Homogeneous Differential Equations

[@TheOrganicChemistryTutor]

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congrats on making it to JHU!

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[@ayishanuhu4955]

I just realized you don't have a video on exact differential equations! Kindly make one. Others might benefit in the years to come. Thank you.

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Agree

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www.patreon.com/MathScienceTutor its in the description

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[@allanokoth1312]

accurate

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Please I need a full playlist of differential equation tutorials.

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x2

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[@klarizamadrazo6852]

Thank you so much for helping me all the way from trigonometry to calc 1-3, chemistry, physics, and differential equations. You're truly an angel. Anyways can you please make merch that actually says "The Organic Chemistry Tutor"?!

[@MShazarul]

Here's a summary of first order homogeneous differential equation. 10% calculus 990% algebra Right, that's about FOHDE

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100000000% brain

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[@crismoreno9533]

Keep making differential equations videos the whole book lectures if possible. I have been able to learn solid calculus 1 and 2 because of you it would be highly appreciated.

[@arsh1357]

Have you graduated whats going on now?

[@szaman4180]

i have the same question lol@@arsh1357

[@Mathefusions18]

The equations must have same dimensions for them to be homogeneous. Great work. Thanks.

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did you pass the course?

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I was facing numerous problems in D. Es but this guy made my life simple

[@georgesadler7830]

Please watch Professor Leonard Differential Equations Playlist. This professor is awesome.

[@Zwalter]

I am really confuse where should you place the "c" because it does matter on where you would place the c because you will get a completely difference answer. Lets take for example on you last example. The general solution I got was: lnx^2 = e^(y/x) +c Now when I computed the value of c, I got 0 = 1 +c c = -1 Instead of e^(y/x) = lnx^2 +c 1 = 0 + c c = 1 Please do clarify on this one because I am having confusions. Thanks!

[@georgesadler7830]

Professor Organic Chemistry Tutor, thank you for a powerful analysis on First Order Homogeneous Differential Equations. This is where Techniques of Integration in Calculus Two is used for finding solutions to Ordinary Differential Equations. Differential Equations is an extension of Calculus. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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[@daltonbruce2053]

Great work but you can also write e^c as k, because e^ to a constant maybe a constant, but it's not the same constant "c". It's different so you have to call it something different.

[@nickvrvilo996]

Just want to say thank you for the explanation and work, you're fast paced yet clear enough to where I can follow along. Keep it up!

[@allahdinosoomro1447]

Can anyone explain y= dy/dx?

[@sjwdestroyer6336]

My "professor" tried to teach us this in two minutes by doing one problem and moving on to the next section.

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Same case my professor just gave a example and moves to bernoulis equations...thanks to this man.

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​@@mrx57503:52 3:52

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[@luker.6967]

Can you explain why we stop at abs(x) = c(x-y)^2? Why is that the desired form? Does it just happen to coincide with applications in the sciences and therefore this is useful for university students?

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[@mrfeliscatus3236]

Question, at 2:14 when dividing both sides by x, why is 2x(xdv+vdx) simplified to 2(xdv+vdx) instead of x(xdv+vdx)?

[@mohhasan9215]

Left side is constant, he only took out the x from the right side as a co_ factor and replace it with 1

[@xogmaster]

the x's cancel so ur left with 2. see: (2x)/(x) = 2

[@amimo2556]

Thank you times a thousand man the three examples are up to the point and the explanations are clear and precise. Now i know how to solve homogeneous differential equations.

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[@winner1338]

0:44 How do you come up with this equation? y = vx Why not, y = x + 1 or anything else?

[@eashanmathur2030]

Thats just the special v-subsitution you need to memorize for homogenous equations. You want to see if the function can be a function of y over x. v=y/x is the same as y=vx

[@winner1338]

@@eashanmathur2030 ok, thank you

[@hitsubitsa5039]

Just think of "v" as a variable that allows x to always have the same value as y for all values of y.

[@briandutch7403]

Win Ner the goal is to substitute v into the equation so you can turn it into a separable equation in terms of x and v and solve it like that. Then you just substitute your v=y/x back into the equation for the answer

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[@ekowdadzie8779]

Is there a video on exact differential equations?

[@zaynumar0]

+The Organic Chemistry Tutor Hello sir , are you going to upload a video in regards to solving second order differential equations?

[@janjansen2091]

the second example was a second order DE

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[@aytakshams7595]

can someone please explain the first example isn't that a linear first order equation , so can we use the integration factor method?

