thank you for this! my professor always explains stuff in an unnecessarily complicated way and this got me to really understand what's going on and not just blindly following how he does it.
Finally the best video of variation of parameters . Some of the videos have so much confusion but this one is great & the way he is teaching is awesome 👍👏
comin back for final tomorrow, and I just realized, it's the first component of a row vector, then the second, then we take the derivative. It's way more straightforward than I remember XD. The 1 and 2 just tell us which is being replaced. and then it's, albeit barely, like we're integrating the unit component (very not accurate, prolly but it's easy to see it like that). From there we dot u and y for our particular, and all is well.
This is one of those I think would do me a lot of good to include on my study guide. Luckily it is pretty intuitive I think. Just the combos of u's and y's and stuff. Honestly, it's not bad... But yeah, do you have any videos where you derived this? It reminds me of cross products were we have two vectors and (The order may be different in the actual derivation, and that may or may not be important) where W is the z component, W1 is the y component and W2 is the x component. From then I don't know where we're going, or like what we can glean from it all? Right now it is one of my least favorite things in mathematics, a formula I have to memorize without really understanding the parts, Heck, the quadratic formula that is notorious for being that isn't even that at this point, x value of the maximum plus and minus the roots! Honestly, I wish I could visualize what an integral in 3 dimensions really was... I guess I could look it up if I'm that curious, but AFTER studying for the midterm is complete!
I took an exam a few minutes ago.This was my exam question , exactly the same . Online exams are very intresting :) Thanks for this video. I will pass the lesson at the end, thanks to you sir :) ..
Good question. Let's do this problem accounting for the constants and find out why. Given y" + y = sec(t) Homogeneous solutions: yh1 = cos(t), yh2 = sin(t) yh = A*cos(t) + B*sin(t) Wronskian: W=1 Cramer Wronskians: W1 = -sec(t) W2 = +sec(t) Integrals to find yp: yp = -cos(t)*integral sec(t) *sin(t) dt + sin(t)*integral sec(t) cos(t) dt yp = -cos(t)*integral tan(t) dt + sin(t)*integral 1 dt yp = -cos(t)*(-ln(|cos(t)|) + C1) + sin(t)*(t + C2) Distribute: yp = cos(t)*ln(|cos(t)|) - C1*cos(t) + t*sin(t) + C2*sin(t) Add the homogeneous solution, and gather the like terms, to find the complete general solution: y = (A - C1)*cos(t) + (B + C2)*sin(t) + cos(t)*ln(|cos(t)|) + t*sin(t) As you can see, the constants C1 and C2 will ultimately combine with the arbitrary constants A and B from the homogeneous solution. Keep it simple, and let C1 and C2 both equal zero. We're left with: y = A*cos(t) + B*sin(t) + cos(t)*ln(|cos(t)|) + t*sin(t)
It has no elementary solution. There is a way to use this method, to solve a differential equation in general, in the form of y' + y = f(x), and here's how: Solve the homogeneous part, by the meaning of homogenous for 2nd order diffEQ's: yh' + yh = 0 yh = C*e^(-x) Now, find the Wronskian (W) and Cramer Wronskian (W1) of this homogeneous solution. The Wronskian is just the solution itself, and the Cramer Wronskian is just f(x). Thus: yp = e^(-x) * integral f(x)/e^(-x) dx This means the general solution is: y = C*e^(-x) + e^(-x) * integral f(x)*e^x) dx An example that requires more than undetermined coefficients to solve, that can be solved in elementary functions, uses f(x) = tanh(x). integral tanh(x) * e^x dx = e^x - 2*arctan(e^x) Thus, the general solution to y' + y = tanh(x) is: y = C*e^(-x) - 2*arctan(e^x)*e^(-x) + 1
No, because the other methods only work, when your non-homogeneous part of the given equation, is one of the forms that either annihilates when differentiated, or loops on forms of itself when differentiated. In other words, exponentials, simple sine and cosine trig, polynomials, constants, and linear combinations of the above. Exponentials and simple sine and cosine, are functions that loop back on forms of themselves when differentiated, while polynomials are functions that annihilate when differentiated. Functions such as logs, non-whole numbered powers of t, secants and tangents, are functions that would require variation of parameters. And very few of these are even possible to do with solutions in terms of elementary functions.
