Yes, there are easier ways, such as just using the formula, but this way (as well as the method involving matrices) kind of gives you some background on how to derive the formulas.
For anyone watching, after he gets the system of equations that he has to solve, you can use cramer's rule to get v1' and v2' and then integrate to get v1 and v2. It's a slightly faster way. Great video though. I love this guys channel.
you are a lifesaver guy!!! I have been hovering multiple videos about this topic, none of the videos had any clear and confidential explanation, rather than yours. they just be there for the purpose of collecting subscriptions.
Is variation of parameters ever easier than linear integration? Here’s the same problem solved directly: y”+y=tan t y”*cos t+y*cos t=sin t y’*cos t+y*sin t=C1-cos t y’*sec t+y*tan t*sec t=C1*sec^2 t-sec t y*sec t=C1*tan t-ln|sec t+tan t|+C2 y=C1*sin t-cos t*ln|sec t+tan t|+C2*cos t
Issue: What is a differential of an irrational argument? Let a= some rational approximation, and A be the irrational number itself (if that makes sense). Then A - a > dA and there is no way a + dA > A
plotting this solution reveals a weirdly lumpy periodic function with the period of 2*pi. The lumpiness is kind of interesting if the coefficients of sine and cosine are relatively small. picture: i.imgur.com/Pnu2Yua.jpg
Hi blackpenredpen , I didn't realize why can I find the harmonic solution by solving a polinom? (sorry if my English is incorrect, I don't good in English)
Given: y" - 2*y' - 3*y = 3*sinh(2*x) - 12*cosh(2*x) For this example, I'd use the Laplace transform, and assign arbitrary initial conditions. This can also work using the method of undetermined coefficients, because sinh(2*x) and cosh(2*x) are both a linear combinations of e^(-2*x) and e^(2*x), so you can assume both of these as your ansatz of the particular solution for undetermined coefs. Let y(0) be u, and let y'(0) be v. Also let Y(s) = £{y(x)} £{y"} = s^2*Y(s) - v - u*s £{-2*y'} = -2*s*Y(s) + 2*u £{-3*y} = -3*Y(s) £{3*sinh(2*x))} = 6/(s^2 - 4) £{12*cosh(2*x))} = 12*s/(s^2 - 4) Thus: (s^2 - 2*s - 3)*Y(s) - v - u*s + 2*u = 6/(s^2 - 4) - 12*s/(s^2 - 4) Shuffle initial conditions to the right: (s^2 - 2*s - 3)*Y(s) = 6/(s^2 - 4) - 12*s/(s^2 - 4) + v + (s - 2)*u Change right side to common denominators: (s^2 - 2*s - 3)*Y(s) = (6 - 12*s + (u*s - 2*u + v)*(s^2 - 4))/(s^2 - 4) Isolate Y(s): Y(s) = (6 - 12*s + (u*s - 2*u + v)*(s^2 - 4))/((s^2 - 4)*(s^2 - 2*s - 3)) Factor denominator and expand/gather numerator: Y(s) = (u*s^3 + (v - 2*u)*s^2 - (12 + 4*u)*s + 8 u - 4 v + 6)/((s - 2)*(s + 2)*(s + 1)*(s - 3)) Partial fractions: Y(s) = A/(s - 3) + B/(s + 1) + C/(s - 2) + D/(s + 2) From experience, we know that the two solutions whose denominators are factors of the characteristic polynomial in front of Y(s), they will be the two solutions whose output depends on initial conditions. As such , C and D are the two coefficients that will be from the particular solution, and won't depend on initial conditions. Heaviside cover-up works great for finding them. So we might as well just leave A and B alone, as unknowns for the general solution. C = (u*2^3 + (v - 2*u)*2^2 - (4*u + 12)*2 + 8*u - 4*v + 6)/((2 +2)*(2 + 1)*(2 - 3)) = 3/2 D = (u*(-2)^3 + (v - 2*u)*(-2)^2 - (4*u + 12)*(-2) + 8*u - 4*v + 6)/((4 + 4)*(4 + 1)*(4 - 3)) = -3/2 Y(s) = A/(s - 3) + B/(s + 1) + 3/2/(s - 2) - 3/2/(s + 2) Take inverse Laplace: y(x) = A*e^(3*x) + B*e^(-x) + 3/2*e^(2*x) - 3/2*e^(-x) And we can consolidate our particular part of the solution as a single sinh: y(x) = A*e^(3*x) + B*e^(-x) + 3*sinh(2*x)
my teacher solved it in more easy way
DONT DISRESPECT THE MAN HES FUCKING AMAZING
KEEP ON DOING YOUR THING BPRP
Yes, there are easier ways, such as just using the formula, but this way (as well as the method involving matrices) kind of gives you some background on how to derive the formulas.
