at 5:49 shouldn't we have a negative sign in front of u2' ? I don't know if I'm making a mistake with calculation but I tried multiple times and I'm finding the same result after and after
I am not really good at integrals but I got u2 as [ln(sinx+tanx) - sinx +c] I can't seem to understand what kind of algebraic manipulation to get in the form mentioned in the video. great video btw. Your channel really helps me with studying engineering! keep up the good work
5:47 I am confused shouldn't u2 prime be -tan(x)sin(x)? If u1 prime is sin(x), then u1prime(sin^2(x))+u2prime(cos(x)sin(x))=0 would yield u2prime as -tan(x)sin(x), not positive tan(x)sin(x), right?
Absolutely incorrect assumption to derivatives in y' equals 0. The reason, why we can do it, is convertion of this second order ODE into the system of first order ODEs. Substitute u = y', and got linear system of ODEs: y' - u = 0 u' + y = tanx This implies solution in form (y, u), and here variation of parameter implies construction y' - u = 0 contains only derivatives. This isn't an assumption, just result of substitution of variation into system of equations.
My favorite way to solve ODEs and PDEs is by setting them up into Green's Form (set the g(x) to a delta δ(x-ξ) ) then take either a Fourier or Laplace Transform of the Green Function, afterwards inverse transform then your result is the Kernel you integrate multiplied to the g(x)!!! It's so beautiful.
I feel like you get a better feel for the general pattern with the third order version of this. Basically there are n-1 constraints and the last step takes care of the non-homogeneous thing. It's been a while but yeah, the Wronskian matrix and linear independence is huge here, "invertibility goes brrrrr" as the kids say.
I wonder if you'll read this but if you do, is there a good resource in the internet where I can find problems and detailed solutions in math, for free preferably. I feel like textbooks do questions that cover the topic but they don't push the knowledge like exam questions do. thanks
Hello Dr. Trefor! You're videos have been inmensely useful for me all the past year, first in calculus and now in ODE's. So thank you! I was thinking that if you could format the content of the videos into slides and put it in the description as a pdf it would be very helpful. Because I usually take a photo with my ipad of a moment in the video, paste it in my note taking app and then scribble some comments over it. So having a pdf with all the contents would be much faster. I know there is a textbook but it just isn't the same.
Fantastic video. I did not know it was possible to turn the inhomogeneous second order linear DE into two first order DEs like this. Very cool! For the integral of sec(x) (which is similar to sin(x)tan(x) as you pointed out), I much prefer the form ln|sec(x)+tan(x)|. It's equivalent but imo looks nicer because 1) it doesn't have the 1/2 in front, and 2) it's much easier to differentiate and gives you an intuitive sense for "why" that is the integral of sec(x). You get a very nice fraction: (sec^2(x)+sec(x)tan(x))/(sec(x)+tan(x)), which cancels down nicely.
tanxsinx = sin²x/cosx = (1-cos²x)/cosx = secx - cosx cosx can be integrated separately secx = (sec²x - secxtanx)/(secx - tanx) The numerator is the negative of the derivative of the denominator. So integrating, we get -ln|secx-tanx| = (-1/2)ln((1-sinx)²/(cos²x)) = (-1/2)ln((1-sinx)²/(1-sin²x)) = (-1/2)ln((1-sinx)/(1+sinx)) Thank you for this method of solving DEs, I did not know that :). I'm actually amazed it worked so well. Definitely will be using it in my exams.
Thanks for liking this comment btw, I was mostly just messing around but I would like to say that your videos are the best maths videos around. It really gives off that maths professor oozing with knowledge vibe that I can't find in other UA-cam videos. I love how you linked it to other aspects of calculus taught and really made me understand it at a deeper level, and make the learning process a lot more interesting and fun
@@DrTrefor ooohh... so you got to pull out all the tricks you have when facing an integral? sounds like a wild west, to know which one has a closed form. similar feelings with nonlinear DEs
and when it *is* impossible, are there other ways to look into it further without going straight to numerical computing? (series expansion comes to mind)
Thanks. Great Video. I especially like your discussion of the constraint. It's always just stated in text discussions without any explanation of why it's safe to do that.
