Start learning today, and be successful in your academic & professional career. Start Today!

• ## Related Books

### Linear Equations

Linear Equations (PDF)

### Linear Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Lesson Objectives 0:19
• How to Solve Linear Equations 2:54
• Calculate the Integrating Factor
• Changes the Left Side so We Can Integrate Both Sides
• Solving Linear Equations
• Further Notes 6:10
• If P(x) is Negative
• Leave Off the Constant
• The C Is Important When Integrating Both Sides of the Equation
• Example 1 10:29
• Example 2 22:56
• Example 3 36:12
• Example 4 39:24
• Example 5 44:10
• Example 6 56:42

### Transcription: Linear Equations

Hi and welcome to the differential equations lectures here on www.educator.com.0000

My name is Will Murray and I’m very glad that you are going to be joining me, today we are going to start right away with linear equations, that is that first topic of differential equations.0004

There is a special algorithm on that and we are going to learn to solve them, let us jump right in.0015

We are going to learn how to solve linear differential equations, the way you recognize a linear differential equation is that you can put it in this form Y′ of x + p(x) × y(x)=x3.0020

You might be using different variables like you might have a (t) instead of (x), that is ok the methods are exactly the same.0034

I want to make a couple of notes about the form here, the word linear does not necessarily mean what you think it would mean.0042

You can still have lots of functions in here that are not linear, linear means that we think of y and Y′ as the variables, not x and y.0053

But you think of y and Y′ as the variables and you think of the p(x) and the x3 even though they are functions of (x), you think of those as being the coefficients.0061

They might be something like cos(x) or ex or anything like that, but the point is you think of those as being coefficients.0072

If you look at this equation, the linear form is (Y′) + p × y =q, if you think of y and Y′ as the variables and you think of (p) and cubed being coefficients then what you have there is a linear function in terms of Y′ and (y).0079

That is what we mean when we talk about a linear differential equation.0103

If you have something in front of the Y′(x), if you have a coefficient right here, the solution method that we are going to learn in the next slide does not work until you get rid of the coefficient.0109

What you have to do is divide both sides by that coefficient to get rid of it and give your self a nice, clean Y′(x) with no extra coefficient.0123

Make sure to divide any coefficient here in front of the Y′ before we start the algorithm that we are going to learn on the next slide.0133

Divide away any coefficient here first, that is a very common mistake that some of my own students make when I teach them differential equation.0150

I will give them an equation where I have a coefficient here and they will go through and use the algorithm that we are about to learn but it would not work because they forget to divide that coefficient away.0162

Let us start to learn that algorithm, how you solve linear equations is you calculate something called the integrating factor which means you look at the coefficient of the y(x) which is the (p).0172

You look at the p(x) and you integrate it, you take its integral and you raise (e) to that power.0186

So it is e to the integral of p(x) and we call that the integrating factor i(x) and we multiply that by both sides.0194

A lot times this makes the equation get a lot more complicated, let me show you why we want to do that.0202

What we have here on the left hand side, we have i(x) × Y′ and then here we have y × a very large mess but what this mess turns out to be is the derivative of i(x).0210

When you take the derivative of something, the derivative of (e)u, d by d (u) of (e)u is just (e)u × the derivative of (u).0231

Let me just write d of eu is e u, d(u).0247

What we have is that is the eu right there, this is the derivative of that (u) because the derivative of the integral of p(x) is equal to the original p(x).0256

That sounds a little complicated, I think it will make a lot of sense after we jump in to some more examples.0273

Do not worry about it too much right now, the important thing is to notice that we have i(x) × Y′ and (i)prime(x) × Y.0278

If I write it this way (i)Y′ + (i)′ (y), that exactly what we would have gotten if we have taken the derivative of (i) × (y) using the product rule.0290

We are really exploiting the product rule from calculus 1 to solve this kind of differential equation.0300

What that means is you can integrate both sides, when you integrate both sides you can get i(y) on the left and whatever you get on the right.0308

The point is we are going to take it from there and we are going to solve for (y).0318

I think this will make more sense after we do some examples, so if it is still a little fuzzy right now, do not worry because we will work through some examples together and you will definitely get the hang of it.0322

As I said from the previous side, after you integrate both sides, you are going to get i(x) × y(x) × y on the left.0333

Remember I said you will get i(y) on the left and you will get whatever you get by doing this integration on the right.0341

You can then solve for y(x) by diving both sides by this i(x).0350

That is the plan for these differential equations, after we do some examples I think it will be a lot more clear.0357

There are couple of notes that I want to make before we jump in into some examples, let me just remind you of the original form we are solving here.0367

Y′ + p(x), (y) is equal to q(x) that is the form that we are solving.0377

If p(x) is negative, meaning if you have a negative sign right there, you want to make sure to include that when you are finding i(x), that is a very common mistake that my students often make when I’m teaching differential equation.0386

Remember that the integrating factor i(x) is e to the integral of p(x) d(x).0398

If there is a negative sign here, then that is a part of the p(x) and you have to include that in here, but a lot of people do not notice that negative sign and they forget to include it.0408

Trust me that will really ruin all the calculations after that, make sure you include the negative sign.0418

Another related mistake that is very common is resolving this e to the integral.0425