So why are you watching dis....
Go n study to your teacher......dont show off man!
Watch my channel for more easy way..
Hey guys chill alright? It's completely normal that there are faster ways to solve a question, don't get so butthurt.
For anyone watching, after he gets the system of equations that he has to solve, you can use cramer's rule to get v1' and v2' and then integrate to get v1 and v2. It's a slightly faster way. Great video though. I love this guys channel.
you are a lifesaver guy!!! I have been hovering multiple videos about this topic, none of the videos had any clear and confidential explanation, rather than yours. they just be there for the purpose of collecting subscriptions.
For those who want the final solution: y = C1 cos(t) + C2 sin(t) - cos(t) ln|sec(t) + tan(t)|
Spencer Key yay!
Problem: find the Fourier Series.
Excellent work as always. By the way it was easier to substitute v2 derivative instead in the first equation, exploiting the zero.
Can you do one using wronskians and integral? Ex: u_1 = int((y_2(t)*g(t))/W(y_1,y_2))
I understand this now with 1000% clarity.
you my friend deserve a medal. You will reach 1 mil subs
It happened, on November 11, 2022.
太谢谢你了，我考前复习正好看到这个问题不懂，你这个办法也挺好的，用包含矩阵的积分公式似乎对我来说难以记忆，如果真忘了那我只好推导了哈哈老师说我们时间很够。
THANK YOU!!!! PLEASE KEEP IT COMING I HAVE CALC 3 IN THE FALL AND LINEAR ALGEBRA IN THE SPRING
you are the best!!!!
#RespectFromSouthAfrica
Is variation of parameters ever easier than linear integration? Here’s the same problem solved directly:
y”+y=tan t
y”*cos t+y*cos t=sin t
y’*cos t+y*sin t=C1-cos t
y’*sec t+y*tan t*sec t=C1*sec^2 t-sec t
y*sec t=C1*tan t-ln|sec t+tan t|+C2
y=C1*sin t-cos t*ln|sec t+tan t|+C2*cos t
Very helpful. God bless you
THANK YOU BlackPenRP
Excellent video! Thank you so much.
y"+y=tan(t)
I love all your videos!
blackpenredpen blue pen? there is something wrong with this video... :)
raees khan thanks!!
Finally i understand this. Thank you!
Issue:
What is a differential of an irrational argument?
Let a= some rational approximation, and A be the irrational number itself (if that makes sense).
Then A - a > dA and there is no way a + dA > A
Just Amazing Brother..!!!
why dont u just plug in V2' which u just solved
Very Helpful
After finding V'2 why can we not plug it into one of the equations to acquire V'1??
you can
Yep, you can get it many ways. Whatever floats your equation bro
Thanks you
You and your methods are amazing and easily understandable also....
Thank you so much!
Cool. We've only learnt how to solve linear second order differential equations by inspection so far.
Thanks! would love to see more examples for variation of parameters!
Thank you, my teacher want us to do it this way and not the wronskian method
Good job I love it so much
plotting this solution reveals a weirdly lumpy periodic function with the period of 2*pi. The lumpiness is kind of interesting if the coefficients of sine and cosine are relatively small. picture: i.imgur.com/Pnu2Yua.jpg
#RespectfromIndia
6:18 - 6:22 was epic
طريقة الحل جدا مفهومه شكرا
I have to do it via wronskian
Have to? That's the easier way. I always prefer a formula.
This is the safest solution possible. So, why not? haha
Where is C from the particular solution in the final answer?
he makes me laugh when he laughs lol
i love your videos
Thanks so much
Is it easier to use wronskian couple with cramers rule?
thank you
why not use the wroskian approach
hello there how to do that when i have more than 3 equations? are there techniques to quicken the solving?
@blackpenredpen
hi thnx a lot. And how about this diff. eq. y''+5y'+11=tan(x)?
🙏🙏sir u r amazing
Hi blackpenredpen , I didn't realize why can I find the harmonic solution by solving a polinom? (sorry if my English is incorrect, I don't good in English)
Very helpfull. How Can i solve this equation please? y''-2y'-3y=3sinh(2x)-12cosh(2x)
Given:
y" - 2*y' - 3*y = 3*sinh(2*x) - 12*cosh(2*x)
For this example, I'd use the Laplace transform, and assign arbitrary initial conditions. This can also work using the method of undetermined coefficients, because sinh(2*x) and cosh(2*x) are both a linear combinations of e^(-2*x) and e^(2*x), so you can assume both of these as your ansatz of the particular solution for undetermined coefs.
