^ 3x since this one is 2. We can't say equal to z like that. So what are we going to do?

[01:58:11]We are just going to put everything inside our equation.

[01:58:16]everything inside.

[01:58:18]I'm coming. You are going to So, I can do this, right?

[01:58:27]I can do this.

[01:58:36]finish you know how to generate from a given constant from a given function let's say I give something like y = c1 a 2x + c2 x 2x + c3x this will be a third order equation let me This second order equation the number of arbitrary constants in any function they give you like this is the order is number of times and the order that you get. So this one you differentiate it twice and you get the second order.

[01:59:25]Can you do it?

[01:59:30]Y prime = what? C1 E^ X + 2 C2 E^ 2 X Y prime because there C1 U X right plus what C2 X so the point is about eliminating constants subract so it means y - y = this 2 c2 c2 e^ 2 x.

[02:00:16]So look at looking at if I if I subtract let me see the if I subtract y 2 and this y prime - y1 = this will go so this is two c 2x >> which is twice of this you get what I'm saying >> which is twice of this that means y prime - y prime = 2 y1 y prime - y prime = 2 y prime - 2 y bring together what's y prime this comes become minus so thatus this one comes here so this is So that's >> Yeah, you can use the operator.

[02:01:29]>> Yeah, you can use the operator of this one.

[02:01:32]>> This one is dus.

[02:01:35]So open the bracket square - 2 - 3 + 2 y. So y this is chapter this chapter two.

[02:02:02]Yeah this is chapter two.

[02:02:08]So please do we understand the way I've been doing this look like me of the simple now this one you subtracting 5 y from let me write everything y prime is what 2k1 prime e^ 2 + 4 k1 U 2x + 3 k 2 prime e^ 3x + 9 k^ 3x.

[02:02:54]So what we are going to do next is subtract this y prime subtract 5 y where our y this y this one five of it subtract it from this so that will be minus 5 5 * 2 10 - 10 k1 e^ 2 x right - 5 * 3 is 15 - 15 k2 to power 3x.

[02:03:29]So what next? We going to add 6 y plus 6 k1^ 2x + 6 k2 x. You understand how we did this?

[02:03:44]So what are we going to do next? Compare this K1 this K1 this 10 - 10 for 3 K2 K2 K2 15US 15 cancel it's canceling out so it means 2 K1 2X there's a prime here plus 3 K2 prime u what else remains everything else has gone is equal to >> is equal to s² now don't forget the initial this one that we made k1 prime >> u^ 2x + k2 prime u^ 3x is equal to z so we can now solve this together so this very so I'm now going to solve this together and say k1 prime e^ 2x = what - k2 prime u^ k x which means k1 prime or let me just put it inside.

[02:05:11]>> So if I put this inside here I'm going to - 2 k1 I get - k2x + 3 k2 e^ 3x right = x²us this it >> so it remains k2 prime e^ 3x = x² which is k2 prime = x² - Do you get what I did?

[02:05:45]So we process 3x.

[02:05:55]So this is integration by part.

[02:05:58]So we do integration by partation by part.

[02:06:10]Do you even understand what?

[02:06:12]It's too long.

[02:06:16]It's too long for you.

[02:06:28]>> There's one method they call it. I don't like it.

[02:06:33]No, it's for integrating this one.

[02:06:35]Whenever you have a product and you want to integrate the product of two functions that are not this is an product, this is an exponential product.

[02:06:42]We have the N I A C E.

[02:06:49]>> You remember this inverse trigonometric and exponential function. So that's the other.

[02:06:57]So is your this is your exponential.

[02:07:04]Yes. U by integral of UV = U minus integral that's formula of integation by part but instead of by part some people use a method called the method I me I differentiate integrate so we have the function is s² x² e - 3x you differentiate this you get z 2x 2 you integrate this you get whatever e - 3x - e - 3x / 3 + 1 9 e - 3x and the last - 1 27 e - 3 e x that's integration + 1 yes exactly so they now start multiplying through diagonally but it does not always yield correct answer maybe because I don't method So let's integrate this at least let's integate this together we are not going to use so we use for >> no I want to get my final answer I want >> this one after differentiating this you understand so I proceed to this This y prime I wrote it this one y prime everything I wrote it 5 y prime minus 5 of y prime that I initially did what was the y prime that's 2k >> this one >> so this so 2k y 2k 1 so I wrote the two five of it that means time 10 * 5 5 * 2 is 10 >> you subtract raing - 10 of this - 15 of this I wrote - 10 - 15 do you get it?