A lot of times when you take the integral you get the natural log or something, my students sometimes see e to the –natural log or something, I wrote cucumber here, e to the -natural log of cucumber.0433

My students will say “oh I know that e and natural log cancel each other out, so e to the- natural log of cucumber is just –cucumber”.0444

That is wrong, let me and let me show you why it is wrong.0453

If you have e to –natural log of cucumber, then that is really the same, remember the rules of natural logs as e to the natural log of cucumber to the -1 power.0456

The rules of natural logs say you take the numbers on the outside and they turn into the exponents.0476

That is e to the natural log of cucumber-1 which is 1/cucumber, which is just 1/cucumber.0481

That is a very common mistake that I see students make all the time, they want to cancel (e) and natural log but they do it so quickly that they do not notice that negative sign is in there.0496

They just try to pull that negative sign outside, it does not work, this is how it works out as 1/cucumber, you got to be really careful about stuff like that.0508

Another note that I want to make is that when you are doing this integral to find the integrating factor, (e) to the integral of p(x), you can leave off the constant there.0518

It does not really matter if you add an arbitrary constant or not, you might as well leave it off, you do not have to worry about the constant there.0526

However, later on in the problem when you are integrating both sides of the equation, the constant is very important, I can not stress this enough, I need to highlight that part right there.0535

The constant is very important when you are integrating both sides of the equation and it is also very important that you do it exactly at the moment when you do the integration.0546

When you do the integration that is when you have to add the constant.0556

The reason you have to be so careful about that is after you do the integration, you usually have to do a couple more steps of algebra to solve for (y).0560

You have to keep track of that constant as you go through those algebraic steps, sometimes that constant gets tangled up in the equation.0570

This is a very big difference from when you first learned how to do integrals back on calculus 2 or even in calculus 1.0578

You could do the whole problem and then just hack on an arbitrary constant at the end.0588

You could even do 5 problems and at the end you just go back and put +c in every single one of them.0591

That does not work in differential equations, you have to add the constant in the middle of the problem when you do the integration step and then keep track of it after that.0598

If you just go back at the end and tack on a constant at the end, you will get the wrong answer.0608

I think you will see in some examples how that plays out but let us just remember to add the constant exactly when we do the integration.0615

Let us jump in with an example now, we are going to find the general solution to the following differential equation Y′ + xy=x3.0630

Let us remember the strategy here, our generic form for a linear differential equation is Y′ + p(xy)=q(x).0639

That is what we have got here, we got Y′ + a function × (y)=another function.0652

The strategy here is to find the integrating factor i(x)=e to the integral of p(x) (dx), that is always our strategy for linear differential equations.0658

In this case, the p(x) is that (x) right there, so that is the (p), and we have to do (e) to the integral (x)(dx) and that is just e to the x2/2, the integral of (x) is x2/2.0674

When you are doing this integration to find i(x), you do not have to worry about the constant here, so no (+c) necessary at this integration.0690

I’m misspelling necessary here, no (+c) necessary at this step, later on there is going to be another step where we are going to do integration.0704

At that point it is going to be absolutely crucial to include the (+c).0715

What we do with this integrating factor is that we multiply it by both sides of the equation, so I’m going to multiply e to the x2/2 × Y′ + I’m going to write this in the middle.0720

(xe) to the x 2 /2 × Y= x3 e to the x 2/2.0735

The whole point of this linear equations and this integrating factor strategy is that this left hand side is supposed to be the derivative of y × i(x) using the product rule, let us look at that.0746

I’m saying that this is the derivative of y × e to the x2/2, that was our i(x), it is supposed to be the derivative of that using the product rule.0761

It is always good to check at this step that it worked, if we took the derivative of (y)e to the x2/2, we get Y′(e to the x2/2).0772

(y) × the derivative of e to the x2/2, which is e to the x2/2 × (x).0783

That right there is i(x), I′(x) and that right there is i(x), we do get something that looks like the aftermath of the product rule.0790

That confirms that we did our work correctly and this is still equal to x3 e to the x2/2.0805

We are going to have to solve this and the way we are going to that is by integrating both sides, if I integrate the left hand side, I have the derivative so I’m going to go (y) e to the x2/2.0816

That is by integrating both sides is equal to the integral of x3 e to the x2/2 (dx).0831

Now I have to solve that integral on the right, I think I have to take another side to do that because it is not the easiest integral in the world.0843

Let me just recap what happened on this side here, we started out with this linear differential equation in this form Y′ + p(xy)=x3.0850

Our strategy for linear differential equation is to form this integrating factor where you do e to the integral of p(x).0861

P(x) is just (x) and its integral is x2/2, we do not have to add a constant for that integral.0869

We multiply that by both sides, so we multiply that by all three terms here.0876

The point is that makes the left hand side into i(x) × Y′ + I′ (x) × (y), we get a product rule going on and that is the derivative of (y) × i(x).0882

If we integrate both sides, on the left hand side we just get y × i(x), on the right hand side we have an integral that we still need to solve.0895

That is what I’m going to carry over on to the next slide and we will keep going there.0903

We still have a couple of steps to go on example 1 here, we had on the previous side y × e to the x2/2 was equal to the integral of x3 e to the x2(dx).0910

We have to solve that integral and what I’m going to do is make a little substitution here, I’m going to do (u)=x2/2 and then my (du) would be equal to 2x/2, just x(dx).0930

