And once again, a, b, and c, they are constants, and in this case, we don't want a to be 0, otherwise this term will be gone, right?

再說一遍，a、b 和 c 都是常數，在這種情況下，我們不希望 a 為 0，否則這個項就會消失，對嗎？

So it's not second-order anymore.

是以，它不再是二階的了。

All right, as we can see, we have y, y' which is the first derivative, and this is y'', the second derivative.

好了，我們可以看到，我們有 y，y'是一階導數，這是 y''，二階導數。

And it seems like these are just the constant multiples of the original first and the second derivative, right?

這些似乎只是原始一階導數和二階導數的常數倍數，對嗎？

Do we know a function so that its derivative is just a constant of its original?

我們是否知道一個函數的導數就是它的原常數？

I think so.

我想是的。

We do, right?

我們有，對嗎？

On the side, let me just show you.

在旁邊，讓我給你看看。

If I start with y equals to, well, if I have, let's say, e to some power times x, well, in this case, we use t, by the way, for second-order, use t.

如果我從 y 等於開始，那麼，如果我有，比方說，e 的某個冪乘以 x，那麼，在這種情況下，我們使用 t，順便說一句，對於二階，使用 t。

Let's say we have e to the 3t.

假設我們有 e 到 3t。

Well, if you differentiate that, you get y' and you know this is going to give you e to the 3t, the function part stays exactly the same, but the chain rule says I will have to multiply by 3, right?

那麼，如果將其微分，就得到 y'，你知道這將得到 3t 的 e，函數部分保持不變，但鏈式法則說我必須乘以 3，對嗎？

So let me put it down right here, right?

所以，讓我把它放在這裡，對不對？

And you see, this 3 stays, and this is just the original, so 3 times y, right?

你看，這個 3 保持不變，而這只是原樣，所以 3 乘以 y，對嗎？

And likewise, I can do it again. y'' is going to be, we will keep this 3 times e to the 3t, but then we multiply this 3, right, for the derivative by the chain rule. 3 times 3 is 9, and e to the 3t is the original.

同樣，我可以再做一次。y'''將是，我們將保持這個 3 乘以 e 到 3t，然後我們乘以這個 3，沒錯，根據鏈式法則求導。3 乘以 3 是 9，e 到 3t 是原數。

And you see, whenever you have an e to some power, the first derivative, the second derivative, they are just going to be the constant multiple of the original, right?

你看，只要有一個 e 的冪次，一階導數和二階導數就會是原數的常數倍數，對嗎？

So what this is telling me is that this will suggest, let me just put this down, this suggests that for the function, we should have the form y equals to e to some power times t.

是以，這告訴我，這將表明，讓我把它寫下來，這表明對於函數，我們應該使用 y 等於 e 的某個冪乘以 t 的形式。

I don't know what this number should be, earlier I just used 3, right?

我不知道這個數字應該是多少，剛才我用的是 3，對嗎？

So in general, let me put down r times t.

是以，總的來說，讓我來寫下 r 次 t。

And once again, for the second-order situation, we usually use t because there are a lot of applications that impose time and things like that.

再說一次，對於二階情況，我們通常使用 t，因為有很多應用都會強加時間之類的東西。

Anyways, this is my starting.

總之，這就是我的起點。

First of all, we begin by saying y is equal to e to the rt power, and the idea is that I'm just going to go ahead, differentiate this twice, and then plug in.

首先，我們先說 y 等於 e 的 rt 次方，這樣做的目的是讓我先做兩次微分，然後再輸入。

Hopefully, we can squeeze out some conditions that will help us to solve for this kind of differential equation.

希望我們能擠出一些條件，幫助我們求解這種微分方程。

All right, y' is going to be e to the rt times r, let me just put r in the front, and then y'' is going to be, this right here repeats, so r e rt, but then we multiply by another r, which will be r times r, which is r².

好了，y''將是e到rt乘以r，讓我把r放在前面，然後y''將是，這個在這裡重複，所以r e rt，但是我們再乘以另一個r，這將是r乘以r，也就是r²。

And now, let me put all this into their corresponding.