[02:09:39]+ 6 y. Y is k1 k1 2x + k2 3 x that's y. So + 6 of this y + 6 k1 of this k2 of this. Do you get this y1 2 + k2?

[02:10:03]Do you get? So you started looking at it. K1 K1 and K1. If you add these two together, that's 4 - 10. 6 + 4 10 10 - 10. You cancel everything out. 6 + 9 15 - 15 cancel everything out. So everything is collapse into 2 K1 + 3 K2.

[02:10:26]Wait.

[02:10:30]Okay. Yeah, everything is correct. Can I continue?

[02:10:36]So what we want to do now is to integrate this to find our K2.

[02:10:48]So I want to integrate this to get my K2 K2 equals to when it use this formula integral of UV= U minus integral of VU whatever. So here we are using L I A.

[02:11:07]So our is our U. The one that comes first is the one that comes next. U DV is my U. This is my DV.

[02:11:18]U= what? X². So it means d u = 2x dx.

[02:11:25]My dv.

[02:11:29]So my v= e - 3 Yes. So this my this my So I've listed everything out.

[02:11:49]The first thing you need to know is your which one is your is the one that comes first. Whatever this one comes first. So this is my u this is my v everything here is my d with this dx everything is my so u = x² = >> you know I got you now >> I differentiated 2x dx so my dv = to my dv = to e - 3x dx I will now integrate both sides which means my p = to integrating this one we A - 3 X - 3 is coming down which means it must meetus and three on ground which will cancel it out.

[02:12:43]If it's coming the minus coming down must meet minus which turn into plus and it must meet 3 is one which cancel it out. That's how I integrated it. Do you get? So I have my u I have my du I have my v I have my d. So now let's integrate it.

[02:13:01]U V U what's my u e^ - 3x - 3us integral of v du what v 1 / 3 minus is there so I can bring it out of v - 3x d u d u is 2x 2 can come out x can come out dx integration.

[02:13:31]Yes, actually there was one that we try try. So you will get to a point where you have to represent this one by I especially using trigonometric trigonometric returning themselves. You differentiate cos you get sin x cos give you know that kind of return. So you have to replace it with i. So just bring just one crazy like that.

[02:14:00]So k2 = x² - 3x - 3 + 2. Now I want to integrate this one. My u is x u = x. So it mean = dx.

[02:14:18]My d is e minus dv = to v - 3x dx which means my v= - 3x - 3 I can write it 2 / 3 u what my u x vus 3us integral of v is - 1 so this becomes 1 e - 3x dx we just so it means k2 equals don't exceed the answer k2 = x² e - 3x - 3 + you so 2 this 2 x e - 3x x / 9. This one is coming here + 2 / 9. If we integrate e - 3x, you get e - 3x - 3. So this - 3 will turn this one to minus. This one can come out and turn this one to 27. You get what I did this.

[02:15:38]You know what? Let me just 27. So this is my K2.

[02:15:44]>> This is my K2. But I'm feeling for >> Koditra.

[02:16:10]How much?

[02:16:13]>> Let's use it. We areing K1. We need to get K2. We are going to go through the same process to get K2. After getting K2, we are now going to write it as Y = K1 2X + K2X.

[02:16:30]So this is K2. Now look at this. Look at this. K2= we going to multiply K2 by 3X. So I'm now going to get k2x.

[02:16:42]Let me write k2x = iip.

[02:16:48]This one will go away. So I have x^2 - 3.

[02:16:54]This one will go away - 2x 9.

[02:16:58]This one will go away. - 27 - 27.

[02:17:06]Can you see?

[02:17:09]That's it.

[02:17:11]So we get we supposed to get this back.

[02:17:17]We finished or just jump.

[02:17:29]Now look at this scan method.

[02:17:32]Look at scan. So we can just do it on your own and just finish it up.

[02:17:44]>> Now look at this scan.

[02:17:49]If you can memorize, you can click.

[02:18:02]>> Now the meaning of rosan is the meaning of rosan is rosian is the determinant of y1 y2 y1 prime y2 prime. Do you get how it is?

[02:18:17]You just write y1 and y2 then different it for a two dimens just two. So you just write y1 y2 but if it is longer than that you can have y3 but it is always going to be a square matrix.

[02:18:38]If this is three this must be three. So we now have y1 double prime y2 prime y3 prime y3 prime. So now find so this one is just a 2x2 matrix.

[02:18:59]Do you get how I did this one? So since this is just a 2x2 matrix, it means the determinant of this matrix determinant is this time this minus this time this.