I’m going to write this integral as, I think I’m going to take that x3 and write it as x2 and will give me another (x) to put over here with the (dx).0949

This is eu and now x2 is 2 × (u), so I’m going to pull out 2 out here and change that x2 to a (u), eu and x(dx) is (du).0961

I still need to do integration by parts to solve (ue)u, let me set up integration by parts to solve (ue)u.0976

I have a little short hand technique for integration by parts, it is called tabular integration.0986

I thought this on some earlier educator videos for the calculus 2 or calculus BC series.0991

If you go back and look at those calculus BC lectures here on www.educator.com, you will see a section for integration by parts.0998

That is where I thought this section on this trick for tabular integration to do integration by parts quickly.1007

What I’m going to do is take derivates of (u), derivative of (u) is 1 and derivative of 1 is 0, then I’m going to take integrals of eu, that is just eu and eu again.1014

Then I write little diagonal lines here and put alternating signs positive, negative down the diagonal lines.1028

Then I just multiply down the diagonal lines, I still have a 2 on the outside and I have (u)eu - eu, that is still equal to (y)e to the x2/2.1035

I want to really emphasize something right here, this is the step where I did the integration and this is where I have to put a (+c), that is really crucial here, I will put a (+C) right now.1051

Let me really emphasize that, must add (c), we can not just add (c) at the end, so we must add (+c) when we do the integration, not earlier and not later.1064

The reason we have to do it then is because the next step is to solve for (y) and that is going to involve some algebra that will going to change the equation around a little bit.1092

And the (c) is going to be tangled up in with that algebra and we need to let that happen, we can not just tack it on at the very end when we are all done.1102

What we have here is I need to get things back in terms of (x), 2 × (u) is x2/2, e to the 2/2 – e to the x2/2 + (c).1110

I would like to divide both sides by e to the x2/2 because I want to solve for (y) on the left.1133

If I divide both sides by e to the x2/2, I’m just going to get 2 × in the inside here x2/2 – 1 + (c).1140

Now dividing by e to the x2/2 is the same as multiplying by e to the –(x)2/2.1153

I get (y) is equal to, if I distribute that 2 I get x2 - 2 + a constant × e to the –x2/2.1161

That as far as I can take this one, I’m going to stop there, that is called a general solution right there.1173

The general solution always has an arbitrary constant somewhere in it and the way you would find the value of that arbitrary constant will be the U’s initial conditions if you were given an initial condition.1179

In this case, we are not given an initial condition, we can not find the arbitrary constant, so U’s initial condition if given.1191

In this case, we are not given one to find the exact value of (c) and if you do not have an initial condition then you just leave it in terms of the general solution like what I did here.1209

What we really found here is the general solution, that as far as we can take it on this one.1225

The next example we will have an initial condition and you will see how we are going to plug that in the end to find the value of (c) but in the mean time let us go back over what happened with this example.1236

This is the integral that I got from the previous side, this is all work that we did on the previous side, you can check back on the previous side if you do not remember where this all came from.1247

We solved it to the point of getting (ye) to the x2/2 is equal to this integral and this is a slightly messy integral so I made a little substitution.1259

(u)=x2/2, of course I have to make my d(u) as well and then because I have an x3 here, I factor that in to x2 and an (x).1266

The point of that was I want to segregate out an (x) to be my d(u) here, so that x(dx) became the d(u).1278

That x2 is 2 × (u), that gave me a 2 and a (u) here, we have eu.1283

Just to do (ue)u I have to do a little integration by parts problem.1294

I have this tabular integration method that I covered back in my lectures in the calculus BC section here on www.educator.com.1299

You do derivatives of (u), so this are derivatives and integrals of eu, we do integrals.1308

You cross multiply them along with these little alternating signs plus, minus, plus and so we get (ue)u - eu, substitute back in (u) as x2/2.1316

We are trying to solve for (y) that means I have to divide both sides by e to the x2/2.1331

That is what I did here, I cancelled off my e to the x2/2 here and here.1338

Now with the (c), I have to multiply by e to the –x2/2, that is the same as dividing by e to the x2/2.1342

I just distributed this 2, cancelled of the ½ and we got x2 - 2 + (c)e to the –x2/2.1350

At this point, if you had an initial condition, you would use it now to figure out the value of that arbitrary constant.1359

If you do not have an initial condition which is what happened with this problem, you will just stop and you say that is your general solution.1366

In example 2, we have to find the solution to the following initial value problem cos(XY)′ + sin(xy)=cos(x5) sin (x), and the initial condition is y(0)=2.1377

Let me remind you the format for linear differential equation, it has to be in this form (Y)′ + p(xy) is equal to q(x), it got be in that form.1390

In particular, it is not allowed to have anything, any coefficient in front of the Y′, here we do have a coefficient in front of the Y′, we have cos(x).1403

What we are going to do is divide both sides by that cos(x) to get it into the required linear form, we get Y′ and we are going to divide by this.1414

We got to divide both sides here so I get Y′ + sin(x)/cos(xy), now if I have cos5 on the right and I divide by 1 cos, it will go down to cos4(x) sin (x).1427

I’m not going to worry about the initial condition because paradoxically you use the initial condition at the very end of the problem, we would not worry about that for quite a long time.1445

