How to solve linear differential equations

[@luciouskoroma1661]

such a smart guy, i like his facial expression when he's thinking abt a smarter way to the solution. Thanks man, people like u are needed all around the globe. from sierra leone west africa.

[@Navetsification]

My lecturer for engineering computations was making this seem so much more complicated today, this really helped . Cheers

[@DrChrisTisdell]

Best wishes to all in Belgium. I spent a few weeks at Univ. Cath. Louvain and had a great time.

[@yaadyute]

How come the example at 17:30 isn’t in the linear form?

[@PhysHow42]

My god, man! You are brilliant! I understood exactly what to do after 7 minutes, after being completely confused by lectures and assignments. Thanks so much!

[@koober_]

Thank you so much for this. My professor just gave us the formula and expected us to memorize the process; a method I'm not good at understanding. Knowing where the parts come from, as you've shown, is much easier for me to absorb.

[@senamilebuthelezi2066]

i get the linear intergration more clearer now,thnk you,ure the best professor

[@vivvpprof]

6:00 The way I was taught this is simpler: 1) Solve the homogenous eq. dy/dx + 3y = 0 1a) Take apart y and x: -dy/3y = dx 1b) Integrate both sides: (-1/3) ∫ dy/y = ∫ dx 1c) (-1/3) ln y = x + C1 [ln = natural logarithm] [C1 = some constant] 1d) "Exponentiate" both sides exp(LEFT) = exp(RIGHT) 1e) y = C2 exp(-3x) [C2 = some other constant] 2) Now turn the constant into a variable C2 := V(x) and solve the inhomogenous eq. 2a) y = V(x) exp(-3x) dy/dx = dV(x)/dx exp(-3x) + (-3) V(x) exp(-3x) 2b) Plug it into the original inhomogenous eq. 2c) dV(x)/dx exp(-3x) = exp(-x) 2d) dV(x)/dx = exp(2x) 2d) Integrate both sides 2e) V(x) = 1/2 exp(2x) + C3 3) y(x) = 1/2 exp (-x) + C3 exp(-3x) Voila.

[@thezhancrew]

Wish I had you as my prof for my first course in differential equation...my intro to DE course was messed up!

[@Phughy]

I have so much appreciation for the amount of time you put into this. Big respect sir! You helped me with my maths, I have my finals in a few days! This helped me so much. Everything became as clear as water. Greets from a Belgian Engineering student!

[@kocon03]

first video that has actually helped me understand and see the left side of the equation is the derivative of the of the product taken out in product rule form. This just became so much more clear thank you

[@warronfrench8163]

I took DEiffEq 20 years ago. My professor back then was aweful and I only got a C in the course. I feel like I could go take a chapter test on 1st order DiffEq problems and ace it now. I don't have a need for this knowledge, but it always bugged me that this branch of mathematics escaped my understanding. Now I know it wasn't me. Thank you Dr. Tisdell.

[@MrColinnunn]

As a teacher of mathematics for many many years, colour me very impressed by excellent lesson.

[@jarredslimovitch2725]

Dr Chris, I have been searching for hours looking for an explanation as clear and concise as yours. Thank you very much for clearing these concepts up! Keep up the incredible work, Jarred

[@keiajackson7446]

Great ! only person that i actually understood ! We need more teacher like you

[@mohmd252]

الله يوفقك ويبارك لك ويسعدك يارب يادكتور كريس فعلا انت قامة وهامة في العلم واستفدنا كثير منك ياربي يسعدك ويحرصك ياغالي، فعلا إحنا بأمس الحاجة لأمثالك لتنير الطريق لطلبة العلم، فا بالعلم تنهض الأمم وتتقدم وتتطور أكثر وأكثر

[@mohab.m.metwally]

دكتور كريس كدا مش هيفهمك :)

[@mielu_gazos]

Audio for this video is only on left channel. Maybe u can change it to a mono channel, so it could be heard in both speakers. Thanks!

