Variation of Parameters - Nonhomogeneous Second Order Differential Equations
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- Опубліковано 29 вер 2024
- This Calculus 3 video tutorial explains how to use the variation of parameters method to solve nonhomogeneous second order differential equations.
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So many The Organic Chemistry Tutor chrome tabs open..
warryen cramming for finals, so many tabs of this dude open I opened a whole new chrome to make it easier to see...
This just summarised a week of lectures, equivalent to like 15 hours
My professors from the University need to take a lessons from O-Chem Tutor for how simple his videos are. Quite sad and ironic that many of my Upper Division Professors were unable to adapt to Zoom and online teaching
Professor Organic Chemistry Tutor, thank you for another exceptional video/lecture on The Variation of Parameters in Calculus Three/ Introductory Differential Equations. The example in this video is a well-known problem in all of Introductory Differential Equations. Differential Equations books used Wronskians, to solve this exercise. Although Wronskians was not mentioned in the video, it is also used to solve the problem. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
I got an Amy Schumer Tampax ad in the middle of this video 😫
Can i consider those equations as undetermined coefficients?
No that has a seperate video i guess
from struggling over calculating empircal formulas in grade 11 to struggling over calculating ODEs in 4th year university, the struggle never ends but you remain the GOAT. Keep up the good work brother.
we have to solve ODEs in our first year 🥲
@@priyanshusharma1812 second year for us
Yo why they got us doing this in our second year :(((
Same@@priyanshusharma1812
@@priyanshusharma1812 😂😂 ouch
It's the second yesr for us
How the Condition at 2:57 came? Is it Pre decided or Taught in Some else lecture
Plz tell *Professor* ⚠️⚠️⚠️⚠️⚠️⚠️
Just in Time! We have our DE final in less than 2 weeks and I HATED variation of parameters. You are the best! :)
waw mine is tomorrow
mine is in 4 hours 😁
Mine in 2 hours😅
mine is in 5 days, goodluck to everyone in the future
mine is in 29min, damn what am I doing with my life?
My professor sucks at teaching this, you're out here saving lives right now.
This inspires me to drop out of engineering school.
Engineering school presents an easier format of DE compared. No need to write proofs.q
To me is crazy how me pay thousands of dollars to universities to take these courses, but we always end up coming back to the organic chemistry tutor to actually learn for free.
You said it.
I wish there were more sample equations, these are so helpful!!!
The differences in teaching the same material differ from both teacher to teacher, but also country by country. In the Netherlands, they taught us this method in relation to the Wronskian and lin algebra. It is nice to see this different perspective as well to solidify the material thought to me.
sounds like the Netherlands teaches it stupidly
😂😂😂
They taught us with the wronskian too
we in india also been taught in the wronskian method
Yikes looks painful for a beginner to grasp
this whole thing is confusing
Now when you add Power Series, Bernoulli Equations, UC Functions, Laplace Transforms, Separable Differential Equations, Integrating Factor, you end up with a very hard class.
I was waiting for the second example bcs the first one have imaginary nums LOL
Men you're my university education saviour. I owe you lunch.
You are too fast slow down 😭for people to digest in the equations
At 1:29, why do we pick r1, and not r2? Could it have worked either way, meaning we can have two differing yc's? One where beta is 1 and another where beta is -1?
Why don't you use Wronskian formula to slove this?
Think he does without explicitly saying so. When he does the whole Product Rule of Primes, isn't it basically Wronskian?
thanks so much for this, much easier when you explain step-by-step rather than comparing textbook to bare-minimum "example" vids teacher provides 👍
These tutor and some more visualisation based chanels are a gem to society.
The integral of -tan(x) is -ln|sec(x)| is it not?
It is. With properties of logs, you can show that it is also +ln(|cos(x)|)
How does the u primes add up to zero when the sin and cosines aren't second derivatives
yea
Please make a video on how to solve DE’s with power series next!!!!!!!!
Have you thought about doing another example of one of these but using the Wronskian method? It has another concept to understand, but it makes problems like this take a bit less work and makes them a little easier in my opinion.
this was great content for teaching the topic and all, BUT THIS IS THE MOST BORED IVE EVER BEEN EVER!!!!! Talk about something cool like cars or women... NOT MATH!!!! NERDDD!!!!
What is the software you use to make these videos? The drawing platform.
smoothdraw4
Not among your best tutorials imo
This lecture was confusing for me
for the first equation you made y_c wouldn't you have sin(-x)? since you have to still use -1i no?
Wronskian method is easier
v1' = w1 / w
v2' = w2 / w
integrate to get v1, v2
yp = v1y1 + v2y2
If there was a cot x instead of a sec x, then how can we find the values of U1 and U2 in elimination process.
not to be that guy but equations are not milk. there's a difference between homogeneous and homogenous.