Let us remember our general strategy for linear differential equations which is that you use this integrating factor i(x) is equal to e to the integral of p(x) d(x).1454

In this case, my p(x) is the coefficient of (y) so that is p(x) right there.1470

By the way, you could rewrite this as tan(x) if you wanted, I do not think that is going to be useful for what is coming up.1475

I do not think there is any particular benefit for writing this as tan(x) but it would not be wrong, that is the p(x) right there.1483

What we get here is the integrating factor i(x)= e to the integral of p(x), so e to the integral of sin(x)/cos(x)d(x).1494

In order to integrate that, I think I want to do a little substitution, of course you could remember what the integral of tan(x) is.1508

Certainly that would be great if you did remember it, but I’m not going to assume that you remember it because we are going to solve that just with a little substitution.1515

We are going to use (u)=cos(x) and then d(u) would be –sin(x)d(x), this is (e) to the integral of, there is a negative sign there so I have to put –d(u).1523

d(u) is sin(x) d(x) with a negative, cos(x) is (u), e to the negative integral of d(u)/(u) and that is e to the –natural log of (u), which is e-natural log of u is cos(x).1541

Now a couple of points to mention here, first you do not have to add the arbitrary constant when you are finding the integrating factor.1559

You can omit the (+c) at this point, it is ok, later on it will be very important to add on that arbitrary constant when we do another step of integration later, but here you do not have to worry about it.1567

Next thing to be very careful about is, let me write this in red to show that it is wrong.1580

A lot of my students would look at this and say this is equal to e to the- natural log of cos(x), the e in the natural log is cancel so this is just –cos(x).1587

That is wrong, I want to make it clear that you do not want to do that, instead what you want to do is you write that as e to the natural log of cos(x)-1.1596

Remember that negative becomes an exponent.1614

That is just cos(x)-1 which is 1/cos(x), that is the actual integrating factor, that is also the same sec(x).1617

That is the integrating factor not –cos(x), that would have really stirred you wrong.1637

Let me go back to the original equation here, we have Y′ + I think I’m going to write it as tan(x), tan(xy)=cos4 x sin(x).1642

Remember the point of the integrating factor is you are going to multiply both sides by that integrating factor, so multiply both sides by sec(x).1658

We will get sec(x) Y′ + sec(x) tan (xy) is equal to, now remember sec(x) is 1/cos(x), we are really dividing by a cos(x).1667

We will go down to cos3x sin(x) and the point of that is that the left hand side is exactly what you would have gotten if we have taken the derivative of sec(x) × (y) using the product rule.1684

The product rule says derivative of 1 × the other, the derivative of other × 1 and that would have given us sec(x) × Y′ and then the derivative of sec is sec(tan) × (Y).1708

That is exactly what we have gotten with the product rule, have we taken the derivative of sec(x) × (y).1723

The product rule suppose to work for every linear differential equation that is the whole point of this, this is still equal to cos3x sin(x).1731

We are going to integrate both sides here, and on the left that means we are going to get sec(x) just × y without the derivative because we integrated the derivative, we just get back to the original function.1744

On the right we still have the integral of cos3x sin(x) d(x) and we have to wrestle with that integral.1758

I think I’m going to go ahead and work that out on the next slide, let me remind you what happened on this slide.1769

We had this differential equation here but it is not quite in perfect linear form and the reason it is not in perfect linear form is because of this coefficient in front of the Y′.1777

We are not supposed to have anything in front of the Y′ except for a 1, we have to divide a way by that cos(x) so that is why I divided the cos(x) in front of the Y.1789

I got sin(x)/cos(x) and cos5 went down to cos4 because I divide by cos, we got this new formula in differential equation and that is in linear form.1800

We can use our generic linear strategy which is to use this integrating factor which means you look at the p(x) and you try to integrate it and you raise e to that power.1812

To integrate sin(x)/cos(x), I use a little (u) substitution, (u)=cos, d(u)=-sin, so we got –d(u)/u which gave me –natural log(u).1827

I did not have to add (+c) here, that is going to come later on, the –natural log(u) substitutes back to –natural log of cos(x).1839

It is very tempting at this point and it is something that students often do is to try to cancel the e in the natural log.1849

But it does not work because of that negative in between them, you have got to turn that negative into an exponent first then you can cancel off the (e) in the natural log.1856

You will get 1/cos(x) or sec(x) as your integrating factor.1866

The point of that was you multiply both sides of the equation by that integrating factor that is what I’m doing here.1871

On the left I get sec(x) × Y′ + sec(x) tan(x) × y and the point of that is that it is the derivative of sec(x) × y using the product rule.1877

If you use the product rule and expand this out you get exactly these two complicated terms.1889

On the right hand side when I multiply by sec(x), that is like dividing by cos(x), I got cos3x × sin(x).1896

I’m still stuck with that, I still have to integrate that, that is why we are going to jump over on the next slide and keep going.1904

On the previous side, we got it down to sec(x) × (y) was equal to the integral of cos3x × sin(x), this is all coming from the previous side here (dx).1912

I have to solve this integral on the right, I’m going to use a little substitution now, u=cos(x) and my d(u), we always have to find the d(u) when you do a substitution.1928

D(u) then would have to be –sin(x) d(x) and I have the integral of u3 and that sin(x) should be negative.1938

