Non-separable linear ODE y = x+y

In this video, I used the integrating factor to solve a nonseparable linear ODE after expressing it in the standard form

[@shivnathbanerjee5868]

' Never stop learning. Those who stop learning stop living '. Best quote I have ever heard👍👍

[@declup]

Such a cool vibe. Everything's just so crisp and collected. Mr. Prime Newtons, sir, you're the Bob Ross of the blackboard.

[@demetriuspsf]

Amazing! Such a clear way of explaining not only how to solve this case but how to identify and apply the technique for other cases.

[@Mr._Nikola_Tesla]

The Bob Ross of Mathematics! Excellent Work, as always 👌

[@jeroenvandorp]

When solving an ODE becomes your daily relaxation moment. Merry Christmas!

[@clayton97330]

I get it... people have asked me what I do to relax before bed. "I watch people solve differential equations"

[@ensislucis2395]

I am in Calculus BC right now and My teacher said that we would not have to learn inseperable differential equations until college but I was curious so I looked up how to solve them. Thanks for explaining so well!

[@tomaskharoun2262]

Yo wont have to learn this even till after Calc I, maybe in Calc II, but defiantly in Differential equations and Calc III. Good to see that you're getting ahead of everyone. Keep learning!

[@channalbert]

I lalways ove how you develop the algebra, so intuitive! I gave it a thought before clicking the video and i found a lazy way: if you substitute u = x + y, plug it in and solve for u, you get the answer easily by replacing y = u - x; although its always handy to remember the formula for when this doesn't work.

[@brenobelloc8617]

Love this guy. Merry Christmas

[@AcryllixGD]

I did it by letting u = x + y so du/dx = 1 + dy/dx and then solved from there

[@TheOnlyAnonymousMan]

Thank you Sir for your guidance 🙏. Love your videos. First thing that came to my mind was to take u = x+y, but your method must be much safer in other similar cases I guess. The only other thing I'd like to add that I'd normally take , the integrating factor = k * [e to the power of (-x) ] , where k is any constant. The end result here will be the same though.

[@elmekkielfajdouhi7492]

Merci beaucoup

[@klmkt4339]

You are smiling man. 😊

[@PrimeNewtons]

I am!

[@darthtardis5465]

If you use the u-sub method for (x+y), you end up solving (du/dx)=1+u. Once you separate and integrate each side, you are left with ln|1+u|=x+c. You can rewrite this as e^(x+c)=|1+u|, or more simply, c*e^x=|1+u|. Furthermore you can resub the (x+y) in for the u to eventually arrive at the final answer of y=((+or-) c*e^x)-x-1. Does anybody know why the answer in the video can ignore the abs. value function around 1+u by assuming only the positive version? I am curious, as I know we used different methods and still got very similar answers.

[@alessandrotinaoui3428]

I think it's because ultimately it's the c that determine the sign of c*e^x so you can just "absorb the +- in c". Similar to e^(X+c) = e^c*e^x = c*e^x

[@user-3bs8jd83js]

At 9:20, we can also see the LHS as ∫ 1 d(ye^-x). We know that integral of 1 with respect to anything is the same that thing! That is, ∫ 1 d(ye^-x) = ye^-x. That's how the integral sign and the d cancels out!

[@andrewdong3875]

Beautiful handwriting.

[@lukaskamin755]

wow, cool to recollect that long forgotten theory))

[@weo9473]

Bruh this is harder than it looks

[@coreymonsta7505]

The thumbnail is a lie is why

[@coreymonsta7505]

He kinda forgot to explain what mue is and also what even P is

[@kengored_alt]

​@@coreymonsta7505he said that he'll show how to obtain the integrating factor (mue) in another video As for P(x), it's a function of x, just like Q(x), and he uses the P to compute mue, if you watched the video, you would've noticed that

[@weo9473]

​​@@coreymonsta7505 yeah I was confused too. It's like sin = A+B 💀

[@Tisakoreann]

this was amazing!

[@mosespeters5546]

Hey, love your videos! Wanted to know what was the music you used for your intros?

[@PrimeNewtons]

I made the music myself.

[@mosespeters5546]

Hey ​@@PrimeNewtons, oh that is pretty cool! Whats the name of this one you have made? Btw Happy new years!!

[@PrimeNewtons]

@@mosespeters5546 There's no name. You can call it the Prime Newtons Naija Jam

[@yoyogameryt4557]

Sir we can solve this by taking x+y =t then differentiate both side of x+y=t wrt x then find dy/dx and then proceed further

[@flavioing1]

Very good!

[@pedrojesus4967]

Thanks for your work

[@Penguinzo0000]

Thanks a ton!

[@benhassineiem]

Why not just put u=x+y and then go from there. I think it would be simpler, right?

[@KELVINYEMOAHKWARTENG]

Ermm how would it have been done

[@atifny6263]

Sir how to solve x^2 * y' = 3x^2 + y^2 * tan^-1(y/x) + xy ?

[@FilosofisChannel]

Your voice are goods

[@domanicmarcus2176]

, I think that there is 1 accidental mistake.. The final part should be C, multiplied by e raised to negative x and not positive x. Thank You so much sir for all your hard work.

[@timer570]

No, he got rid of the other e's by multiplying by e^x. You could also think of it as dividing by e^-x, which is the same as multiplying by e^x. I hope that makes sense.

[@RicardoPerez-zc4ej]

Gracias sir

[@tapansaharia7115]

Sir,Can you please explain the calculus??

[@Calcufast001]

Great. But is there a way we can solve analytically the DE. dy/dx=x²+y²

[@holyshit922]

It is quite difficult equation to solve This is Riccati equation and you should reduce it to the linear equation but second order and with non-constant coefficients Then try to use power series solution If you can find particular solution of Riccati equation you can easily reduce it to Bernoulli or linear first order but in your Riccati equation you probably get Bessel equation after reduction to the linear second order equation If you have equation y' = p(x)y^2+q(x)y+r(x) you use substitution u(x)=exp(-Int(p(x)*y(x),x))

[@jyotiprakashchowdhury7252]

You present things jovially!Nice but the impressions on the writing board should have been bold and conspicuous.

[@klmkt4339]

Good man

[@amtep]

Can the constants in the answer be folded into just y = -x + c ?

[@nanamacapagal8342]

No, because the C is attached to a function of x.

[@amtep]

Aha I was blind :) it wasn't a constant

[@memotto123]

8:27

[@programmingpi314]

Neither the title nor the thumbnail are correct.

[@programmingpi314]

The question on the title is a lot easier. x=0

[@kyokajiro1808]

technically d/dx=x+y implies d/dx 1=x+y which means 0=x+y so y=-x, dy/dx is not d/dx

[@methatis3013]

The question is not d/dx=x+y The question is dy/dx=x+y

[@kyokajiro1808]

@@methatis3013 in the thumbnail and the first few frames of the video it's d/dx=x+y

[@methatis3013]

@@kyokajiro1808 ah, true

[@Harrykesh630]

Easy!!!!!!

[@sajuvasu]

First

[@disgracedmilo]

if you know basic diffeqs this takes less than 2 minutes (0.5 if you're indian)