Solving Differential Equations with a Composition (Obvious) Substitution (Differential Equations 22)

[@buddhakonwar3791]

If someday in future I achieve something in mathematics then I'll give all the credit to you for making my base strong enough to deal any kind of mathematical problem. Thank you so much Sir.

[@husamobaid859]

I can't thank you enough. Your efforts are changing people's lives for the better.

[@scorpionedge]

Best prof on the planet! Never stop this incredible talent of yours!

[@GaryTugan]

Hi Prof Leonard! I found your vids a few months back. They've been extremely helpful in reminding me of specific problem types and their solutions since its been a LONG time since I got my engineering degree from VA Tech (1987). Yeah, so it's been awhile, and since I am now working with students in math and engineering and the like, I like to keep myself fresh in the know. So, thanks for these vids!! In watching this particular vid, I found an interesting point in the 2 different ways you went about solving the integral. At 22:35, you've just shown the u-sub method. And then you go into the other really cool way. In doing both methods, I noticed a difference in the two answers... U-sub gives "v+1- ln(v+1) = x + C", but the 2nd method you showed gets the same thing, but without the +1 part on the left hand side. I realize you can bring the 1 you get when using the u-sub method over to the other side to join the constant and it simply makes a new constant. But still I am most intrigued as to why the two different methods give seemingly "different answers". I imagine you noticed this as well?

[@lanefrasier3890]

I was wondering the same thing

[@GaryTugan]

@Erik Hamidullin that was my evaluation also as to why it works out. But it still doesn't exactly explain why the two methods create different answers that are a constant (out of phase) so to speak.

[@Frostasy]

I've learnt so much from you over the years. You're helping me so much in my EE program.

[@lacelulaqueimplota3955]

I swear to God I will pay you back the day I am able to. Unfortunately, right now I am a poor bastard. I cannot thank you enough Professor Leonard. Thanks for everything man. :')

[@aaronguerrero592]

So this is like "Reduction to Seperable"... AWESOME!

[@geethapillai4701]

Never knew it was possible to understand calculus to this extent, thank you so much Prof Leonard ^^

[@darkcry2682]

your making the world a better place sir, thanks for yours videos

[@CHRISTIANZEGARRA2018]

Question: How far are you in the course of differential equations? Up to this video is this considered half way through the course or beginning? I’m taking differential equations next semester. Couldn’t have passed calculus 1-3 without your videos.

[@habeebijaz5907]

It's only 20 %

[@Peter_1986]

@@habeebijaz5907 He said in a comment section for one of the earlier videos in this playlist that the Laplace Transforms "will be covered later" or something like that.

[@AmitKumar-oq6qu]

I will meet you one day just to say you "THANK YOU SIR" .......

[@supernovae9990]

Will you upload solving homogeneous and non-homogeneous differential equations using power series technique? Also the Laplace transform ?

[@alrayyaniQtr]

I've noticed you've been stressing yourself over this course. Dont burden yourself with a big promise such as a complete course on DE. The load will be stressful and you'll hate cause it will feel like a job. Just make a video on a topic when you feel like it. You'll find yourself actually enjoying it not stressing it.

[@ProfessorLeonard]

Thank you for that :) I've taken a small step back and it's made a difference. Family is happier too!

[@alrayyaniQtr]

@@ProfessorLeonard glad to hear it :)

[@itsamarmot]

@@ProfessorLeonard Hey I noticed it's been a while since your last Diff Eq video but I want to reiterate what an impact you've made on so many people. I've always had a rough time with math (especially DiffEq), often getting frustrated and spinning my wheels. It's been an uphill battle with my engineering classes until now. I'm so grateful to have found your channel, you do an unmatched job of explaining these difficult concepts. You literally change people's mindset about math and that is very, very unique. None of my professors have even come close to explaining concepts this well. Please keep doing what you're doing. I'd be overjoyed if you complete the DiffEq playlist. The world seriously needs more educators like you.

[@diyartok2518]

@@itsamarmot so he hasn't completed yet ? I thought that is all

[@thomasjefferson6225]

@@itsamarmot hes caused me to love a subject that I loathed before. DE's are fun now!!!

[@mausplunder5313]

really nice video... i think u can drop the abs value on the last answer because sqrt(v)+1 can never be less than zero.

[@theoendre6484]

the substitution in the last example is better to be v^2= x+y+1, it will make the integral a lot easier, you just need to be careful when taking the derivative of v^2 after solving for y, also be careful at the last step when you want to replace the v's, because your sub was v^2 & not v. Hope this will help

[@garethpienaar3119]

24:47 When I use the u sub method for that integral, I end up with the whole integral equating to (v+1 - ln(v+1). And this is different to the (v-ln(v+1)), which is what Professor Leonard got. Which one of these answers is actually correct, because when I google it, I keep getting different answers.

[@marklando303]

Was wondering this too. It's both correct, because when simplifying the C constant can absorb the 1. Hope this helps

[@garethpienaar3119]

@@marklando303 Oh okay, thank you very much, now I understand.

[@MikoPellas]

ran into the same problem, worked out in the end anyway using the constant

[@volkanulker6432]

can we use trig sub for 33:00 ? We can rewrite equation as 1/ v ^(1/4)^2 + 1^2 and gets arctan

[@r1e2y4]

I'm sorry but I'm a bit confused about the sub technique @22:08 if you let u=v+1 and after you integrate you will get u-ln|u|=x+c1. Replacing u with v+1 I'm getting (v+1)-ln(v+1)=x+c1 and putting back v=(x+y), I end up getting (x+y)+1-ln(x+y+1)=x+c1 which is different from your answer doing the other technique. Someone help.