I will put negative out here, -d(u), that is –u4, using the power rule over 4.1949

I just did my integration, now is when I have to add a constant, it is very important that I add it right now, -u4/4, which is -1/4 × changing (u) back to cos(x), this is cos4x + the constant.1959

This is still equal to sec(x) × (y), now I want to solve for (y), I want to divide sides by sec(x), that is the same as multiplying by cos(x).1983

I will multiply both sides by cos(x) to cancel off my sec and on the left I will just get (y) and on the right I will get -1/4 cos5x + (c) × cos(x).1995

That is why we had to add that (c) is because the algebra that we do afterwards to solve for (y) starts to tangle up the (c) into the equation here.2011

If we did not have an initial condition, we have been done because what we found right here is the general solution.2021

But this time we have an initial condition, y(0)=2 and we are going to use that here, y(0)=2, I’m going to plug in 2 for (y) and 0 for (x) -1/4 cos5 of 0 + (c) × cos(0).2027

Cos(0) is 1, 2 is equal to ¼ + (c) and my (c) if I move that ¼ over the other side, 2 + 4 is 9/4.2046

I’m going to plug that back in to my general solution right there and I get y=-1/4, cos5x + 9/4 cos(x).2058

That is my complete solution to the original initial value problem, it solves the differential equation and it matches the initial condition.2076

Let me remind you of the steps here, it was on the previous side where we got this down to the point of having sec(x) × (y)= the integral of cos3x sin(x).2087

To solve that integral we used a little (u) substitution, (u)=cos(x), du=-sin(x), because of that negative, that is where that negative comes from.2100

Then we get an easy integral of u3, that means u4 + a constant, on the left hand side I have this sec(x) × y.2108

To solve for (y), I’m going to divide by sec(x), that is the same as multiplying by cos(x), I multiplied both sides by cos(x), that is why this 4th became a 5th here and the (c) got multiplied by cos(x).2119

Now I use my initial condition here, plug in 0 for (x) and 2 for (y), so 2 is -1/4 cos50 + (c) cos(0).2134

Cos(0)=1, 2=-1/4 + a constant, move the ¼ to the over other side I get (c)=9/4 then I take that and I plug it back in to my general solution right here, that is where I get y=-1/4 cos5x + 9/4 cos(x).2145

Example 3, we are given a linear differential equation and we are just asked to find the integrating factor.2173

We do not have to solve this one all the way through, what we have to do here is just find the integrating factor.2178

Remember the general form for linear differential equation though is Y′ +p(xy)=q(x).2187

In particular, you are not allowed to have anything in front of the Y′, any coefficient here, we do have a coefficient right here, x+ 1.2197

What I’m going to do is divide that away, I’m going to go Y′ – 1/x + 1y=1/x + 1 × sin(x).2206

The way you find the integrating factor, you do (e) to the integral of p(x)(dx), the p(x) here is whatever in front of the Y and the key thing here is do not forget the negative sign.2223

The negative is part of it, so that is all p(x) right there, this is e to the integral of -1/x + 1(dx), now that is (e) to the -, now the integral of 1/x + 1 is the natural log of x + 1.2237

You do not have to worry about a constant when you are finding the integrating factor.2261

You can not cancel the (e) and the natural log because there is a negative sign in between them so you have to write (e) to the natural log of x + 1-1.2266

Now you can cancel the (e) in the natural log, you will get x + 1-1 and that is just the same as 1/x+1.2278

That is all we are asked to do for this example, is to find that integrating factor and we found it, we are done.2291

Let me go over those steps one more time, the key thing here is I remembered the generic form for a linear differential equations.2299

It got to be Y′ + p(xy)=q(x) and that only works if the coefficient of Y′ is 1.2307

Here, the coefficient of Y′ was x+1, I just divide it away, I just divided both sides by that, that is how I got it into this form Y′ -1/x + 1y = 1/x + 1 × sin(x).2314

Whatever that is in front the Y is your p(x) but you got to include that negative sign.2330

I plugged that into my generic integrating factor formula including the negative sign and that is an easy integral.2337

The integral of 1/x + 1 is just natural log of x + 1, but there is still that negative sign so I can not cancel the (e) and the natural log yet.2343

Instead I make that -1 in to an exponent and now I can cancel the (e) and the natural log, I end up with just 1/x + 1.2351

In example 4, we are given a linear differential equation sin(XY)′ – cos(xy)=x2 and we want to find the integrating factor i(x).2366

Let us remember our generic form for our linear differential equation, Y′ + p(xy)=q(x).2377

In particular, you can not have any coefficient in front of the (Y)′ so we got this sin(x) here, we got to get rid of it by dividing that away.2387

Divide that and get Y′ – cos(x)/sin(x), y=x2/sin(x), and now I can use my form for the integrating factor which is i(x)=e to the integral of p(x)(dx).2397

That p(x) means whatever in front of the y but it includes the negative sign so go ahead and put that in there (e)cos(x)/sin/(x)(dx).2424

To solve that integral, I could think of that as cot(x) but I do not think that is particularly useful here, what I’m going to do is a little substitution.2438

I’m going to use (u)=sin(x) and d(u) is –cos(x)(dx), (du) is just positive cos(x) because the derivative of sin is +cosine, I was looking ahead at my negative sign there but the derivative of sin is just +cos.2449