Jackson Shirley Clark Anthony Davis Matthew
Really helpful, thank you!
You're not just telling stories, you're saving lives
i just want to graduate and then probably will end my life i am just not eligible for this world, thanks for making my miserable life easier, liked and subbed so helpful
Hey, don't say that. We all have our own problems going on in each of our own lives. But we can find hope in the midst of suffering. Faith in God, and specifically in His Son Jesus Christ gets me through tough times. Jesus died for our sins and rose again from the dead, so I know that by believing in Him I am forgiven for all of my past mistakes and free from the guilt of my sins. We can live with the assurance of life beyond the grave by believing in Christ. Please reach out to me if you need someone to talk to. God bless.
Thanks so much sir
Ur taking tiz degree home
Why in 2:41 you write down that condition U1'y1 + U2'y2 = 0?
Thanks God blessed you
Amin
bless your beautiful soul for making variation of parameters much easier to understand!
how did you obtain the c1cosBx+c2sinBx on 1:45
When solving homogenous equations, there are three types of solutions. One of them is complex roots. That’s the given equation for complex roots whereby the only thing that changes is the alpha and beta. The layout of the equation is same in complex roots..
1 year university student here. Why tf we need to solve this?
Hardest way to solve any question
Sir where is the first order? I can't find it anywhere😭
Bro is the only one who is holding me up in university
I watched 6 ads in 11:35 minutes like how much time did i waste ?
Isnt this supposed to have a wronskian involved somewhere? im confused :S
Yes u r correct
I am just an average dude trying to go for math and physics and these videos have been helpful to keep me up with my peers!
2:49 WHY that condition is applied? What's the reason?
I know that is helpful to find U1 and U2 but why?
please i still dont know, does anybody know?
@@sanjaysivapragasam3795 Both U1 and U2 are unknowns, we have to eliminate them and write in terms of Y1 and Y2. Two unknowns, we are free to apply to conditions. That's is why it is chosen in such a way that it will eliminate the unknowns.
Beautifully explained in book SL Ross
Is that condition u1"y1+u2"y2=0 for all Variation of parameters, or was it just for this question?
Assuming that u'y1+u2y2=0 because we hve two unknown i.e u1 and u2 but our original differential equation is 1 equation , So inorder to get an equation for the the unknowns
5 seconds into this video:
Nah bruh, imma drop out
Can I ask why we are multiplying by wind and cosx at 7:28?
Please and thank you!!
I know I'm super late in your response, but the answer to your question is Systems of Equations. After multiplying the sinx and cosx, you're able to cancel the u1'cosxsinx.
At the integration step , you integrated without adding a constant c , wouldn’t that create a problem in the solution?
Thank you.
According to me there should be arbitrary constants in there. When I plug in random values those also yield the same solutions.
Exactly what I thought but then I realized, when you add yp and yc, the constant term is added from there already, so just assume it as a total constant
This man is doing God's work
Wronskian Gang rise up.
Yeah i am gonna repeat this course
I have been waiting for this. THANK YOU 👊
Watching this in 2024, your videos never get old. Thanks boss
Why was I not subscribed already? I know your channel for a few years now and I knew how good quality the content was.
Pfff, at least it's corrected now.
yo i don't understand wtf
Did I miss a video on this? Or is this his only video on VoP?
is the calc 3 playlist complete?
Thank you
no wronskian, this is undertermined constant method hahahahaha
anyway ur ans is almost similar to wronskian so i will giv it a go
since c1=u1 and c2=u2 according to 2:54, can they be written as c1=lncox and c2=x in the final solution?
They're not equal. You must replace the constants with functions of x for variation of parameters.
This video is sponsored by UA-cam Prime (u_2'). :)
the shortest shortcut 🥵
God bless you OCT, I'm writing a D.E paper in 30 minutes.
Man makes complex things so easy to understand. Thanks
Thanks😁
Omo
Extremely helpful. How would I solve an equation in the form ay''(t) + by'(t) + cy(t) = dx'(t) +ex(t)?
Rearrange it to standard form, which is a trivial step:
ay''(t) + by'(t) + cy(t) = dx'(t) +ex(t)
a*y"(t) + (b - d)*y'(t) + (c - e)*y(t) = 0
Now it's just a second order equation with constant coefficients. b-d becomes the new B, and c-e becomes the new C.
a*y"(t) + B*y'(t) + C*y(t) = 0
Assume an ansatz of y(t)=e^(r*t).