[@DrChrisTisdell]

+Leonard Gabriel Apostoaie Wish I could. When we recorded it we used two microphones: one on my shirt and one on the camera. The camera one acts as a "back up" just in case the shirt mic fails, but (as you say) is not as loud as the other one.

[@ssebitaabawamala6861]

You are the boss Sir...I finally get the idea at 0242 am

[@menensa]

😂

[@ochiengomwanda6952]

I'd rather watch your videos than attend my lecture's class... So sweet, simple... Thumbs up

[@KommaKameleon]

You've transformed something that seemed terrifying into something that is simple and manageable. Thanks a lot.

[@ripuncleted]

You made this more clear than anyone else. Thank you so much.

[@enadmhd]

oh man, you really helped me a lot. i am from saudi arabia and i will share your videos with my classmates.

[@self_amit7889]

Sir,,i am from India,,,i just love your class very much,,,,

[@castroh]

2:20 is written in the paper: ¨where P(x) and Q(x) is a continuous function of x." As far as i know 1/x is not a continous function of x

[@samirhachemi2447]

Concise and right to the point . Efficient and smooth pedagogical approach . Thanks. ( do yo have any book ?)

[@AR-Race]

I'm going to try and start posting "how to" maths video's but i'm not sure what people would want to see. Im studying a masters in mathematics so anything up to that sort of level really.

[@YungPatron]

Yo u still down to post videos? :p

[@tavocholo18]

Thank you for the video, you're an awesome teacher!

[@alil8419]

ur such a good tutor , u have my gratitude sir...

[@egumawilliam3104]

wow, this is great. I do appreciate the role of mathematical modelling of phenomena in real world science and engineering

[@AtseyohanneslV]

thank you so much you explained it more than my teacher did.....I am good to solve problems now thank you!!!

[@inverse_functor]

I'm just now understanding how incredible mathematics is.

[@omnibrain8]

I am glad to be your student.Thank you for this video.I like your explanations. You are a great teacher.

[@jeremymwilliams]

I happened to notice that the first example can be done by forcing seperation of variables with two iterations of a substitution method as follows:dy/dx + y/x = x let m = y/x→ y = mx which the chain rule impies:dy/dx = m + x dm/dx and with the substitution we have:2m + x dm/dx = x rearranging:dm/dx = 1 - 2m/x the next iteration is:let n = m/x→ m = nxdm/dx = n + x dn/dx and with the substitution we have:x dn/dx = 1 - 3n which seperates as:1/x dx = 1/(1 - 3n) dn and integrates to:ln|x| = - (1/3)ln|1 - 3n| + C ln|x| = - (1/3)ln|1 - 3m/x| + Cln|x| = - (1/3)ln|1 - 3y/x²| + C- ln|x³| = ln|(x² - 3y)/x²| + C- ln|x³| = ln|(x² - 3y)| - ln|x²| + C ln|1/x|= ln|(x² - 3y)| + CC = x(x² - 3y) +++++→ y = (1/3)(x² - C/x)+++++

[@sibonisomdlalose8061]

thanks Mr nice video, however, is it right to integrate a function without dx or dy next to it? e.g i saw you integrating x squared and e to the exp.. 2x without dx next it. please give me valid reason or understandable reasons.

[@charliec8434]

+SIBONISO MDLALOSE d/dx[xy]=x^2 , put dx on the other side so it becomes d [xy]=x^2 dx , integrate both sides, left side is left with [xy] since if u integrate a derevative it cancels, so [xy] = 1/3(x^3) + C and further simplified y = 1/3(x^2) + C/x

[@surrendherify]

+John Doe derivatives ARE fractions in infinitesimal calculus and they work just fine, plus they are much more intuitive then the epsilon-delta business of real analysis. the reason separation doesn't work with second order diffeq's is not because it yields an incorrect result,but because it's impossible to evaluate the resulting integrals- d^2x is just the differential of dx ie. d(dx), so the integral of (-1/x)d^2x cannot be computed as it is, as -1/x is not a function of dx but x

[@surrendherify]

***** maybe i'm being ignorant here, but i've never yet encountered a situation where the derivatives couldn't be treated as fractions. Could you show me some examples?