We have (e)-integral of (du)/u, that negative sign stays there but the integral (du)/(u) is –natural log(u).2481

That is (e)-natural log(u) was sin(x) and we got this negative sign in between the (e) and the natural log, I can not cancel.2499

A lot of students are very tempted to cancel the (e) and the natural log and say “that is –sin(x)”, but it is not.2511

Instead, you use that negative and make it into an exponent that is (e)natural log(sin x) to the -1.2518

Now you can cancel the (e) and the natural log, you got sin(x)-1 and that is equal to 1/sin(x) which is cosec(x).2527

By the way, sin(x) of -1 is not the same as arcsin or inverse sin(x), this is really an exponent so it really means 1/sin(x).2544

Be very careful here but this is not the same as sin -1 (x) or arcsin(x), it does not mean the same thing as either one of those, that is another frequent mistake that students make.2554

We are done here because we are just asked to find i(x) and we found it, let me recap the steps there.2570

We wanted to get the equation into linear form Y′ + p(xy)= q(x), in order to do that, I have to divide away this sin(x).2577

I divided away and I get Y′ – cos/sin(y)= x2 /sin(x) and in the integrating factor you do (e) to the integral p(x)(dx), where the p(x) is whatever is found in front of the Y.2587

You got to include that negative sign there, that is all p(x).2601

To do that integral, I did a little substitution (u)=sin(x), d(u)=cos(x)d(x), we got –(du)/u which is –natural log(u).2607

That is substituting back, that is –natural log(sin x) now I can not cancel the (e) and the natural log yet because I still got the negative sign in between them.2618

I will make that an exponent and now I can cancel the (e) and the natural log and I will just get sin(x)-1, that is not the same as inverse sin or arcsin, it is really sin(x)-1.2627

That is 1/sin(x) which is cosec(x).2644

On example 5, we got to solve the following initial value problem t + 1y′- 3y=t and y(1)=2.2652

Let us go ahead and get that into linear form because it is not quite in linear form yet.2664

Remember linear form is Y′ + p(t), y=q(t) I have got to divide away that t + 1 to get it in that form.2670

I’m going to go Y′ – 3/t + 1y=d/t + 1, I’m not going to worry about that initial condition for quite a long time yet, you do not have to worry about that.2682

My integrating factor, that is my strategy for all linear differential equations is (e)integral of p(x) or p(t) in this is case, which is (e)integral of -3/t +1 p(t).2696

Remember I got to include the negative sign when I’m finding the integrating factor.2716

This is (e)-3 d(t)/t + 1, I use a little substitution (u)=t + 1, d(u)=d(t), then I just got 1/d(u)/(u), this is (e) -3 natural log (t +1).2724

This is (e)natural log(t +1) raised to the -3 power, now I can cancel my (e) and my natural log.2751

That is t + 1-3 which is just 1/(t + 1)3.2764

Now the point of that integrating factor is you got to multiply it by both sides of the differential equation.2776

We will multiply that by both sides of this form of the differential equation, I get 1/(t + 1)3 × Y′ – 3/(t + 1)4 × y= t + 14.2785

The point of that is that this left hand side is always equal to the integrating factor × y if you took their derivative using the product rule.2811

It is always a good exercise to check whether that left hand side really is the derivative of i(t) *y2824

Because it is a derivative what we can do is integrate both sides and we will get i(t) × y= integral of t/t +14 d(t).2833

My i(t) remember was 1/t + 13 × y = integral of t/t + 14 d(t).2851

I have to solve that integral and that looks a little bit unpleasant, there is a nice little trick you can do here, which is (u) substitution.2867

(u)=t +1 and then d(u) would just be equal to d(t), I see I’m going to have a (t) in the numerator that I have to deal with.2873

That would be t=(u-1), if I solve my substitution for (u), this is equal to the integral of (u-1) × d(u) all over (u)4.2884

I can separate that out in to (u)/(u)4 which is (u)-3 - (u)-4(du).2901

Now I can integrate using the power rule, the integral of (u)n(du), learned this back in first semester calculus is (u)n +1/n + 1.2915

If (n) is -3, (n) + 1 is -2, this is (u)-2/-2, now (n) is -4, (u)-3/-3.2929

I just did the integration, now is the step when I have to add the arbitrary constant, not sooner not later, it got to be right now.2949

(+c) here, if I turn that back in to t + 1, I’m going to pull this -2 out and make it -1/2, t + 1-2 + 1/3, t + 1-3 + c.2957

This is all equal to 1/t + 13 × (y), and now I want to solve for (y) so I’m going to multiply both sides by t + 13 on both sides here.2984

On the left I get (y) = on the right – ½, (-2 + 3) is just 1 so, (t + 1) + 1/3.3004

T + 1-3 × t + 13 is just 1 + (c) × (t + 13).3020

I’m going to keep going with this in this next slide so we can solve for the constant but let me remind you what we did in this side.3029

I had to get it in this generic linear equation form which means I have to get rid of this coefficient of (t +1).3037

I divide both sides with (t + 1) and this was my new form of the differential equation.3044

I looked at whatever was in front of the Y and I call that p(t), I have to integrate that.3050