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I'm speechless. Cryatal clear explanation

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[@운동뿌셔]

hello! thank you for your great videos! 9:37 why there is lxl? can't we just write x instead of lxl? + I am not sure where do I have to stop changing the equations. any tips for that plz?

[@ichantongohan5951]

See the x on the right side of the equation. The x in C(x-y)^2. Imagine the value of that x is a negative number. Let's say -1. And the value of y is any number. Let's say -4. So C(x-y)^2 is now equal to C(-1-4)^2 = C(-5)^2 = 25C. The point is, no matter the value of x and y is, when you square their difference the value is always positive. Since the value of C is not yet known, it is always implied as positive. So regardless, the value of the x on the left side of the equation must always be positive. That's why it is written as |x|.

[@Scarabola]

@@ichantongohan5951 i still dont get it :(

[@alesiatsalikou8189]

Hello sir , i want to thank you for your helpful videos on chemistry , physics and maths. I think you made a mistake in the video at 6.00 ... c isnt equal to lnc but to lne^c. Then you can continue the same 😁

[@nboris31]

nope

[@tonito5588]

this is a year late but since lnc can be a value for any constant, it is the same as replacing c (which is a variable representing a constant).

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same here

[@greatsea]

folks need to be careful at 5:34. The -1 is to be considered inside the LN(), e.g., ------> ln [ ( 1 - v )^(-1) ]

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Yes. This threw me off

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[@me.5622]

My solution: 2xdy= (x+y)dx --> dy/dx = (x+y)/(2x) = 1/2 + y/(2x) Let v = y/x --> y = x * v --> dy/dx = v + x * dv/dx (product rule) Now we know: 1. dy/dx = 1/2 + 1/2 * v 2. dy/dx = v + x*dv/dx --> 1/2-v/2=x*dv/dx --> int(1/x) dx = int(2(1-v)) dv, (int = integral) By using algebra we therefore get: y = x + c*sqrt(x) (we might as well let c be positive because it's arbitrary anyway, for now) Initial condition: y(1) = 0 y(1) = 1 + c = 0 --> c = -1 Hence: y(x) = x - sqrt(x)

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[@goodtime8007]

What did you find x in the first problem and y in the second problem?

[@nathaniel4146]

Please explain the proper positioning of C. Because, I got lnY + C in the second problem and that would be a problem. I cant transfer it to the other side cause it would be a negative which will not allow the exponent rule of when exponents are added, the variables are multiplied. Its confusing, from the beginning. Again in the second problem, why is that XY dx is in the left side?

[@saujanyapoudel7378]

You can transfer the C to the other side and still have the positive sign. Since C is an arbitrary constant, the negative of C can still be treated as positive of C

[@saeedhemeda]

Excellent, can i ask about the program of writing used in explanation.

[@OfficialFoliLucker]

This is very helpful in so many ways! Thanks a lot

[@sb27380]

For the last question when we get to: e^(y/x) = 2ln(x) -2C Can we not do: e^(y/x) = ln( x^2 ) -2C , sub in (1,0) e^(0/1) = ln( (1)^2 ) -2C e^(0) = ln(1) -2C 1 = 0 -2C C = -1/2 so, e^(y/x) = ln( x^2 ) - 1/2 instead of simplifying e^(y/x) = 2ln(x) -2C into: e^(y/x) = ln( x^2 ) + C , sub in (1,0) e^(0/1) = ln( (1)^2 ) + C e^(0) = ln(1) + C 1 = 0 + C C = 1 therefore(as shown in video), e^(y/x) = ln( x^2 ) + 1 and why?

[@mugtabaaladhmawee8417]

I just want to say from the bottom of my heart, I love you and I will always love you.

[@juniorjay001]

Why are we able to multiply left side by x but not the right hand side at 8:00? Doesn't that change the equation?

[@eshawn____2373]

The idea is that x/x is the same as 1, therefore, there is no need to multiply the right side by 1 again. It will yield the same answer. He just did that to uncomplicate the left side.

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i'm following this

[@snekz341]

i'm dropping my first pay as a donation as soon as i graduate, you deserve it.

[@rosemariesuplico5526]

THANK YOU. I HOPE I CAN ANSWER THE QUIZ ON HOMOGENEOUS.