[02:19:08]So it means w = what? y1 y2 prime minus y2 y1 prime. That's our room.

[02:19:19]Do you get it?

[02:19:21]So the room scan for this number one.

[02:19:23]This is our y1. y1 = what? C1 u 2x. You don't need y2 = 3 x.

[02:19:36]So now let's find w= y1 e^ 2x dot different 3 e^ 3xus what's y2 e^ 3x do differentiate the first one 2 e^ 2 x so what do we get 3 e^ 5x >> - 2 e^ 5 X what 5X= 5. So you not y particular our y our answer straight away equals to - y1 integral of y2 + y2 integral of y1 What is the shortcut for >> a shortcut for you What I said is the So please just plug everything inside.

[02:21:54]No, >> don't forget this. I already function that >> the function that makes this equation >> y1 y1 y1 2x y^ 3x my x² X plus what was my Y2? E^ 3X integral of what was my Y1? E^ 2X^ 5 X.

[02:22:45]So this one will now simplify to y = e x² e - 2x dx + e^ 3x integral of x e x. So these are the two things that will be our part.

[02:23:19]Sorry for There's no other.

[02:23:46]>> Is there any other system?

[02:24:06]system.

[02:24:23]>> Calm down.

[02:24:27]I did at least 40.

[02:24:32]>> Yeah, maybe 40.

[02:24:46]is kind of that one is for those. That one is for those.

[02:25:02]Yes. If you did what they did as you can finish it off.

[02:25:16]>> Let's finish it off.

[02:25:18]>> So what? So you have done the three of them.

[02:25:22]>> And which one is the best?

[02:25:27]>> Actually this is the best method. This is the most beginner friendly method >> until you reach this particial functions and trigonometric functions like cos x and sin x is only these two So what we use for this one is scan is the best for this is best to use.

[02:26:12]So let me just clean this up. I can go on your phone.

[02:26:39]So integrating x² - 3x e - 2x dx.

[02:26:59]So what's my unit? x².

[02:27:02]My vx - 2us integral of v is e - 2x + 1 / 2 x² v du is going to be 2 x dx I just want to can you integate by please can you using this Wait.

[02:27:42]I saw direction.

[02:27:58]>> 70 questions.

[02:28:06]So using partial integration integration there's partial integration integration by part is once you have these two parts like this two what's it called two functions together like this so you going to make one your following this method following this this one this Yeah, you have like you don't really need to but to be causing problem why instead of this one to be reducing this one we integrating this one it will just be increasing instead of be increasing want to reduce our problem we still must follow this pattern my u = x² which means my du = 2x dx x my dv = to e - 2x dx. So which means I'll integrate both sides and say my v = e - 2x over now - it must mean something to counter it so that everything can come back to normal which is 2. The minus comes to cancel the plus the two come down each other so it's still normal. Do you get what I said?

[02:29:26]You want to integrate e - 5x. You want to integrate dx. So this will be e - 5x - 5 is coming down. So it must or you can use what's called the u method. You can use substitution method.

[02:29:48]Let coffee 5x want to integrate e^xation.

[02:30:08]>> So let's continue. So we are using u minus v.

[02:30:13]So integral is equal to u v. What my u x^2 What's my v e - 2x - 2us integral of v du what's my v e - 2x integration I can bring the constant out so this will turn to + 1 / 2 now what's my v du this my v what's my du 2x that means I can write x and bring two after so this will cancel each other you Yes.

[02:30:46]>> Yes.

[02:30:48]This is still a product.

[02:30:50]I use my part again. So this one is let's integrate this one now. So integrating this one. We now have this u u = x du = what? dx. Thal.

[02:31:05]So u = dx. D = v - 2x dx. Which means my v = what?

[02:31:14]Integrate it eiation - 2 just come down. So it become - 2 e - 2 down is differentation. Integration is over exponential function. You want to integate or differentiate exponential function. Let's say we have a no.

[02:31:45]If you differentiate this, what do you get?

[02:31:48]>> If you differentiate this, you get a power.

[02:31:53]If you have a x integrate, you have a power. If you integrate, there's a lot to control.

[02:32:08]Let's finish this. So now let's integrate this one. Now we are going to plus.

[02:32:16]Now what's u u is x x e - x - 2x sorry / 2us integral of v du v e - 2x / 2 >> du v du. What's du dx? So this is sorry -4.

[02:32:43]So minus to cancel it out. Please get what I'm doing. So now it means I have this plus this plus integral of this.