Sir thank you so much for making the video public. I'm following your content past 3 months; extremely helpful for competitive exams. Thank you once again :)
It is an interesting topic, and well explained, but why is the notation so 'weird'? I am going to assume there must be a historical reason for it. Do any of the videos explain the historical reason we prefer y'' + y = tan(x) to what to me intuitively seems cleaner y'' + y = tan or perhaps even better y''(x) + y(x) = tan(x) And prefer y = u1*sin(x) + u2*cos(x) to y = u1*sin + u2*cos Is it because it is trigonometric functions that their notation differs? Also is the multiplication (and addition) involved meant as the product of the functions or a product for some particular x (and then repeated for all real x in the "zero to pi domain") ? When I sat in mathematics classes many years ago we were taught that one of the most important things was to introduce/specify the ingredients (and I largely still try to adhere to that advice).
Ya I hear your point. I suppose I’m habituated to always write (x) explicitly for trig functions, but for generic ones I only write (x) explicitly if needed and omit it for shorthand convenience.
@@DrTrefor Cool. I will chalk it up to personal preference then. That is perfectly fine, I was worried there was some kind of established principle at play. 🙂
Your explanation is brilliant. I searched for hours trying to understand this but only found methodologies w/o explanations. I could see clearly Cramer's rule being used, worked out what the matrix equation must be but could not understand where the matrix equation (Y.U' = G) was coming from. I now feel that I at last understand the method of variation of parameters completely. btw in 8:02 I think your u2 has the wrong sign. We used a different method when we solved this kind of d.e. some 40 years back. We would have operated on tan(x) with the operator 1/(D²+1)
Professor Bazett, thank you for analyzing and solving Non-Homogeneous Differential Equations using Variation of Parameters. This topic is messy from start to finish.
@@DrTrefor Do we know that though? I know its true for homogeneous equations. Non-homogeneous adds a complication and Ive never seen a proof that only two linearly independent solutions exist for this case. Dont get me wrong, I believe you. It just isnt as obvious to me as it is for linear homogeneous equations.
Guys, I'm searching for a site or an app for graphing differential equations. Can you help me? I want it to approximate or show the actual graph of the solution. Thanks
Thanks for the videos you have been a huge help! What is an appropriate way to deal with online tests for proof heavy classes? I'm a proof class and enjoy it, but I don't even want to write closed book, no Google, tests because I feel like I'm at a disadvantage if I'm honest. I feel like it would be fair to allow open book, but test on the explanations rather than the answer. Google isn't always perfect and I find the people who break it down sometimes do it in a way that's hard to explain. Kudos to them, but annoying when you get stuck on the same part of an induction proof as a mathematician in our textbook though!
Given a 2nd order linear homogeneous diffEQ in the form of: a*y"(t) + b*y'(t) + c*y(t) = 0 The standard solution method is to assume an Ansatz of y(t) = e^(r*t). When you take derivatives, you'll accumulate r in front of e^(r*t) each time. After factoring, you end up with: (a*r^2 + b*r + c)*e^(r*t) = 0 Since e^(r*t) can never permanently be zero for all t-values, the polynomial of the variable r in front of it, must be zero. So you are interested in solving for r, which is a quadratic formula. If you get two distinct real solutions, you have overdamping. The solution will be a linear combination of e^(r1*t) and e^(r2*t). If both r's are negative, you have two exponential decays added together. If one or both of the r-values are positive, you'll have an unstable system that eventually grows indefinitely as t gets large. Example: y"(t) + 6*y'(t) + 5*y = 0. In the special case that you get a single real solution, you have critical damping. Due to the overlap, the solution will be a linear combination of e^(r*t) and t*e^(r*t). This is the case that decays the fastest, without overshooting its target. Example: y"(t) + 6*y'(t) + 9*y = 0. For a complex conjugate pair of solutions for r, you end up with underdamping. The imaginary component will be the frequency of the sine and cosine terms, and you'll have a linear combination of the two of them. Common to both terms, you'll have an exponential decay, with a decay constant equal to the real component of the solutions for r. It will have exponential decay when r is negative, and exponential growth when r is positive. Example: y"(t) + 6*y'(t) + 13*y = 0. For purely imaginary solutions for r, you'll get no damping. The solution is a linear combination of sine and cosine. Example: y"(t) + 9*y = 0.
Hello, Dr. Bazett! Do you have a video on how to solve a higher-order differential equation (say degree three) to get the particular solution using this method?
