Separable differential equations introduction | First order differential equations | Khan Academy

[@kurkwelch773]

Why does this one guy literally know EVERYTHING

[@SoWe1]

wouldn't ask him about pottery, pretty sure

[@kerr1221]

@@SoWe1 He's an engineer, so he actually probably could tell you a bit about pottery.

[@SoWe1]

@@kerr1221 you're not particularly skilled in any manufacturing field yourself, are you?

[@kerr1221]

@@SoWe1 I was a machinist for 10 years, so you could say I am. In school for EE now. Depending on where someone goes with their courses, learning about pottery and ceramics at some point is a possibility. We talked about them in my strength of materials class. Edit: At least in my program, engineers are exposed to a fair amount of hands on/manufacturing. Ofc that doesn't make them as good as people who do it for a job, but they aren't clueless.

[@dalilpebble]

@@kerr1221 r/iamverysmart

[@dogeeatsveggies]

fucking support this channel everyone i hope in 2050 this guy will be recognized by MORE people and he makes it into history books and he becomes a fucking legacy.. not only sal.. but the etire khan acad team..

[@irinakhlebanova7249]

Salman Khan

[@dangerouslykeety]

Totally 100% agree

[@jackrotter9434]

I understand you much easier than my Russian native diff eq professor haha!

[@victorfrancis919]

Khan Academy savior of all exams!

[@MwansaTaizya]

He has is own way of making people understand within minutes 🥺 thank you for being the best tutor

[@KemalCetinkaya-i3q]

sweet 🥺

[@whitwat]

Idk why but your voice is incredibly calming and love your explanation on these concepts

[@thealexgalaxyoriginal]

2:34 "Little more space" lol

[@zayedacademy988]

watch our videos ,we are recreating his videos in a great way

[@gregoryfenn1462]

How can you multiply by dx? dx isn’t a number, it’s just a notational convenience for small changes (eg over a small amount of time). Can you explain that logic?

[@mokkes7340]

I know that I'm a little late... but remember that (dy/dx) represents a infinitimally small change of y over x, or the slope of a curve. So from a physical perspective, you can rearrange the formula to say that for every change in y (dy) it is equal to a change in x (dx). Hence why the transformation is very useful to describe physical phenomina.

[@thestructuralanalysismatri590]

In beams problem we can apply this call separable differential equations , and I have come across with hinged beam ,where moment is zero at the hinge, and also x=L, so (L,0) is a coordinate point, initial condition , given information to find a particular solution ( M=Mx) moment function to a higher degree order differential equation which it only takes place within a member and linked with constrain of the structure, the use of principle of superposition will help us to analyse redundant forces by setting derivative forms.I would like to share my experience....any emai?

[@pujawaraa]

Hello i'm from Indonesia.. I see this because of the work of the lecturer. but I don't understand because I'm not very good at English. Thankyou❤️

[@adipamungkasofficial]

Mee too

[@exol3024]

thank you for this sir!

[@renatoedaojr.1205]

In the first example, are we going to do integration by parts in -xe^-x^2 dx? I'm kinda confuse. Notice this one please. Thanks!

[@SilverArro]

dusky _gaming No. He does u substitution there, although he does it quickly and without going through the whole process since it’s easy enough to do in your head. He’s assuming you’re very familiar with integral calculus since it’s absolutely a prerequisite for a Differential Equations course.

[@earonfazonela7912]

No. It is exponential integration. Letting u= -x^2 therefore the du= -2x. That is why khan multiply it to 1/2.

[@earonfazonela7912]

No. It is exponential integration. Letting u= -x^2 so therefore the du= -2x.

[@ascaniuspotterhead2484]

Does someone know why y*dy integrated is equal to (1/2)y ^2 ? Im very confused because usually y integrated should be equal to (1/2) y ^2. but why is there the same solution with and without dy?