[@bloodyadaku]

When you take the derivative of your substitution, shouldn’t dv/dx technically be a partial derivative since it’s a function of two variables? Is it ever going to make a difference? Obviously in these examples it didn’t matter because the x’s and y’s were separate but if you had something like v=sin(xy), y=arcsin(v)/x, then when you take the derivative your dv/dx would then have to be a partial, no?

[@BoyleFrancis]

Going to the fitness center worked out...thousands of more subscribers.

[@borisstrunjas8612]

Professor do you can divide videos and make playlist 1 and 2 for calculus 2 and 3? I need transcript because english my second language. I learning math and english langauge through your videos:-)) Thanks a lot!

[@mesutbakhtyari]

Definitely you gonna helps me a lot next semester

[@Peter_1986]

So basically this is a more general version of the "v = y/x" scenario?

[@ProfessorLeonard]

a very similar idea

[@bonbonpony]

So, once we get such an implicit solution, how can we verify it now that it is indeed a solution to our original differential equation? :q I mean, if it's not solved for y, we can't calculate its derivatives and substitute them back into the ODE to check if they satisfy the equation :/ Is there any other way to verify this? (Maybe something involving implicit differentiation?)

[@karimaessatty1168]

thank you professor .. I

[@Pete1711]

Very helpful, thank you!

[@santiagoblandon3022]

You should have a course to teach teachers how to teach :D

[@b071139]

I like it !!! Take it to the other level !!!!💪🏽💪🏽💪🏽💪🏽

[@AlphaNiner-yk8kh]

That moment when the marker died tho lololol 23:05

[@nicksardelli890]

if you do a u sub then you get v + 1 - ln|v + 1|. So there's an extra + 1 in the equation

[@yashrajjaiswal6051]

Eagerly waiting for it

[@Festus2022]

why did he get 2 different answers for the integral ofV/V+1: using u-sub (@ 22minutes)-->V+1- ln(V+1), vs. the alternate way --> V - ln(V+1). Presumably, the +1 by the u-sub method is incorporated into the C of the alternate method.

[@haakonkk]

Can you make a video on the wave and heat equation?

[@Peter_1986]

You could always check out Chris Tisdell's channel. He has several videos on PDE equations - including things like D'Alembert's Equation and applications of the Fourier Transform etc - and he shows the same genuine passion for teaching as Professor Leonard, and is just as likeable.

[@oorjitgupta]

in the root(x+y+1) problem, taking the entire root as v, then squaring both sides, and THEN differentiating makes the problem a cakewalk

[@wesseljooste414]

In the first example problem, when working with 1/v^2 + 4 , is it incorrect to divide the bottom by 4 to get 1/4 (v^2 /4 + 1) ?

[@shaheer_ghazi]

That is not incorrect. Going along with that step, v^2/4 can be rewritten as (v/2)^2. So, the integral would be tan^-1(v/2) multiplied with a factor of 2 to cancel out the chain rule of the v/2. You would have gotten the same exact result as the Professor so your method is just correct as the Professor's.

[@nikolozperadze4887]

Awesome!

[@aaronguerrero592]

36:25 ... answer as to why there is an extra +1 in the first option (u sub) 22:08

[@rohanpradhan5669]

Yeah, I have the same problem

[@rohanpradhan5669]

v+1-ln absolute value of v+1

[@cheesecake7434]

23:22 Pardon my french but holy shit that's beautiful

[@godblessamerica9313]

Yo proffesio love ur video thank you for dummin it down for students. I have not seen u in a while

[@kaizen9451]

God bless America he’s not dummin it down, he’s explaining it well!

[@godblessamerica9313]

@@kaizen9451 we all have our opinion to you it appears his explaining it well. I say his dummin it down. Teachers cant even explain it as he does

[@dk-xn9el]

Sir u a really a good teacher but what about differential eqns of second order you didnt talk about them

[@ProfessorLeonard]

We will get them

[@dk-xn9el]

Like when because we are soon to finish diff eqns this semestre

[@dk-xn9el]

And l really like to watch yo videos ,the way u teach is too good

[@DemiHalf]

can't we just not seperate anything and take the integral of 1/(x+y+1)^(1/2) dy? no? ok.

[@hafizsahib4351]

Please upload lectures of complex variable and transforme

[@duckymomo7935]

the words obvious and trivial are hilarious in math.

[@Peter_1986]

I hate when a math book keeps saying "obviously" when everything is a confusing mess, lol.

[@duckymomo7935]

@@Peter_1986 apparently r^2cos(θ)sin(θ)

[@Peter_1986]

​@Mi Les "Obvious" is a bold statement, but it becomes obvious when you realise that the term sin(θ)cos(θ) can never have a value greater than 0.5, since sin(θ)cos(θ) = 0.5sin(2θ)

[@Peter_1986]

+Mi Les I noticed that I didn't mention that the maximum value for the left-hand side could be either +0.5r^2 or -0.5r^2, depending on which angle you pick (+45° or -45°), but either way, both of them will give a value that's less than r^2.

[@shubham1999]

Sir, can you tell me how to use the method of multipliers to solve linear PDEs?

[@taekwondotime]

If you do the u-substitution at 22:00 and you get u - ln|u| + c, how does it equal what you get using the other method? Using the first approach, u = v+1, so you get: u = v+1 ---> v = u - 1 du = dv = u - ln|u| + c = v+1 - ln |v+1| + c whereas the other method gives you: v - ln|v+1| + c What happens to the +1?

[@taekwondotime]

EDIT: Never mind. He answered my question later in the same video! :)

[@nicksardelli890]

dividing by v^2 + 4 was strange but ok

[@spanishlanguageeducational3737]

Nice shirt.

[@marquez2390]

U natty brah?