[@teebiz8694]

thanks sir .... the main issue is to understand is that ..... after multiplying both sides by the integrating factor (IF) the left hand side is is differential coefficient of the product of 'y' and the IF ..... this one thing confuses most students .... this must be emphasized ..... regards

[@MrAlbashiri]

I wash my prof was as organized as you. thanks for the amazing video.

[@arkadiuszjarosz9859]

Great Video with clear explanations!

[@er.shailendrasingh7731]

its a very nice video..explain the things very clearly

[@ampofogyekye5372]

this tutorial has really helped...thanks

[@shantvfunnyvelogs6085]

Thank you sir for giving this lecturer it's very helpful.from pakistan

[@DrChrisTisdell]

You are welcome. If you found this useful then please take a look at my new ebook. The link is in teh description.

[@FuzzyAason]

May I ask this for the final question? Why didn't we multiply t^2 on the RHS? Cos t and leave it as it is?

[@KommaKameleon]

What exactly do you mean? At what point in solving the equation are you referring to? And please be a little clearer.

[@iamben28]

for the 4th example, d/dx(t^2.y) = Cost. (t^2 , is not shown. did he not multiply the integrating factor to the RHS ? )

[@RonewaNematswerani]

:) thanx so much for the help, but i have a question :isnt that square suppose to be cube on the final answer since we integrated X square?

[@ennajariahmed4517]

100000 thanks from morroco casa blanca

[@User12345fan]

You're a great professor

[@TheYls3]

Thank you so much. I am completely understand these stuff because of you :)

[@WithASideOfFries]

Solid, solid video. Thank you for the clear information.

[@sanjubhati5476]

Thanks for it

[@jakedixon889]

I wish my teacher was an eighth as clear as you are. And that's coming from a yank. Good stuff doc.

[@reubenwilliammpembe667]

Thank you Sir you are the best!!! #RespectFromSouthAfrica

[@wraywray]

best tutor in the world xx

[@khaledabdelsalm1667]

Thank you Dr. Chris

[@jkzhou7021]

thanks o lot,sir,you have got a new subscriber!!

[@mohab.m.metwally]

thanx Dr chris it was a very useful lecture

[@nickcooley2857]

best vid on ode linear !

[@sree4448]

tat is such a cool explanation.... thank u so much chris.

[@Kyleser1BlogspotCa]

This was SO helpful! I can't thank you enough!

[@DrChrisTisdell]

I am very glad to know this video was useful to you.

[@shivdurgamandirjamalpur3096]

Very informative video thanks dear 🙏🏼

[@balikakamble1433]

thank you so much sir. it is so important and useful

[@bilalsiddiqui314]

Balika Kamble Yaaaaaaa

[@MukulChaudhary07]

Best Explanation!

[@bigmacmonster]

You just saved my life

[@otienojohn6334]

Quite good and well done....

[@javiermendez7080]

at 23:39 if we're integrating with respect to "t" shouldn't the result be ((t^3)/3)y ??

[@AbirProchchhod]

Javier Mendez No. Because we're integrating [d/dt (t^2 y)], that's not yet simplified or derived. And for that reason, only the operator (d/dt) is removed.

[@RealationGames]

14:50 How can that be right? How is this valid: x*(dy/dx)+y=(d/dx)(xy)

[@a7med3la2]

it is valid since (d/dx)(xy)= derivative of first term (1) multiplied by 2nd term (y) Plus derivative of 2nd (dy/dx) multiplied by first (x) ^^

[@halarsanjrani8583]

great work sir...thanks alottt

[@kristycho4620]

Hi, I tried to open your link for your ebook for this topic, but it also sent me to a bookboom page saying "Not Found". Is there a way I could get a copy of your ebook still? Thank you!