(a*r^2 + B*r + C)*e^(r*t) = 0
Solve for two values of r, and construct a linear combination
r1 = (-B - sqrt(B^2 - 4*a*C)/(2*a)
r2 = (-B + sqrt(B^2 - 4*a*C)/(2*a)
In the case that both r1 and r2 are real and distinct:
y(t) = K1*e^(r1*t) + K2*e^(r2*t)
In the case that r1 = r2, call it r:
y(t) = K1*e^(r*t) + K2*t*e^(r*t)
In the case that r1 and r2 are complex conjugates:
y(t) = e^(Real(r) * t) * (K1*cos(imag(r)*t) + K2*sin(imag(r)*t))
why u1'y1+u2'y2 = 0 only why it cannot be a any function.
HERO
What application you used to make these videos ?? Please any one can tell me
what a beautiful problem and method
I understand this more than my professor's lecture .
PLEASE MAKE AN INTEGRATION VIDEO
Please what app do you use for this recording?
this is insanely hard compared to separation of variables etc
My first Diff Eq exam is in one hour
but there is a formula to find u1 and u2 directly without this process. Can you share that
I think it is the Wronskian method
Thank you~ You are a life saver~
UR No.1 in the world 👏👏
Can Anyone tell me how to write Yp ?
your video's are really nice
literally the goat
(8) clear
Big love Mr...
what if the given differential equation is of order greater then 2?
The methods of second order diffEQ's still works, you just need to solve a cubic formula instead of a quadratic formula. You'll also need second derivatives to construct the third row of the Wronskian, and you'll evaluate sub-Wronskian determinants.
As an example (I'll just show an arbitrary function g(t), since it's very difficult to come up with examples that can be practical to do):
y'"(t) +7*y"(t) + 14*y'(t) + 8*y(t) = g(t)
Assume a solution of y=e^(r*t), for the homogeneous equation
(r^3 + 7*r^2 + 14*r + 8)*e^(r*t) = 0,
Since all the signs are positive, this means there cannot be any positive roots, unless accompanied by an equal and opposite negative root. The rational roots theorem, tells us that -8, -4, -2, and -1 are all possible roots, as well as their positive counterparts. Trying -4, -2, and -1 will produce our three roots.
This means our homogenous solution is the following combination:
yh(t) = A*e^(-4*t) + B*e^(-2*t) + C*e^(-t)
Calculate the Wronskian of these three solution components. The underscores are just placeholders to help me align.
| ___e^(-4*t), ___e^(-2*t), e^(-t) |
| -4*e^(-4*t), -2*e^(-2*t), -e^(-t) | = W = 6*e^(-7*t)
| 16*e^(-4*t), 4*e^(-2*t), e^(-t) |
Then we evaluate the Cramer's Rule Wronskians, by replacing each column, with a column of zeros and a 1 in the final row:
W1 =
| 0, ___e^(-2*t), e^(-t) |
| 0, -2*e^(-2*t), -e^(-t) | = e^(-3*t)
| 1, 4*e^(-2*t), e^(-t) |
W2 =
| ___e^(-4*t), 0, e^(-t) |
| -4*e^(-4*t), 0, -e^(-t) | = -3*e^(-5*t)
| 16*e^(-4*t), 1, e^(-t) |
W3 =
| ___e^(-4*t), ___e^(-2*t), 0 |
| -4*e^(-4*t), -2*e^(-2*t), 0 | = 2*e^(-6*t)
| 16*e^(-4*t), 4*e^(-2*t), 1 |
The particular solution will become:
yp = yh1(t) * integral g(t)*W1/W dt + yh2(t) * integral g(t)*W2/W dt + yh3(t) * integral g(t)*W3/W dt
Implementing what we have:
yp = e^(-4*t) * integral [g(t)*e^(+4*t)/6] dt + e^(-2*t) * integral [1/2*e^(+2*t) * g(t)] dt + e^(-t) * integral [ 1/3*e^(t) * g(t)] dt
The general solution would therefore be:
y(t) = A*e^(-4*t) + B*e^(-2*t) + C*e^(-t) + e^(-4*t) * integral [g(t)*e^(+4*t)/6] dt + e^(-2*t) * integral [1/2*e^(+2*t) * g(t)] dt + e^(-t) * integral [ 1/3*e^(t) * g(t)] dt
I'll leave it as an exercise to you, to come up with a non-polynomial / non-trig / non-exponential case of g(t), that makes these possible integrals to do.
i love you
LOL i lasted till the 8 minute mark before i was lost lol the power of an Eng Sci student :')
Have you tried Davis' lectures? It's SO helpful
@@qingyuanwu davis explains things so well. i LOVE his teaching style
but u can use the formula just plug in
Nice explanation
Thank a lot
You saved me
7:27. you dont really explain why you chose to multiply by sin or cos
He's using the algebraic method of solving a system of equations, the Elimination Method. First step in this method is to multiply the two equations in a system by terms such that a term repeats, therefore, you can eliminate the term by adding/subtracting both equations.
@@tyromejenkins2442 Wow thank you, that was a great explanation