Let y(0) be u, and let y'(0) be v. Also let Y(s) = £{y(x)}
£{y"} = s^2*Y(s) - v - u*s
£{-2*y'} = -2*s*Y(s) + 2*u
£{-3*y} = -3*Y(s)
£{3*sinh(2*x))} = 6/(s^2 - 4)
£{12*cosh(2*x))} = 12*s/(s^2 - 4)
Thus:
(s^2 - 2*s - 3)*Y(s) - v - u*s + 2*u = 6/(s^2 - 4) - 12*s/(s^2 - 4)
Shuffle initial conditions to the right:
(s^2 - 2*s - 3)*Y(s) = 6/(s^2 - 4) - 12*s/(s^2 - 4) + v + (s - 2)*u
Change right side to common denominators:
(s^2 - 2*s - 3)*Y(s) = (6 - 12*s + (u*s - 2*u + v)*(s^2 - 4))/(s^2 - 4)
Isolate Y(s):
Y(s) = (6 - 12*s + (u*s - 2*u + v)*(s^2 - 4))/((s^2 - 4)*(s^2 - 2*s - 3))
Factor denominator and expand/gather numerator:
Y(s) = (u*s^3 + (v - 2*u)*s^2 - (12 + 4*u)*s + 8 u - 4 v + 6)/((s - 2)*(s + 2)*(s + 1)*(s - 3))
Partial fractions:
Y(s) = A/(s - 3) + B/(s + 1) + C/(s - 2) + D/(s + 2)
From experience, we know that the two solutions whose denominators are factors of the characteristic polynomial in front of Y(s), they will be the two solutions whose output depends on initial conditions. As such , C and D are the two coefficients that will be from the particular solution, and won't depend on initial conditions. Heaviside cover-up works great for finding them. So we might as well just leave A and B alone, as unknowns for the general solution.
C = (u*2^3 + (v - 2*u)*2^2 - (4*u + 12)*2 + 8*u - 4*v + 6)/((2 +2)*(2 + 1)*(2 - 3)) = 3/2
D = (u*(-2)^3 + (v - 2*u)*(-2)^2 - (4*u + 12)*(-2) + 8*u - 4*v + 6)/((4 + 4)*(4 + 1)*(4 - 3)) = -3/2
Y(s) = A/(s - 3) + B/(s + 1) + 3/2/(s - 2) - 3/2/(s + 2)
Take inverse Laplace:
y(x) = A*e^(3*x) + B*e^(-x) + 3/2*e^(2*x) - 3/2*e^(-x)
And we can consolidate our particular part of the solution as a single sinh:
y(x) = A*e^(3*x) + B*e^(-x) + 3*sinh(2*x)
y''+4y=cos(2x)
You can use undetermined coefficient method
just use wronskian instead and easier and workds
lit outro music
What about y''+y'=tan(x)
thank's
is this still cauchy euler method applied?
taking this adv ODE in year 3 mechanical engineering, will take adv PDE next year hope u can cover the topics in adv PDE love
We can all be happy that the original question is y"+y=tan(t) and not y"-y=tan(t).
That is correct!
I'm struggling along with y''-y=sinh(2x)
Why there is no the coefficient C_3 at cos(t)*ln(abs(sec(t)+tan(t))) in the answer?
Particular solution.you don't need a c3.
I am not getting how he takes y1=cost and y2 = sint. Please explain that
Why y1' +y2' = f(t)/a is valid? Is there have any theorems?
you can watch the previous video about the variation of parameters, introduction, and idea, in case you are still alive.
In what video do you explain the y_p equation?
Alright thanks!
can you help me how to solve y''+y= sec (theta) tan (theta). im confused when it has (theta)
Great vid as always! By the way, would this method also work with higher order derivatives (f.ex. y''', y"" etc)? Thanks!
Yes it does!
if sin^2t+cos^2t=1 then v1'+v1=2v1'? isnt it?
👏🏽👏🏽👏🏽👏🏽👏🏽
ive just got the y1 for the yh, how i can solve it???
My equation is a 3rd order nonhomogeneous. How can I solve it ?
You forgot to write y2v2 its sinx(-cosx)
Day orange6 ?
He did, it cancelled out. :)
what's y'' is it the second derivative of y?
yes
how do you become this much smart
한국인처럼 생겼군요 ...... 김태형 아시 잖아요
what is the sect ???? THANKS
sec(t) = 1/cos(t)
Not sure if title of video is quotation marks or 2 of '
niklas schüller lol
It's 2 of '
Cooll
My teacher makes use of matrix to make it look easier and faster.
wakanyanya
At the end u forgot to ad y2u2
uau
Thank you very much. Much appreciated.
thank you