[@godpain8298]

Love you sir 😍

[@Surviving_With_Nthangeniph]

hi in the first example we actual have p(x),which is equal 3x^0.so our e^3x is not miraculous .nice lecture doc

[@vaas695]

excellent explanation

[@dsalazar108]

Thank you so much! It's exactly what I needed.

[@gendrygaming9062]

isn't it that the derivative of e^2x is just 2e^2x? why is it (1/2)e^2x at 7:30?

[@gendrygaming9062]

oh yeah noob mistake its integral not derivative. haha

[@omprakashchaurasia7761]

thank you sir this helps me a lot..... again thanks....

[@syedusamamanzoor1838]

Dr. Tisdell! thanks a lot sir.

[@tanzulalfiya4628]

Plz make more videos on it.

[@DrChrisTisdell]

Very good luck with your studies!

[@ThePepinator]

Can someone help me on this. I have a question on the rearregament step at the end, for example at time 21:08 of the video, and don't understand that step.

[@BaSs_HaXoR]

this is amazing. thank you. 😀✌️

[@dduqueinvest]

Thank you, you are great!

[@SurajSingh-pw9ew]

sir I have one doubt...while finding integrating factor why we don't write+c while applying integration

[@patrickmoloney672]

e^c is a constant. You can just add it to the constant of integration.

[@londelakea8630]

ths is marvellous lecture

[@黃士哲-y6j]

Thank you professor!

[@akshey26]

you need to lecture at my university!

[@samuelgordon8950]

Thank you so much! Very helpful!

[@n6151h]

Very, VERY hepful!

[@stinkiekillerdeclerck8964]

Thanks for making this video :)

[@khaledabdelsalm1667]

RealationGAMES it is valid when you apply The Product Rule for Derivatives

[@valor36az]

what a great lecture

[@HEHE-hr7df]

the video clearly shades light!

[@davebocking1750]

You are my hero.

[@mohab.m.metwally]

why u did not write the constant in the integral factor after integration

[@nathanpavlidis1520]

this guy is awesome

[@mbisetakobana]

Thank you Dr.

[@spartanmsufan1]

Amazing video

[@tshifhiwathovhakale9385]

ithink you dont explain fully,how can you jas choose e^3x...one has to intergrate to gt tht one

[@beingsajid110]

Thanks for ur honest guestures.,.,.,

[@omarfarukh8200]

Sir please fix the quality of the Sound

[@gfferre302]

Is this a first year calculus vid?

[@rodericklenz5030]

C Park It could be part of a first year calculus class, but ODEs are complex enough to make an entire course on their own as well.

[@secretarygeneral]

Congratulations!

[@shakroka3174]

can you solving equation cos(X).dy/dx +y = sin(x)

[@PrajodChemmarathilPrasad]

+shakroka Divide both sides by cos(x), dy/dx+y/cos(x)=sin(x)/cos(x) Remembering trig, 1/cos(x)=sec(x), sin(x)/cos(x)=tan(x) dy/dx+ysec(x)=tan(x) Use integrating factor e^(\int P(x) dx), P(x)=sec(x), Q(x)=tan(x) \int sec(x) dx=ln(tan(x)+sec(x))+C e^(\int P(x) dx)=tan(x)+sec(x) Solution, y(tan(x)+sec(x))=\int tan(x)(tan(x)+sec(x)) dx To solve the integral , we expand the brackets, ytan(x)+ysec(x)=\int tan^2(x)+sec(x)tan(x) dx We know, \int tan^2(x) dx=tan(x)-x+C, and d/dx(sec(x))=sec(x)tan(x), so we have, ytan(x)+ysec(x)=tan(x)-x+sec(x)+C You can handle from there.

[@balunkaba3629]

Thanks for assistance

[@WithASideOfFries]

27:08 OH COME ON HOW CAN YOU TEASE ME LIKE THAT ARE YOU SERIOUS WTF BRO

[@huldaimeldapokolo]

Thank you very much.. I