現在，讓我把這一切對應起來。

Here we have a, y'' is r² e to the rt, and then we add it with b, so we put on plus b, y'' is r e to the rt, and then we continue, plus c, y is e to the rt, and this right here is equal to 0.

這裡有 a，y''是 r² e 到 rt，然後與 b 相加，所以我們加上 b，y''是 r e 到 rt，然後繼續，加上 c，y 是 e 到 rt，這裡等於 0。

And now, can we kind of squeeze out a condition?

現在，我們能擠出一個條件嗎？

Let's see what can we do.

看看我們能做些什麼。

Every term has e to the rt, so we can factor it out, of course, e to the rt, and that will give us ar² plus br plus c equals to 0.

每個項的右邊都有 e，所以我們可以把它因式分解，當然，e 到右邊，就得出 ar² 加上 br 加上 c 等於 0。

All right, this is an exponential part, right?

好吧，這是指數部分，對嗎？

Exponential function e to the something.

指數函數 e 的東西。

This right here, we know, is never 0, isn't it?

我們知道，這個永遠不會是 0，不是嗎？

Never 0, let me just spell it out much better.

Never 0，讓我說得更清楚些。

So, when we have this quantity times that quantity, since e to the rt is never 0, so you can either divide it out, or you can just forget about it, because we just want to focus on this.

是以，當我們有這個量乘以那個量時，因為 e 到 rt 永遠不會為 0，所以你可以把它除掉，或者你可以忘掉它，因為我們只想關注這個。

So that means we must have this part, we must have, let me just write it down, we must have the situation that ar² plus br plus c equal to 0.

是以，這意味著我們必須有這一部分，我們必須有，讓我寫下來，我們必須有 ar² 加 br 加 c 等於 0 的情況。

In fact, this right here is the condition that we need, because from here, this is pretty much just a quadratic equation, isn't it?

事實上，這裡就是我們需要的條件，因為從這裡看，這幾乎就是一個一元二次方程，不是嗎？

Quadratic equation in terms of r.

以 r 表示的二次方程。

From here, we can solve for r, and we can just plug into the r here, and we can generate the building blocks of the solution.

從這裡，我們可以求出 r，然後把 r 插到這裡，就可以生成解決方案的構件了。

And I'll show you guys what I mean by that.

我會告訴你們我說的是什麼意思。

And before I show you guys an example, let me tell you guys that this equation here has a name.

在給你們舉例說明之前，讓我先告訴你們，這個等式是有名字的。

This is called the characteristic equation, and this is also called the auxiliary equation.

這叫做特徵方程，也叫做輔助方程。

And now, let's go ahead and solve this for y'' minus 5y' minus 6y is equal to 0.

現在，我們繼續求解，y''減 5y' 減 6y 等於 0。

The first step is we have to take this and change it to its corresponding characteristic equation.

第一步，我們必須將其轉換為相應的特徵方程。

And check this out.

再看看這個。

Earlier, when we start with a y'', we will end up ar², right?

先前，當我們以 "y''"開頭時，最終會得出 ar²，對嗎？

So y'' corresponds to r².

所以 y'' 相當於 r²。

That means right here, the first term is going to give me 4r².

也就是說，在這裡，第一項的結果是 4r²。

And then we just continue.

然後我們繼續。

Minus 5, and the y' corresponds with r, we change that to r right here, and then minus 6.

減 5，y' 對應的是 r，我們在這裡將其改為 r，然後減 6。

If you have y, we didn't have any r, right?

如果你有 y，我們就沒有 r，對嗎？

So this is just minus 6.

是以，這只是負 6。

This is it.

就是這裡

And now, we just have to solve this quadratic equation and do it whichever way that you would like, and I will factor this out for you guys.

現在，我們只需求解這個一元二次方程，並按照你們喜歡的方式來做，我將為你們計算出結果。

And I'll show you guys with the tic-tac-toe factoring, and to do so, I want to ask myself, what times 4 gives me 4r²?

我將向你們展示井字派因式分解，為此，我想問自己，4 乘以 4 得到 4r²？

And let me tell you guys the correct combination, which is 4r times r.

讓我告訴你們正確的組合是 4r 乘以 r。

And what times 4 gives me negative 6?