I got to integrate -3/(t + 1) that is easy with the (u) substitution, (u)=(t + 1) so I get -3 natural log(u), natural log(t + 1).3056

Before I can cancel my (e) and my natural log, I have to take that -3 and turn it into an exponent then I can cancel the (e) and the natural log.3069

And I will get (t + 1)-3 or 1/(t + 13).3078

I took that 1/(t + 13) and multiply that by both sides 1/(t + 13) and that is how I get this line here 1/(t + 13) Y′, 3/t + 14y and t/t + 14.3081

The point of that integrating factor, this is always true for all linear differential equations, the left hand side turns in to the derivative of a product.3106

It is the derivative of the integrating factor × y, the right hand side you will never going to know what it turns in to but we integrate to undo that derivative.3116

We got the integral of the right hand side t/(t + ¼), the i(t) is 1/(t +13).3126

On the right hand side I have got to integrate t/(t + 14), my clever idea there is to use (u)=t + 1 and d(u)=d(t), plug that in for (u) here, plug (u -1) in for (t).3136

That now separates out into two nice powers of (u), well fairly nice powers of (u), they are negative.3150

I just remembered the power rule from calculus 1 here, I plug that in with n=-3, n=-3, and n=-2 on the right and that was the step where I did my integration.3156

I have to add a (+c) right at that step then I changed my u’s back to (t + 1) and I just brought this down 1/(t + 13) × y.3174

In order to solve that I multiplied both sides by (t + 13) and that is how on the right I got -1/2(t + 1)1 and 1/3 (t + 1)0 and (c) × (t + 13).3187

We are not done with this yet because we have not yet incorporated the initial condition, what I’m going to do is take this equation over on the next slide.3202

We are going to use it to solve the initial condition and then we are going to be done.3210

Let me remind you of the equation that we had on the previous slide.3216

Our solution so far is (y)=-1/2 × (t +1) + 1/3 + (c) × (t + 13).3222

Now I’m going to use my initial condition y(1)=2, I plug in 2 for (y) and 1 for (t), so 2=-1/2 × if (t) is equal to 1 then I will get a 2 there + 1/3 + (c) × 23.3238

2=-1 + 1/3 + it looks like it is going to be 8(c) there and if I move that -1 over, I get 3 =1/3 + 8(c).3261

If I move that 1/3 over, 3 is 9/3, so 3 – the 3rd is 8/3 is equal to 8(c) and I see I will get (c)=1/3.3279

I take that (c)=1/3 and I plug it back into my general solution and I get (y)=-1/2 *(t +1) + 1/3 + 1/3 × (t + 13).3292

I can do some algebra here, I could maybe expand out (t + 13), maybe expand out -1/2 (t + 1).3313

I do not think it will be dramatically better so I’m going to leave that in that form, I have figure out what my arbitrary constant is, I have solved this as far as it is really useful to go.3322

Let me remind you of the steps here, this came from the previous slide, this is our solution from the previous slide.3338

But it was not a complete solution yet because we have not incorporated the initial condition.3344

We use the initial condition to find the value of the arbitrary constant (c), I’m plugging in from here I get y=2, from here I get t=1.3349

I plugged in y=2 there and t=1 in here and here, that is why my (t + 1) turned into two’s, that simplified to -1, 1/3, 23 is 8, I just did a little algebra to simply down and figure out that (c) was 1/3.3361

I take that and plug it back into the general solution, that is where I got this 1/3, not this 1/3, but that is where I got this 1/3 and the general solution is the arbitrary constant.3381

My whole thing turns into just the general solution with the arbitrary constant given the value of 1/3.3393

On example 6, we are going to find the general solution to the following differential equation (XY)′ + 3y=cos(x) with (x) being positive here.3403

That is going to be a linear differential equation but we need to do some manipulations to get it in to the right form.3419

Remember that form for linear differential equation is Y′ + p(xy)=q(x).3424

In particular, there is not any coefficient in front of the Y′, which is what we have here, I’m going to divide that away to start, Y′ + 3/(xy) = cos(x)/x.3434

That is why we are given the condition (x) is positive, it allows me to divide by (x), I do not have to worry about dividing by 0 there, that is really all we use that condition for.3450

Now I’m going to use my integrating factor strategy for linear equations, my i(x)= (e) to the integral of p(x) d(x) and my p(x) is whatever is in front of the y.3462

That is the p(x) right there, in this case the p(x) is positive 3/x, I get (e) to the integral (3/x d(x)).3482

The integral of 3/x is just 3 natural log(x), I do not need to add (+c) when I’m just finding the integrating factor.3497

A lot of people would like to say that is 3x because it is (e) to the 3 log (x), not true.3506

You can not cancel (e) to the natural log until you get rid of that 3 in between them, so I’m going to make that into an exponent on the (x).3512

(e) to the natural log of (x)3 and this is just x3.3519

What you do with that integrating factor is you multiply both sides by whatever you got.3526

I’m going to multiply both sides by x3 and I’m going to get x3 Y′ + (3/x × x2) is 3x2, y= x2 × cos(x).3533

The point of using that integrating factor is that the left hand side is supposed to look like a derivative.3549

In this case, it looks like the derivative of x3 × y, the reason for that is because if you expand out the derivative of x3 × y using the product rule,3557