[02:32:56]So it means if I integrate this one after integrating this I have x² e - 2x - 2 plus integration of this which is this x e - 2x / 2 + 1 / 2 into bracket let's integrate now e - 2x - 2 are we so minus this one can come here and this minus and this to do you get what I did so this is my plus we need Let's see.

[02:33:44]>> Yeah, we are doing this eus.

[02:33:52]So cancel this one outus 2x. So u cancel this cancel this cancel this you know u 2x * u - 2x cancel so the answer is now x^ 2 - x 2 + 1 + c e - 2xant integration that's the first integration So now after we do it now together everything so we going to get our Y after Y.

[02:34:54]So answer y are like the same method but there are three different methods inside >> after we are done it's going to collapse everything Yeah.

[02:36:08]>> So I don't know if you'll be able to Everything everything we have It's just not everything.

[02:36:51]>> All right. So, so these are our differential operators.

[02:37:06]These three we call it differential operators.

[02:37:10]Are you please can you hear what I'm saying? This our differential oper is just operating. It is functioning as any other meaning of this is what dx that's meaning you're inside the operator operator techn so it means dy give us what d y do you get if I write d² it means what d² dx² the second. So if I write the square y it means d square y d x².

[02:37:56]Now if I write dus of y what does it mean? Open the brackets d y - y.

[02:38:06]Do you get what I'm doing? So it means this is what dy dxus y. Do you get?

[02:38:14]So that one is just an introduction to now look at this function. Let's say we have something like - y = b x where inverse operator now comes in is that I can say my y my y particular my y particular e^x over you get you see that multiply is just like you operating with normal mathematics Can you see that e^x - 1? They give you an equation. For example, the ones we doing before, you can just use this. So we have gus one um get what I'm saying?

[02:39:14]So anytime you have something like this now you are going to use there are five formulas you have to there are five you have to I don't know this whenever we have something like this 1 / 3 - x is always so that e power x e^ Let's say let's replace one. Do you get what I'm saying? Let's replace it. So power malus x dot.

[02:39:57]So this is the expansion.

[02:40:00]You get what I'm saying? So if I want to solve this equation now so it is equal to this one is same as saying e^ x integral of e^ - x do x dx you get what I now simplify this and say e^ x you see So you integrate this what do you get? x x^ y prime - y = x r² - 1 = 0 = 1 = - 1 C1= X.

[02:41:16]So my this is XY.

[02:41:38]- x y2 = x - x. So my different sheets actually technique.

[02:42:20]So if we are doing let's say we are doing something like e^ x over d + 1. So this one will have something like e power - x integral of e^ x do x dx. Do you get what I'm doing this? So you first neglect the you first like you take the negative of the one that is here e - x e^ x this one is that >> no this is let's say + 2 so I'll write this as e - 2x 2x what was on top do you get what I'm doing now please so we now have e - 2x integral of e^ 3x dx. So this will give me e - 2x. If I integrate this, I have e^ 3x / 3. So if this multip Yeah.

[02:43:47]>> Yes.

[02:43:50]>> Yes.

[02:43:59]>> Let me use another one. Let's see.

[02:44:02]Please. There are a lot of things you have to memorize. This is e^ x over d + um let's say g or square.

[02:44:22]So what do you do for this one? You separate it.

[02:44:25]special.

[02:44:28]So this is x / dus 2 d + 2 right?

[02:44:34]You get what I mean? So we can solve it directly by saying you can solve this directly by saying that we first do 1 / dus 2. So we first do e^ x + 2.

[02:44:51]Do you get do you get what I just said? So we now do this one. So we now say 1 / gus 2. So this one now e^ - 2x integral of e^ 2x 3x dx.

[02:45:08]So that's 1 g - 2 of e - 2x integral of e^ 3x dx which is 1 g - 2 of e - 2x dot this is e^ 3x 3. If you multiply this together same x so this one we now do it again we repeat that method.

[02:45:36]If you don't want this long method, you can use partial fraction.

[02:45:52]What's it called? You can >> Yeah.

[02:46:08]>> Yeah.

[02:46:10]>> Yes. Yes. Should we do it?

[02:46:13]>> Yes.

[02:46:15]So we have a / dus b / + 2. So this is now a d + 2 a + b dus 2 must give us 1.

[02:46:34]So a and d all the functions of d are zero. A D + B = 7 2 A - 2 B = 1 A + B = 0 A = - 2 - 2 B - 2 B = 1 - 4 B = 1 which means B = - 1 4 that means a = 1 / 4. Do you get that?

[02:47:17]>> So I can now put it here and say 1 4 - 2 + b - 1 4 g + 2 everything.