那麼 4 乘以負 6 又是多少呢？

And once again, let me tell you guys the correct combination, which is negative 2 plus 3 in these boxes like this, in this order.

讓我再一次告訴你們正確的組合，就是負 2 加 3，像這樣按順序放在這些盒子裡。

Well, I will convince you guys this is correct.

好吧，我會讓你們相信這是正確的。

To do so, you cross multiply 4r times negative 2, which is negative 8r, and then you take 3 times r, which is 3r.

為此，您需要將 4r 乘以負 2，即負 8r，然後將 3 乘以 r，即 3r。

It is correct because negative 8r plus 3r is negative 5r, so this is correct, isn't it?

它是正確的，因為負 8r 加 3r 等於負 5r，所以這是正確的，不是嗎？

Right?

對不對？

Okay, so that's good.

好吧，這樣很好。

And now we have to read the answers correctly from the boxes here.

現在，我們必須從這裡的方框中正確讀出答案。

To read the answer, you go across.

要閱讀答案，請穿過

So the first factor is going to be 4r plus 3, and the second one is r minus 2, and we still have equal to 0, of course.

是以，第一個因數將是 4r 加 3，第二個因數是 r 減 2，當然，我們還是等於 0。

All right, for the first one, we know r is equal to negative 3 over 4, and for the second one, we know r is equal to 2.

好了，對於第一個問題，我們知道 r 等於負 3 大於 4，而對於第二個問題，我們知道 r 等於 2。

And you see, we end up with two different r values.

你看，我們最終得到了兩個不同的 r 值。

In this case, we will end up with two different building blocks for the solution, e to the first r, which is this, and times t.

在這種情況下，我們最終會得到兩個不同的解塊，第一個 r 的 e，也就是這個，以及時間 t。

The second one is going to be e to the 2t, right?

第二個將是 2t 的 e，對嗎？

So let me just write down the building blocks for you guys first.

所以，讓我先為你們寫下構件。

This right here is going to give me e to the negative 3 over 4t, and the second one is going to give me e to the 2t.

這裡的是 4t 負 3 的 e，第二個是 2t 的 e。

And I'm going to explain to you guys what do I mean by building blocks.

下面我將向你們解釋我所說的積木是什麼意思。

First of all, let me just show you guys that both of them will satisfy the original differential equation, and just for simplicity purpose, let me just check this one for you guys only, okay?

首先，讓我告訴你們，這兩個方程都能滿足最初的微分方程，為了簡單起見，讓我只為你們檢查這一個，好嗎？

And you can do this on your own.

你可以自己做這件事。

So let me just put down this right here.

所以，讓我把這個放在這裡。

Let me just say this is the second one I get, so let me put down y2. y2 equals to e to the 2t, and then what I have to do is differentiate this, which is going to give me 2e2t, right?

y2等於e到2t，然後我要做的就是微分，這樣就得到了2e2t，對嗎？

And then I'll do it again. y2 double prime, which is 4e2t, right?

y2 雙質數，也就是 4e2t，對嗎？

And now I will plug in all this into the original, and you see it will give me 4, right?

現在，我把所有這些都輸入到原始數據中，你看會得到 4，對嗎？

And y double prime, which is that, so we multiply by 4e2t, and then minus 5y prime, which is that, which is times 2e2t, and then minus 6y, which is that, e to the 2t, like this.

y 雙質數，也就是我們乘以 4e2t，然後減去 5y 質數，也就是乘以 2e2t，再減去 6y，也就是 e 到 2t，就像這樣。

Do we end up with 0?

結果是 0 嗎？

Yes, we do.

是的，我們有。

This is 16 minus 10, which is 6, and then minus 6, of course, everything ends up to be 0, so 0 is equal to 0, so it checks.

這是 16 減 10，也就是 6，然後再減去 6，當然，最後所有的結果都是 0，所以 0 等於 0，所以它可以檢查。

This right here definitely works, isn't it?

這個絕對有效，不是嗎？

And you can do exactly the same thing for this, but you can also take my word for it, it will work out nicely as well.

你也可以用同樣的方法來做這個，不過你也可以相信我的話，效果也會很好。

All right, now, these are just the function parts.