[@joekerr7643]

more vids about differential equations pls i like the way you teach

[@sangbryan6028]

9:40 |x|=..... 19:36 y=.... What is actually looked for in the solutions?! I'm really confused

[@PM-vo6po]

i went to comments just to find this... i'm so angry rn. it confused me so much.

[@lenyin3588]

c square comes to c?

[@oosmanbeekawoo]

But you don't get it. Differential equations are not solved with a number. They are solved with an actual equation! At school, they gave you "x + 2 = 4", then asked, "what is x?" Now, they give you "dy/dx = 4", and ask, "what equation, when you differentiate both sides, you get dy/dx = 4?". The answer (in this case) is y = 4x y = 4x is the "solution". Differentiate both sides you see. Easy you say? (Don't forget the integration Constant in our solution, also 😉). This time, you just had to integrate both sides, but sometimes you Can't integrate both sides-x and y are simply not separable. Therefore you have to look for other methods. THAT is what studying Differential Equations is all about! Hope makes sense!

[@airsoftmaster76]

Thanks my guy. Feeling confident

[@lkdjlkjd]

This gave me anxiety. You did great, but this is tedious lmao

[@faithsiphuma9488]

youre right

[@thecircle.2563]

Why does the final answer have an absolute x

[@ΚωνσταντίνοςΛαζαρίδης-ξ9ι]

26:30 Shouldn't we use ln in the last equation and seperate y/x so it stays only y=....?

[@thegreathaitian6323]

Bro has a patreon for those who are feeling generous

[@beergnieer1794]

Thank you sir, for you wonderful discussion, I hoped that I passed my midterm exam by watching and listening to your video.

[@alterez6471]

Studied from this channel in my 10th now I'm back.. Currently doing Btech

[@jorgechavez1570]

why does e^-y/x doesn't influence the fact that the equation at 19:55 is homogeneous?

[@rajj5517]

Y/x has degree 0

[@sunmichoi6888]

Thank you so much for the awesome lecture!

[@akiimutunga3957]

Why did you replace y with vx in the first equation?

[@Lloydhansen]

THANK YOU SOLDIER.

[@gur0004]

For the last qtn, I ended up with C as -1 but when I move C to the other side, I get the exact answer as in the video! So is this still correct?

[@ektorpapadimitriou942]

It is very easy, but it takes some time to solve it, thanks for the help

[@jeremymagalona6401]

@ 15:45 why did u add C in the left side with V? Why u did not put it with ln (x)?

[@geraldm13-codmyt78]

im also confused bro, i thought constant C comes after lnx

[@Hackslooi0128]

is there a second order linear homogeneous and non-homogeneous differential equation?

[@alaajihad5403]

Yes, there is 🙂

[@igbatayotaiwo4911]

Good day everyone. Pls which app is he using to make this video? Anyone to assist?

[@bendavis2234]

Hey thanks for making the video. I have a question with the first problem. When I look up the answer on Mathematica and Symbolab, I get the answer: y = x + c*sqrt(x) This looks different from the answer we got in the video. Is the answer in the video just a different form of this answer, but fundamentally the same? Sorry if this seems very obvious and I'm missing something trivial, but I'm confused on the best way to express the equation at the end of the problem. Thank you! Edit: Also one more quick question. When you write y' should we assume that to be the same thing as dy/dx? This could be why I was confused on the last problem.

[@haydenjacobs1712]

If you did not learn the answer yet, yes y' is another form of writing dy/dx = (insert equation here)

[@bendavis2234]

@@haydenjacobs1712 Thanks, that's what I was thinking too. The notation can be confusing in Dif Eq sometimes.

[@hansvangiessen8395]

1/(1-y/x) = C sqrt x (1-y/x) = 1/(C sqrt x) (reciprocal left and right) 1 - y/x = C/sqrt x (1/C = C, new constant) x - y = C x/sqrt x (multiply both sides with x) x - y = C sqrt x (x/sqrt x = sqrt) - y = -x + C sqrt x (subtract x from both sides) y = x - C sqrt x (multiply both sides by -1) y = x + C sqrt x (new constant, C = -C) I think this is better indeed.

[@crismoreno9533]

Question I did it inversely and I got to the point where you got the anti derivative of. 1/x dx = 1/2•e^v. But at the end I got a different answer because I added the arbitrary constant C to the exponential instead of the natural log what is the explanation behind this? Did I just mess up somewhere?

[@ekowdadzie8779]

Is there a video on exact differential equations?