[02:47:34]So I multip and I get 1 4x - 2. So I can you can solve this one.

[02:47:44]You can solve this one either.

[02:47:54]And that's what 610.

[02:48:03]It is 610. Read by yourself.

[02:48:29]Let me write this down please.

[02:48:41]So I'm going to just explain how we use it and maybe I can give an example on each of them. So this is just about I'll just give an example on each of them.

[02:48:51]So the first one you should know technique is 1 / g of x is equal to e^ mx integral of e - mx r x r x dx with respect to x.

[02:49:15]I've given an example thatus.

[02:49:23]So the first one is now the second one of d is a function of is a function like let's say we have square d + 1 y this is a d a function of like it's just like a function of dy you get that's how we write it a function of d is like a function or how just something like this so this one like this is also a function of if you get what I'm saying so I can say g is a function of you get so this is equation you have to px = to This one is used to find used to find the very solution is equal to e^x= the second one the third one of ².

[02:50:46]Of course, dx + q or sin px + 2.

[02:50:57]The same thing as cos x + function.

[02:51:09]It's looking like magic. Maybe another should we do it tomorrow function of d² of u bx and another function of x which can be most ux 1 + So that's transformation.

[02:51:47]I have some questions.

[02:52:01]Please raise his desk on Y= 5.

[02:52:42]This is a function. This is already know how we can open it up. So this y - 3 y + 2 y = 5. So we solve this. So we use operator technique. I y = 5x d² - 3d + 2.

[02:53:13]So for this kind of equation so just compare it. So 5^ this is what I'm saying a function of d a function of t.

[02:53:25]So here your P is your P = 5.

[02:53:30]So P = 5. So what you just do is that anywhere you see D here just input your B which is five. So our Y particular for this one is what? U 5 X 5 which is 25 - 3 * 5 18 + 5 X.

[02:53:51]That's my answer.

[02:53:56]>> This particular >> Why particular particular? So you can find C1 and C2 and that's all + 1.

[02:54:21]= 2x. So looking at something like this, this one we can y = 2x d² + 1.

[02:54:33]Now this one you compare it to something like this. So here your d² you going to transform it to minus p² d² = - 2 that's - y = y = 2x² - 4 + 1 that's 2x - 3 That's our >> just for this one. They give us something like this. You are going You understand?

[02:55:40]>> So if you understand dy d y = u - 2 x + x, right?

[02:55:58]Yes. So the first going to do using this pen.

[02:56:06]So the first thing you going to do is transform it from square to d + so. So anywhere you see g² you going to put dus 2.

[02:56:16]Do you get?

[02:56:19]So this is now so this is d plus d.

[02:56:29]Do you get this square this is the same as this? Let me write this so you can understand y particular = e - 2x + x d.

[02:56:46]So this yp = what?

[02:56:49]E - 2x 1 / I now write this as dracket d². Anywhere I see this square, I put d + p is - 2.

[02:57:03]Anywhere I see this square. Yes, I just don't need to add.

[02:57:10]So our f of x is cos x.

[02:57:14]Our f of x here is cos x. So you now open this one and say we have e - 2x 1 / d² - 2 d + d cos x. So we have e - 2x 1 / 2 - cos x. Now this one now we are not going to use this cos x that we using. So our p is now going to be one.

[02:57:43]So we are now going to use square= square= square= square.

[02:57:54]So we have e - 2x 1 / d² - 1 - d cos x.

[02:58:00]So what are we going to do for this one?

[02:58:04]So this one I can now transpose and I will use conjugate.

[02:58:09]The conjugate of - 1 - 1 + d. So I multiply up and down. So I have e - 2x - 1 + d - 1 + d - 1 - d or cos x. So I can write my cos x here.

[02:58:30]Do you get what I'm doing?

[02:58:32]So this will be equal to e^ - 2x cos x - cos x plus the cos x different cos x which is - sin x - sin x over here if I open this I have + 1 + 1 - d minus d we cancel out - d² + 1 - square. Now don't forget this square is what? - 1. So this square I can rewrite as - 1. So that's - 1. That's + one. So you just finish up like this. So you have e - 2x into brackets - cos x + sin x + 1. So that's not so formulas once you know it.

[02:59:42]>> Thank you sir.

[03:00:13]session.

[03:00:34]This is the only Oh, chin.

[03:01:02]We have come to the end of today's class.

[03:01:06]So you can almost 3 hours 3 hours.

[03:01:21]I think that's okay.

[03:01:38]So this again