好了，現在這些只是功能部分。

What we'll be doing is that, well, remember, when we're solving differential equations, we have that constant, right?

我們要做的是，記住，當我們求解微分方程時，我們有一個常數，對嗎？

And keep in mind, whenever we have the second-order differential equation, we will have two different constants.

請記住，只要是二階微分方程，就會有兩個不同的常數。

Well, do I put on plus C, or divide by C?

那麼，是加上 C 還是除以 C 呢？

I don't know.

我不知道。

The truth is, we multiply the function part by C1 and C2, so this right here is C1, and this right here is C2.

事實上，我們將函數部分乘以 C1 和 C2，所以這裡的就是 C1，這裡的就是 C2。

So, let me just demonstrate this right here on the side for you guys real quick, because now, you see, when I multiply this function by C, so let me just put on C in black like this, guess what?

所以，讓我在旁邊為你們快速演示一下，因為現在，你們看到了，當我用 C 乘這個函數時，讓我像這樣把 C 塗成黑色，你們猜怎麼著？

I will just have to multiply everything by C, isn't it?

我只能把所有東西都乘以 C，不是嗎？

So this right here will end up with C, C, C, C, C, C, C.

是以，這裡的最終結果將是 C、C、C、C、C、C、C、C。

And then, pretty much this term, we'll now have C right here, and this function here, this part here, has a C here as well.

然後，差不多在這一術語中，我們會在這裡看到 C，而這裡的這個函數，這裡的這個部分，也有一個 C。

This right here, just multiply by C, right?

這裡，只要乘以 C，對嗎？

We're just plugging everything accordingly.

我們只是相應地把所有東西都塞進去。

Guess what?

你猜怎麼著？

When I have this C here, I will still end up with 0 on the left-hand side, of course still equal to 0.

當我把這個 C 放在這裡時，左手邊的結果仍然是 0，當然還是等於 0。

The point of that is to show you, these are the places where C should be.

這樣做的目的是要告訴你，這些地方才是 C 應該去的地方。

C1 and C2 are just the multiple of the function part.

C1 和 C2 只是函數部分的倍數。

C1 and C2 are just the multiple of the building blocks of the solutions.

C1 和 C2 只是解決方案組成部分的倍數。

At the end, the overall solution is that you are just going to add them together, and this is it.

最後，整體解決方案就是把它們加在一起，就是這樣。

This is the solution for that. y is equal to C1 e to the negative 0 over T plus C2 e to the 2T.

y 等於 C1 e 到 T 的負 0 加上 C2 e 到 2T。

That's it.

就是這樣。

And the truth is that we can just add them together, and this is the overall solution.

事實上，我們可以把它們加在一起，這就是整體解決方案。

It's because when you differentiate this, right?

這是因為當你區分這一點時，對嗎？

Imagine if I'm differentiating the y, well, I just have to differentiate the first and the second.

試想一下，如果我要微分 y，那麼我只需要微分第一個和第二個。

Differentiate this and differentiate that.

區分這個，區分那個。

Do it again, do it again, do it again.

再來一次，再來一次，再來一次。

Plug in, plug in, plug in.

插電，插電，插電。

You get 0 is equal to 0 once again.

你再次得到 0 等於 0。

So this is the idea, just to summarize this real quick.

這就是我們的想法，簡單概括一下。

All you have to do, just to keep everything simple, get the characteristic equation, solve the quadratic equation in R.

為了簡單起見，你所要做的就是得到特徵方程，然後用 R 求解二次方程。

If you get two different R values, well, e to the first R times T, e to the second R plus T, and be sure you multiply C1 and C2 correspondingly, and then add them together.

如果您得到兩個不同的 R 值，那麼，將第一個 R 值乘以 T，將第二個 R 值乘以 T，並確保相應地乘以 C1 和 C2，然後將它們相加。

This is the general solution.

這就是一般的解決方案。

This is the first situation.

這是第一種情況。

You should watch my next video.

你應該看看我的下一個視頻。

I will show you what happens when I have these two R values being the same.

我將向大家展示當這兩個 R 值相同時的情況。