No solution in terms of elementary functions. You end up having to carry out the integrals of" integral sec(x)*e^x/(e^(4*x) - 1) dx and integral sec(x)*e^(3*x)/(e^(4*x) - 1) dx Both of which have no elementary solution.
thank you for this! my professor always explains stuff in an unnecessarily complicated way and this got me to really understand what's going on and not just blindly following how he does it.
:)
this video was made is 2015 and you're still replying to comments. I really admire that. Thank you for this video. a lot.
You bet😄
Finally the best video of variation of parameters . Some of the videos have so much confusion but this one is great & the way he is teaching is awesome 👍👏
Thank you
Absolute legend, this is way better than how my teacher explains it. Hope you doing well!!
You are one of my new favorites along with the organic chemistry tutor. Apprectiate ya
Thank you from Korea!
Thats him, he is the one that is going to carry me through my course, a truly math sorcerer indeed
thank you!! this is very helpful and easy to understand..
+Marissa Roy glad it helped!
Thank you. So easy to understand.
Wow!! This is an amazing tutorial!! Thank you for posting this!!
+gabrieltnc1 Np glad it helped!
Hi Sorceror, thanks for the video. It just so happens that this is one of my homework problems. Thanks!
awesome,glad it helped!!
This was so helpful! I actually feel ready for my exam, thank you!!
Great! Good luck😄
I'm from Kenya .Thank you very much
Best tutorial ever. Crystal clear
comin back for final tomorrow, and I just realized, it's the first component of a row vector, then the second, then we take the derivative. It's way more straightforward than I remember XD. The 1 and 2 just tell us which is being replaced. and then it's, albeit barely, like we're integrating the unit component (very not accurate, prolly but it's easy to see it like that). From there we dot u and y for our particular, and all is well.
Simply understanding
THANK YOU!
Very much useful video
Thanks a lot brother
No problem 😄
Thank you so much. It was so clear and neat!!
All the best for you :D
np glad it helped!!
Thanku so much. This actually helped
great video
You save my life
👍
this is my exam thank you for the answer 🤙🏻
such a good video
thank you!!!
Thank for you but maybe you missing tan x when you calculate u2
very good, thanks!
glad it helped!!!
Thank you sir
you are welcome1
This vedio gave me 15 out of 15 marks !! And yes am master in cheating 😁😁😁
Thanks bro, this help me a lot!
Can you make a video for y'' + y = sec(x) cosec(x) ? Please I need this so much till tomorrow 🙏🙏
Cool! Ooo, you're seeing my exclusively math account, spoooooky~~~
This is one of those I think would do me a lot of good to include on my study guide. Luckily it is pretty intuitive I think. Just the combos of u's and y's and stuff. Honestly, it's not bad...
But yeah, do you have any videos where you derived this?
It reminds me of cross products were we have two vectors and (The order may be different in the actual derivation, and that may or may not be important) where W is the z component, W1 is the y component and W2 is the x component. From then I don't know where we're going, or like what we can glean from it all? Right now it is one of my least favorite things in mathematics, a formula I have to memorize without really understanding the parts, Heck, the quadratic formula that is notorious for being that isn't even that at this point, x value of the maximum plus and minus the roots! Honestly, I wish I could visualize what an integral in 3 dimensions really was... I guess I could look it up if I'm that curious, but AFTER studying for the midterm is complete!
I took an exam a few minutes ago.This was my exam question
, exactly the same
. Online exams are very intresting :) Thanks for this video. I will pass the lesson at the end, thanks to you sir :)
..
thx
Fucking awesome!!!!!! thank you!
LOL np man
like your work so much
Bro love you !!!
Thanks...
Glad it helped!!
