Added: 3 years ago
From: khanacademy
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  • great teacher, better comedian haha 3:09

  • @UniverselGamer

    A2. Yes. Integrate psi y then add f(x). Get the derivative wrt x. Equate it with psi x to get the value of f'(x) to get f(x).

  • @UniverselGamer

    A1. psi might have terms that 1.) have y 2.) have y plus constant or 3.) or just constants only (which is what we got) When deriving psi wrt x only, all of these are treated as constants, thus all having derivative of 0.  To be safe, you always assume 1.) or 2.) or both is true, that's why you put f(y) instead of C. If f(y) ends up as a constant then you have no problem there.

  • Q1. from "4:40" to "5:30", i don't understand why +C is replaced with f(y). And also i dont understand how when taking the partial derivative of f(y) will equal to 0 wrt x, if y is a constant wouldn't it still be f(y)?

    Q2. if we took the antiderivative of psi(y) would that "+C" be replaced by f(x)?

  • Hey Sal, shouldn't you put "+C" at the end of psi at 6:40 ? Or is it included in f(y) ? Thank you!

  • I'm always astounded at how much more fluidly and coherently Sal is able to explain mathematical and scientific concepts than my professors and TAs. You, sir, are awesome :D

  • shouldnt f'(y)=-1  instead of 1?

  • @manny34711 yeah...

  • at 7:00 shouldnt the derivative for the f(y) be f '(y)(dy/dx)?

  • Your critter was telling you that the answer is -1, not 1

  • waah tnx for posting instructional videos of D.E ... it helps lots!!!!!!!!!!!!

  • khan is legend

  • there is a mistake in taking the differential of M !

  • LOL @ 3:00 , priceless

  • How is M and N exaxt if they differ with a ...1xe... = .... 2xe....

    That's not the same equations.

  • @k88n He wrote a 2 that looks like a 1. His handwriting is a bit inconsistent sometimes since he has to use a mouse to draw everything so it's hard to get precise letters and numbers but I assure you, what he wrote is suppose to be a 2.

  • @Pastafarealist He uses a pen writing tool input, not just a regular mouse

  • @Pastafarealist thanks for clearing that up.

  • :S This is neat and all, but I don't really see what the use for psi is, because I don't really understand what that function means, even though I do get that if you have psi you can solve for y in some cases. And as I understand it, the point of most differential equations is finding y, not some weird function psi. So if I'm not mistaken, you're not really done with the problem yet since you need to solve for y next. Someone please explain, and thanks alot for your videos Sal. cheers from DK.

  • @nejtilsvampe

    I know you asked this a long time ago, but Psi is an intermediate step to solving the Differential Equation in terms of x and y.

    Yes, it's preferred to solve for y explicitly, but sometimes it's hard so an implicit solution is okay.

    For example: Y = X + 1 is the same as 1 = Y - X

  • @nejtilsvamp There may be cases (such as this one) where it is impossible to solve explicitly for y in terms of x, and so the implict solution is the best you can do.

    If Psi turned out to be, say, x+y, then you could have said y = C - x as the solution. But if you get y*sin(x) + x^2*e^y - y = C, you can't find an expression like y = f(x), because there is no way to isolate y.

  • For partial of M ... why is cosx and 2x taken as constants? dont we have to differentiate cosx to -sinx?

  • @tjaistt beacuse you differentiate with respect to y, not x.

  • Dude, you're amazing. I didn't understand how to do this at all 30 minutes ago, and now I not only know how to do it, I know why it is the way it is. <3

  • Thanks

  • are you actually using your mouse to write all that stuff? 8-O

  • Excellent video. However, at 11:12 you said that m is the partial of zi with respect to x and it should be with respect to y, as the derivative of cosx+2xe^y with respect to y is ycosx+2xe^y just like you put down and then integrated M sub y with respect to x...

  • Respectfully, and acknowledging that this isn't the easiest thing to teach, this video is a bit "convoluted" compared to most of your other topics I've watched...

    If you have it in you, you might consider redoing these first couple of Exact Equations videos...

    Don't get me wrong, this and your other stuff is fantastic and you get my compliments, I just think these couple of vids went "off" here and there...

  • brillinat work khan g!

  • OMG you are awesome...

  • is this john mayer?

  • hi are you canadian

  • there is a mistake at 8:56

    he wrots f ' (x)= 1, it should be f '(x)= -1

    ;)

    thank you Mr.Khan for these useful videos

  • @taktak0101 watch the whole video

  • Sal for teacher of the Decade/Century!!!!

  • I understand this better now. Thanks. :D

  • at 1:56 instead of 2x u wrote 1x

    opps lol

  • It is a two!... he says that xD

    it just looks like a 1...

  • i kno he says 2 but he wrote 1

  • its me the only one who gets lost with so many derivatives, anti derivatives, psi's, x's, y's ?

  • Comment removed

  • LOL! "I thought there was a critter in my house or something."

  • LOL!

    "SEE YOU IN THE NEXT VIDEO"

    AKA: If you are watching this and got this far, you are already fucked, and I know you will watch the next one...

  • this is great stuff!

    someone tried explaining this with finding the integral of both fx(x,y) and fy(x,y) and then pulling out a definition for h(y) but this is easy to remember and much more effective KEEP UP THE VIDEOS!!

  • i thought there was a critter in my house for a second....ha!

  • I just want to say what a wonderful job you are doing sal... I passed my calc 2 class with an A and it was partially because of ur videos. I watched all your videos for calc 2 and now i'm watching your differential equations videos. Your a great teacher keep up the good work.

  • AH! I thought about it and I've got it. Because it's a partial derivative and not a regular derivative, taking the integral of the function need only yield a function whose partial derivative is the same as what you were integrating. So it's almost like taking a partial anti-derivative!

  • How come when we take the integral of [the partial derivative of psi with respect to x (Psi sub x)] with respect to x, we can treat the y's constants?

  • but (sinx+x^2 * e^y) is multiplied by y', you didnt isolate it from y'. :/

  • Don't need to.

  • 0 a bit much but started fun XD

  • Imao, this concept is no mystery to me. Because in my linear algebra and vector calculus class, we did something like that. We had to use it to determine whether a vector was conservative or not. Well thanks for the reinforcement. I have to take this differential class over the summer in 9 days so I am getting a headstart.

  • At ~8:50, wouldn't f'(y) = -1 instead of 1?

  • Wait, didn't see that you corrected it.

    I should start watching the whole video before asking questions.

  • do not forget to check that M and N are actually continuous!

  • how do you check continuity? please reply

  • if M and N are polynomic, them they are continuous, you check continuity as you check it in any function or derivate

  • thanks

  • I would just like to thank you! This example problem helped so much it is unbelievable. Everything is explained so well.  Although my professor does something with putting My and Nx together by looking at the common terms. I believe your method is much more understandable. Thanks, again!

  • Wow, i watched the whole EXACT EQUATTIONS tutorial. It was great. I actully understand it now. Way better than my math teacher who cant teach well. Thank You

  • thanks for the help. A great suppliment to my linear class. bad professor. keep up the good work. it really helps

  • Why didnt you solve for y?

    y(x) = −W[ x² ⁄ (sin(x)−1) ]

    Where W is the Lambert W-Function

  • I Think he didnt solve for y because you cant solve for it explicitly.

  • Did you feel inclined to respond to my answer when you didnt even read it? Because I wrote an explicit answer. I guess you just didnt understand it.

  • this is not easy for me to get a grip on but Im appreciating the pedagogical finesse.

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