[@carsonhenderson]

when u take the integral the dy or dx just means you're taking the integral with respect to x or with respect to y so it goes away after the integral has been taken. If you were to see an integral of just dy, you assume you're taking the integral of 1 with respect to dy, so the integral of 1 is just y (+c). The integral of y with respect to dy is 1/2y^2 (+c) as you said.

[@ascaniuspotterhead2484]

Carson Henderson Thank you very much.

[@simplynicolebeauty]

Why can't you cross multiply ?

[@artistiaay5512]

i think its general solution, coz particular solution wouldn't have an independent variable that would vary each time it would have been constant!! If i got it wrong plz tell me what's the difference between particular solution and general solution ??

[@chupapimunanyo2596]

On the integration part you treated it like (e^-x)^2 and at the end you treated it like e^(-x^2) ... Isn't that a big mistake?

[@theepicosityofpizza]

He did no such thing. He treated it as e^-(x^2) in both.

[@ryanhugh9043]

can someone explain the integration part? where does he get the 1/2 to and all the other stuff from?

[@ssyhrr9149]

+Ryan Hugh He used u-substitution, by letting u= -x^2 (of exponential). So, he could use du/2 to substitute -x dx in the question. So basically that 1/2 u asked is coming from du/2. (1/2 S e^u du)

[@AhmedMustafax]

+Ryan Hugh His answer's right, I can't tell where he got the 1/2 from though.

[@marxho]

By sibstitution, u = -x^2, du = -2xdx, -xdx = du/2=(1/2)du.

[@ZainKhan-xx6np]

simple integration review the integration table. You will get the answer its v.Simple

[@nesreenacademy7284]

Thank you very much sir , very clear and helpful, please please I need to know the software that you are working with please.

[@billybobandboshow]

Hi! Do you think you could make a video on Euler's rule for differential equations? I don't fully understand it. Thanks!

[@Dan-bg5fm]

do you understand it now ?

[@LoganBrewski]

Wilfred I sure hope he does

[@Dan-bg5fm]

LoganBrewski lmao

[@nickgiannakopoulos3171]

@@Dan-bg5fm @LoganBrewski I hope he figured it out

[@billybobandboshow]

@@nickgiannakopoulos3171 @Dan @LoganBrewski 13 years later....yes I understand it now lol

[@soccerstudut]

At 5 min 56 seconds, couldn't you also multiply both sides by 2 to get rid of the 1/2 factor on each side and then plug in your point to solve for C. Your C would multiply by 2, but since it is an unknown constant that should not effect the answer. I did that and ended up with the answer of y= + - (e^(-x^2)+2)^0.5. My answer clearly is not as pretty, but I think it is also a correct answer.

[@Electrologia]

yes of course you can do. it will not change the answer but will simplify it.

[@amjedsenpai]

thank you khan academy

[@ankursingh3866]

Nice job good learning system help all out thank u

[@LovinaVargasOfItalia]

Thank you for your "advice".

[@7-j3u]

How do you get exponent of negative x square exponents in math?

[@adampowell576]

It was in the denominator and he brought it up to the numerator making the exponent negative.

[@gwenrees7594]

Great video, thank you!

[@MrJakesterfication]

Looks simple enough!

[@9340Steve]

Does anybody know a really good Calc I level explanation of why multiplying both sides by dx does not contradict the fact that dy/dx is not a fraction?

[@I0lcatz]

because it more or less is a fraction. at least it works exactly the same way, which is why its wrote like that.

[@xlaiaryan6582]

sir which software are you using for these videos????

[@jomembawang]

how does he know the C value at 5:56 is +C (positive) and not -C(negative)? I understand that C2- C1, but there could be the case that C1 is larger than C2

[@superroydude]

Jomembawang He doesn't. C could be either but it being an arbitrary constant it doesn't matter. For example you can write -5 as +(-5) it doesn't matter ( you could have -C if you wanted).

[@gregoryfenn1462]

You could have also written C=(C1+C2)/2 which would be much neater

[@Marantzdv]

Awesome, thanks!

