Nonhomogeneous 2nd-order differential equations
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- Опубликовано: 9 ноя 2011
- Download the free PDF tinyurl.com/EngMathYT
A basic lecture showing how to solve nonhomogeneous second-order ordinary differential equations with constant coefficients. The approach illustrated uses the method of undetermined coefficients. I present several examples and show why the method works.
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Dr Chris, you have no idea how much your lecture helped me. I was prepared to fail my class test but this video literally saved me. Thanks a lot sir. Regards from Singapore.
Same here bro.. From srilanka ❤️✌️
i watched this before my ODE test, very helpful, from Kenya😁😇
Watching this youtube is much much easier than reading a 300 page book on differential equations. Dr. Tisdell is a brilliant teacher and doesn't rush through examples. Very down to earth.
Dr Chris, you have no idea how much your lecture helped me. I was prepared to fail my class test but this video literally saved me. then pleace lecture another class differential equations in some initial points.
This is absolutely amazing, and definitely the best tutorial on RUclips. Thank you so much for this.
This video is literally amazing - following through the examples made the methodology clear enough, but then the explanation at the end w.r.t. where this y sub p has come from brought it all together beautifully. Thank you!
Man, you're doing such a great job! And you explain very well, be blessed!
Dr Chris You are a true genious mathematician. your Lecture has helped us with regards from Soloz, Taps, Chimz and The Skywalker straight outta Zimbabwe.
I think I finally understand 2nd Order ODE's now, the 2nd step was always pretty difficult for me. Thanks for the video, cleared it up for me.
with 1 hour 40 mins and no experience with 2nd order diferentials, managed to get the last bit of my uni coursework in, just on time thanks to this video, thanks allot!
😄 Thank You So Much! This helped me a lot! I can guarantee you that College Students will pass their Math Courses with your help. Same goes for Everyone that created the Math Tutorial Videos in RUclips.
Thankyou Dr Chris, my lecturer is foreign and i have trouble understanding him, thank god you are also australian! You have literally explained this perfectly for me and i understand how to do it! THANKYOU!
You are my saviour, after watching your videos I cant believe I didn't get it before, and that is a testament to how straight forward your explanations are.
Excellent teaching style. Explained what's usually considered a tricky concept in easy and accessible language
The setting out is masterful...the meta-cognitive path to success is masterful and meaningful, even for a math challenged me!
THANK YOU SOOOOOOOOOOOOOOOOOO much!
thanks a lot, Dr Chris ,it is easy to understand when you explain it , many teachers explain this but actually they were unsuccessful
Watching this video is more helpful than listen my prof talking.
Thank You!!!
You sound very confident! That's great. I would suggest you mentioned or stated the C.F. and P.I. for more clarity. But that's fine for an A-Level student. However, I guess you might have covered it in your previous videos. No problem. Keep up the good work ! Thanks so much.
Thank you so much! just when I lost all hope and was about to give up I found your videos!
Sir... THANK YOU!! PLZ MAKE VIDEOS LIKE THESE AND SAVE THE STUDENTS FROM IMPENDING DOOM i.e exams 😀
Thanx a lot for your videos...I haven't been in touch with differential equations for a very long time and this helped me out.
Once again your video teaches me more in 5 minutes than my professor does in the 1 hour and 15 minutes he talks. Thank you so much!
Thank you so much Dr.Chris
You've saved a student
Your videos are absolutely amazing, thank you so much!
thank you so much for posting this video! it was very helpful
petroleum engineering student here, you saved me from my semester 2 exams
Thanks for this video, have a calc final about this in two weeks, your video was really constructive. Where some other videos suggested "guessing" the particular solution, you show me how get this done much more constructively :) Cheers!
you are a saviour :) explained so straight forward thx.
This video helped me a lot, greetings from Norway!
Thank you for this brilliant explanation
Thank a lot Dr. Chris, I learned a lot from this video, your explanations are very clear and precise.
But I still have a question, what about the case when G(x) is a trigonometric pr hyperbolic function ??
Thanks - that was my aim. Sorry for the late reply.
You have helped me tremendously
hi Dr Chris ,
thanks for this video , which help me a lot.
Your equation is a linear, first-order problem and so can be solved using an integrating factor. Can you see what happens if you multiply both sides of your ODE by x?
you are a good man Dr Chris
You're the best Dr. T!
:D
great video Dr Chris
you are so good at explaining things :D thank you.
Sir homogenus equations ka x(√x ka square +y ka square) -y ka square .dx+xy.dy=0
Ka sol
fantastic video u make it easy
Dr. Chris you are effing awesome.
Thank you! Finally makes some sense!
@The04tacom Your suggested roots do not satisfy the characteristic equation, as a simple substitution with reveal. Can you see why my roots are correct now?
Can you just multiply your assumed Yp by x^10 or something so that it's likely to have unique terms at the end? Or does the power of x have to be the minimum possible for it to work?
How do you record these, Doc? I'd like to know. Thank you.
Hi Dr Chris Tisdell, would like to know why is r^2-5^r+6? Supposed to be some substitution by the following; e^lambda(x) , which is why it was reduce to r? And is r an arbitrary constant?