You get x3 × the derivative of (y) + (y) × the derivative of (x)3, so it really does check there.3568

It is always a good strategy to check when you are at this step in any linear differential equation because what it does is it just make sure that you did all the previous work right.3575

It is really easy to make mistakes through this process, it is good to check and make sure that the derivative of x3y really is equal to what you had on the left hand side.3587

In this case is does work, this is still equal to x2 cos (x) and in order to undo that derivative we want to take the integral of both sides.3598

We get x3y the integral of the derivative is just the original thing is equal to x2cos(x) or the integral of x2 cos (x) d(x).3608

In order to integrate that, I have to use parts and I’m going to use my tabular integration trick that I taught you back in the calculus BC lectures here on www.educator.com.3622

You can look this up on this lectures if you like, remember the trick is to take derivatives of the left hand side until you get to 0.3634

The derivative of x2 is 2x, derivative of 2x is 2, derivative of 2 is 0, then you take integrals of the right hand side.3650

The integral of cos(x)=sin(x), the integral of sin(x)=-cos(x), integral of –cos(x)= -sin(x).3655

Then I cross multiply down this diagonals and I put alternating signs plus, minus, plus, and I’m ready to write the answer to my integral.3665

It means you multiply down this diagonal lines, you will get x2sin(x) + 2x cos(x) – 2 sin(x).3677

I have adjusted my integrals, now is the time to add my arbitrary constant, the key is you add it this time and the reason is because we still have not finished solving for (y).3692

As we solve for (y), we are going to do some algebraic manipulations and the (c) is going to be tangled up in there algebraic manipulations.3703

You have to let it get tangled up in those algebraic manipulations or other wise it is the wrong answer3709

In this case, to solve for (y) I’m going to divide both sides by x3 so I get y=x2 sin(x) + 2.3718

It looks like I made a small mistake there I forgot to right down that (x), so it is actually 2x cos(x) – 2 sin(x) + (c).3729

That is all divided by x3 because I was trying to get rid of that x3 on the left hand side.3750

That is as far as I can take it for this one, the reason is because I do not have an initial condition, all I have is the differential equation.3756

If I had an initial condition I would use that to find the constant, I would use that at this step.3768

Let me write that down, you would use the initial condition if it is given to you to find the value of (c).3773

You can narrow that (c) down to a particular number if you have that initial condition but in this case we did not have an initial condition so we just stop here with our general solution.3793

General solution means that it has an arbitrary constant in the answer.3805

Let me recap the steps here, I was remembering my generic form for a linear differential equation, it does not have anything in front of the Y′.3813

It is not a 0 there, let me not suggest that supposed to be a 0 but it just does not have anything there, it is like its multiplied by1.3823

In this case we have an (x) in front of the Y′, the first thing I have to do right away is divide everything by (x).3831

I divide this (x) away, at least away from the Y′ term, that means it shows up in the denominator over here 3/x and cos(x)/x.3837

Now I can use my generic solution method for linear differential equations which is to use the integrating factor, which is (e) to the integral of p(x).3847

That p(x) is the 3/(x), it is whatever in front of the Y if there is a negative sign, you got to include that as well.3857

In this case, there is no negative sign so we have (e) to the integral of 3/x, the integral of 3/x is 3 natural log(x).3864

Before I can cancel the natural log I have to move that 3 over to be an exponent.3875

I get x3 or (e) to the natural log x3 but the (e) in the natural log now they do cancel, I get x3.3880

I take that and I multiply that by both sides of the equation and that is how I get x3 Y′ + 3x2y=x2cos(x).3887

The point of that is that the left hand side is the derivative using the product rule of x3y.3900

We are really using the product rule there and because it is a derivative of something I can integrate both sides.3908

Integral of x3 Y′ is just equal to x3y and on the right I have the integral of x 2 cos(x) which is a little nasty.3915

I’m going to need integration by parts by that but I have my little tabular method that I have learned back in the calculus 2 or calculus BC lectures here on www.educator.com.3925

Here is my tabular integration method, I’m taking derivatives of x2, so I got down the 2x, 2, and 0.3936

I’m taking integrals of cos(x), the integral of cos is sin, the integral of that is -cos, the integral of that is –sin.3945

Then I multiply down these diagonal lines attaching alternating signs as I do it, I’m multiplying alternating signs on.3952

I get x2 sin(x) I see I have 2 negatives here so that gives me + 2x cos(x) – 2 sin(x).3961

It is very important that you have the constant when you do the integration which is right now, I put a constant on there.3972

Because I have x3y, I want to solve for (y) so I just divided that x3 in to the denominator there.3979

If I have an initial condition I would use that to find the constant but since I do not, I just stop here and I present my general solution to my differential equation.3987

That is the end of this lecture on linear differential equations, we got a whole lot more interesting material to cover in differential equations.3999

This is just the beginning of the topic, I hope you will stick around and sign up to see the other lectures on differential equations.4005

We are going to talk about separable differential equations in the next lecture.4013

We are getting in to all kinds of other topics, systems of differential equations, partial differential equations, second order differential equations.4017

We are going to graph slope fields and direction fields, there is all kinds of exciting material for us to explore together and I really hope you will be a part of it.4025

In the mean time, my name is Will Murray and you are watching the differential equations lectures on www.educator.com, thanks for joining us.4034