[@nicoangelogregorio7494]

C = 1/2 though sir... I hope you notice.. Cuz you forgot the x must be equate with zero

[@axelnils]

Absolutely not

[@izzatullokhmakhammadjonov6836]

Hello can you solve the x*y*y'=x^2+1

[@drofeng]

yy' = y dy/dx = 1/2 d(y.y)/dx by the product rule Then 1/2 d(y^2)/dx = (x^2 + 1)/x Or d(y^2)/dx = 2x + 2/x Let u = y^2 du/dx = 2x + 2/x Separate the variables and solve, then substitute y^2 for u

[@pey5571]

thank you!

[@gabitheancient7664]

why can I do that? like, i thougt dy/dx was just a notation, not a real fraction

[@drofeng]

You can write in finite notation, multiply and then take the limit e.g. DeltaY / DeltaX = x^2 Then DeltaY = x^2 * DeltaX DeltaY = y(x + DeltaX) - y(x) dy = lim(DeltaX -> 0) y(x + DeltaX) - y(x) In the limit, dy = x^2 dx

[@RickyPollo]

Sal, there's a minor error in how you work out this problem. Once you separate the equation, you have -x * e^(--x^2)dx on the right. You don't show the steps of your u substitution, but you set u = --x^2. which means that du/dx = --2xdx In order to sub in, you have to multiply the right side by --2 inside the integral, and --1/2 outside of the integral. You forgot the negative sign on the 1/2, so your final solution is the wrong sign. I really appreciate your website, and I'm thankful that you do such a great job with your more complex topics. I'm taking a graduate physics course this fall, after not having done a math course in 15 years (I teach HS physics though), and I've used your site to refresh myself on trig, calc I, and differential equations. Your practice materials are great, especially the practice problems. Thank you so much for this awesome gift you have shared with us!

[@muhammadrafay4743]

i was thinking the same thanks for correcting it

[@in4LifeTime]

It’s actually correct, try using u = e^(-x^2), du = -2x e^(-x^2). He simply manipulates the equation by adding in a 1/2 to get the original expression which was -xe^(-x^2).

[@satioOeinas]

@@in4LifeTime the answer should be +-sqrt(-e^-x^2+c), Sal forgot the - before e^x^2

[@henrymazza4169]

"Boat Sides" by y and dy

[@fortunefolayan2295]

I think there’s a mistake in the “intergration by parts”

[@pcl5636]

Awesome m8

[@yechengyecheng6126]

why say differential equations,not derivative equation

[@halekiroshalemichael517]

very good

[@jferro96]

i thought you couldnt treat dy/dx like a fraction.

[@raullara6082]

It's wrong but also not wrong to do so, but you can think of it as moving what side you will be differentiating. It works out. Depends on the context.

[@Dan-bg5fm]

?

[@jacoblugtu8723]

Raul Lara wait what do u mean wrong and not wrong

[@Colonies_Dev]

I guess the expression is true so you can use algebra?

[@faithelias2798]

God loves you and he wants to save you all

[@atulkumarashish584]

I think we simply can't multiply with dx.

[@madhumithaayyappan5277]

awesome

[@slowbro202]

Hey first of all thanks for all the great videos. Secondly, please rethink your audio levels. With my computer's volume one tick above mute and the youtube volume slider at 1/3 of max, your video still needs to be turned down. When coming from just about any other video where both computer and player volume need to be on full to hear anything, starting one of your videos is physically painful.

[@gideonanningboateng7453]

please try using simple questions for your introductions.....not the one you used

[@RickyPollo]

As far as differential equations go, this is about as simple as it gets.

[@xpredator9525]

I love you

[@philanipraisewell8993]

Lol and I have never studied this stuff but I already know what it is about

[@benbrak3713]

i dnt like how you started

[@1828ali]

kindly change your board color to white. Thanks

[@zayedacademy988]

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[@reececampbell1844]

I love you

[@adengoher4343]

Thank you for this video!