Dear DR Tisdell, thank you very much for your video. If you don't mind me asking, do you tutorial on matrices? Thank you very much for your kind attention.
thank you very much you are a great teacher
Hi Chris, I'm loss in the equating constants particularly on how you would choose the constants wrt to C/D. i.e. 0 - 5C+ 6[Cx+D] = 2x+3. 6C=2 & -5C + 6D = 3. I'm assuming you differentiate away the constant to get 6C = 2 for 1st part, while the rest is about the constants? If that's e' case why not entire - 5C+ 6[Cx+D] = 2x+3?
the coefficient of x is only 6c hence equated to two
Very very helpful. thank you
thank you! it helped a lot
thanks very much! it is very helpful!
The best ever the clearest one
Genius! very helpful
this helped me so much
Hi - the download link is in the description. Best wishes.
Im not sure why my teacher teaches this like it's Rocket Science. Thank You for the video. I was wondering if you have those notes available that you used during the video.
@The04tacoma -1 and 6 do not satisfy the equation r^2-5r+6=0, instead they satisfy the equation r^2-5r-6=0. Note the minus sign! Problems can be a pickle if you don't keep track of your signs ;)
You are the best
that was really a god dam awesome good lecture.....
Hi, this is great video! Thank you! Well, I have one question. How can I solve this problem: y''-y'=10? yp=A, yp'=0 yp''=0
... 0-2*0=10 how to find A? I hope so that question isn't bad as my english :)
thank you so much it was very useful to me but how we get C=10 ? i hope u show more details because we need that details
what a life saver
can you direct me to the lecture you talk about this type of ode
d2y/dx2=1.5(exp3X-y)^0.5 , i am trying to get the "y" on the other side to solve but dont know how to start it
sir,plz explain how to solve y"-3y'+2y=14sin2x-18cos2x. and how can we take the particular solution for this DE
U da real MVP
Sir can u plz ..explain me in the place of G(x) a constant is placed then how we can solve
How about for the R.H.S., it is a constant?
Just the best
Thanks! :-)
I might be missing a rule, but should it your y'' not be 25Ce^4x ?
Why is the solution for the homogeneous equation included at all? I don't understnd how it affects the general solution to the non homogeneous equation if it equals zero.
Because the complete solution is y = y_h + y_p.
If we used y = y_p as the incomplete solution, we would be missing solutions.
We can show this formally.
plug in any solution of the form y_h + y_p into a y' ' + b y' + cy = g(x)
where y_h satisfies: a y ' ' + b y' + c y = 0
and y_p satisfies : a y ' ' + b y ' + c y = g(x)
Plugging in...
a ( y_h + y_p) ' ' + b( y_h + y_p ) ' + c( y_h + y_p ) = g(x)
a ( y_h' ' + y_p' ' ) + b ( y_h ' + y_p ') + c yh + c y_p = g(x)
a y_h' ' + a y_p' ' ) + b y_h ' + b y_p ' + c yh + c y_p = g(x)
rearranging we have
(a y_h' ' + b y_h ' + c y_h ) + a y_p' ' + b y_p ' + c y_p = g(x)
But the left side simplifies to 0 + g(x) = g(x).
We get an identity, so we have shown that y_h + y_p does indeed satisfy a y' ' + b y' + cy = g(x).
I am working the exercises in the eBook for this video. Specifically, Exercise (b) where G(x) = 2x. The Ans has yp = 2 exp(-x) - 4. However, exp(-x) is also in the homogeneous soln. Of course exp(-x) works because any const * exp(-x) is a solution to the homogeneous solution. I believe the actual soln to yp=2x-4. Also exp(-x) is the wrong form for g(x). g(x) should be of the form Cx+D. Using this arrives at the correct answer.
I'm working from the ebook and am stuck on question (B) as well. It seems as if the professor has changed the answer to yp=2x-4 as you suggested, however, I cannot even get that far. What did you use for your yp, y'p, and y"p?
Brit, I do not have the ebook any longer, if you provide the problem statement, I will help. Larry
my teacher can go hell now. Thanks for explaining
THANKS BROOO :)
+Maweth OVO No worries, Mate!
Thanks
Thanks ❤
wat chapter is this
Thanks man
17:21 ME TOOO
Thaaaaanks!! :-)
the videos are great but the explams are same in first and second video and the are simple can you solve hard problems for us thanks
how to solve third order derivative?
goooood stuff
the best
Great video (though it's "aych" not "haych").
How to solve the exact same first example done here on MATLAB? Anyone
19:37
should yp be read as the following
yp''= 2Ce^2x(1+2x)
Hi - I don't think so. Can you show me your working?
Dr Chris Tisdell
well
yp = Cxe^2x
yp' = Ce^2x(1+2x) i.e. Chain Rule
-------------------------------------------------
yp''
so i expand yp' C(e^2x+ 2xe^2x) and differentiate again
using this formula of differentiating (u+v)' = u' +v'
C(2*e^2x + 2[(x)(2)(e^2x) + (1)(e^2x)]
C(2e^2x+ 4xe^2x + 2e^2x)
C(4e^2x + 4xe^2x)
Ce^2x(4+ 4x)
well what did you know i made a mistake :) lol sry abt that
use wolfram alpha
I really like your teaching, but I would appreciate a more clear handwriting!