Thank you so much
np:)
i got sorcereted
LOL!
You did it 😍💪🏻
Why don't we need to put the constant when solving the integral?, like button hitted!
Good question. Let's do this problem accounting for the constants and find out why.
Given
y" + y = sec(t)
Homogeneous solutions:
yh1 = cos(t), yh2 = sin(t)
yh = A*cos(t) + B*sin(t)
Wronskian: W=1
Cramer Wronskians:
W1 = -sec(t)
W2 = +sec(t)
Integrals to find yp:
yp = -cos(t)*integral sec(t) *sin(t) dt + sin(t)*integral sec(t) cos(t) dt
yp = -cos(t)*integral tan(t) dt + sin(t)*integral 1 dt
yp = -cos(t)*(-ln(|cos(t)|) + C1) + sin(t)*(t + C2)
Distribute:
yp = cos(t)*ln(|cos(t)|) - C1*cos(t) + t*sin(t) + C2*sin(t)
Add the homogeneous solution, and gather the like terms, to find the complete general solution:
y = (A - C1)*cos(t) + (B + C2)*sin(t) + cos(t)*ln(|cos(t)|) + t*sin(t)
As you can see, the constants C1 and C2 will ultimately combine with the arbitrary constants A and B from the homogeneous solution. Keep it simple, and let C1 and C2 both equal zero. We're left with:
y = A*cos(t) + B*sin(t) + cos(t)*ln(|cos(t)|) + t*sin(t)
@@carultch Thank you a lot for your answer, finally i can sleep in peace 😀 with variaton of parameters!
Tq
you are welcome!
is it okay if you have two homogenous solutions like y =c1 and yh = costeta +sine teta can you do a wronskian with combined homogenous terms?
Wao!!!
you the best of the best
Thx man
m^2 +m=0? the roots= 0,-1?
i would like a help in this number .....y' + y= sec(x)
It has no elementary solution. There is a way to use this method, to solve a differential equation in general, in the form of y' + y = f(x), and here's how:
Solve the homogeneous part, by the meaning of homogenous for 2nd order diffEQ's:
yh' + yh = 0
yh = C*e^(-x)
Now, find the Wronskian (W) and Cramer Wronskian (W1) of this homogeneous solution. The Wronskian is just the solution itself, and the Cramer Wronskian is just f(x). Thus:
yp = e^(-x) * integral f(x)/e^(-x) dx
This means the general solution is:
y = C*e^(-x) + e^(-x) * integral f(x)*e^x) dx
An example that requires more than undetermined coefficients to solve, that can be solved in elementary functions, uses f(x) = tanh(x).
integral tanh(x) * e^x dx = e^x - 2*arctan(e^x)
Thus, the general solution to y' + y = tanh(x) is:
y = C*e^(-x) - 2*arctan(e^x)*e^(-x) + 1
thanks
thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks thanks
Nice
why didnt you just use sec x as the integral of tan??
şunun türkçesini de çevirin sevabına 🙄
Can you solve this without using variation of parameters?
No, because the other methods only work, when your non-homogeneous part of the given equation, is one of the forms that either annihilates when differentiated, or loops on forms of itself when differentiated. In other words, exponentials, simple sine and cosine trig, polynomials, constants, and linear combinations of the above. Exponentials and simple sine and cosine, are functions that loop back on forms of themselves when differentiated, while polynomials are functions that annihilate when differentiated.
Functions such as logs, non-whole numbered powers of t, secants and tangents, are functions that would require variation of parameters. And very few of these are even possible to do with solutions in terms of elementary functions.
What is solution when Y"-Y=secx
No solution in terms of elementary functions. You end up having to carry out the integrals of"
integral sec(x)*e^x/(e^(4*x) - 1) dx and integral sec(x)*e^(3*x)/(e^(4*x) - 1) dx
Both of which have no elementary solution.
Why was alpha=0?
In the form of a complex number: z = a + bi (a(alpha)= 0 and b(beta) = 1).
z=0+1i