The way that it works for third order equations, and beyond is as follows. Given a diffEQ in the form of: a*y'" + b*y" + c*y' + d*y = g(t) Step 1: find the homogeneous solution components. For 3rd order diffEQ's there will be 3 of them, and you'd solve a cubic equation to find them. I'll call them yh1(t), yh2(t), and yh3(t). Step 2: find the first and second derivative of each one of them. Step 3: calculate the Wronskian, with row 1 of the functions from part 1, row 2 for first derivatives, and row 3 for 2nd derivatives. Step 4: calculate what I call, the Cramer Wronskians, because they come from Cramer's rule. There are 3 of them for a 3rd order system, which I'll call W1, W2, and W3. Replace column 1 in the original Wronskian, with a vertical column of , to get W1. Replace column 2 in the original Wronskian, with the same vertical column, to get W2. Likewise for W3. Then we construct the particular solution: yp = yh1*integral W1/W * g(t) dt + yh2*integral W2/W * g(t) dt + yh3*integral W3/W * g(t) dt The complete solution will be: y = A*yh1 + B*yh2 + C*yh3 + yp It's easier to understand the 2nd order case knowing how it works beyond 2nd order equations, to see where the sign pattern comes from. I've tried to find examples that integrate in closed form to demonstrate this, that you couldn't otherwise do with undetermined coefficients or Laplace/Fourier transforms , but haven't had any success yet.
One example that looks promising, since it's as simple as I can make a cubic with two purely imaginary roots, so the homogeneous solutions are sin(sqrt(3)*t), cos(sqrt(3)*t), and e^(-3*t). It's great to work with sine and cosine, since they commonly reduce to a constant in the Wronskian. y"' + 3*y" + 3*y' + 9*y = g(t) But I've yet to find an example of g(t) that would be challenging enough to require variation of parameters.
TYPO: At 7:56 the expression for u_2 should NOT have a negative out the front!!
nevertheless nice video
But was it actually a type-o? Or was it a momentary concept-o?
I was so confused until I see this comment😂
hah I did all steps to calculate those formulas myself and I didn't understand why I had mistake
Noooooooooo man I just saw your comment after exactly 20 minutes thinking why it has the negative sing !!!! you got met there
at 5:49 shouldn't we have a negative sign in front of u2' ? I don't know if I'm making a mistake with calculation but I tried multiple times and I'm finding the same result after and after
You are correct!
I am not really good at integrals but I got u2 as [ln(sinx+tanx) - sinx +c] I can't seem to understand what kind of algebraic manipulation to get in the form mentioned in the video. great video btw.
Your channel really helps me with studying engineering! keep up the good work
Whoever found out that the constraint that u', v' term is 0 is a genius
5:46 Isn't u_2' = -tan(x)sin(x)? They need to cancel in the equations in order for the right hand side to be zero.
I'm trying to follow this too but it makes no sense lol. You're definitely right, not sure why he hasn't corrected it.
I just got out of a differential equations test, and this exact question was on it! Good to know I got the right answer.
Woah crazy!
This particular equation is also one of the homework problems in the Boyce/DiPrima text.
@@DrTrefor Even I also got this 😂😂
Hey I know it's been almost a year, but I just came across your comment and was curious. How'd the test and the rest of the class go?
@@PunmasterSTP it went well! I got an A in both!
5:47
I am confused shouldn't u2 prime be -tan(x)sin(x)?
If u1 prime is sin(x), then u1prime(sin^2(x))+u2prime(cos(x)sin(x))=0 would yield u2prime as -tan(x)sin(x), not positive tan(x)sin(x), right?
i know im 11 months late but yeah, i also got a negative
@@MegaBeanHead0 Same here
@@jscruz685 same here
Absolutely incorrect assumption to derivatives in y' equals 0. The reason, why we can do it, is convertion of this second order ODE into the system of first order ODEs. Substitute u = y', and got linear system of ODEs:
y' - u = 0
u' + y = tanx
This implies solution in form (y, u), and here variation of parameter implies construction y' - u = 0 contains only derivatives. This isn't an assumption, just result of substitution of variation into system of equations.
My favorite way to solve ODEs and PDEs is by setting them up into Green's Form (set the g(x) to a delta δ(x-ξ) ) then take either a Fourier or Laplace Transform of the Green Function, afterwards inverse transform then your result is the Kernel you integrate multiplied to the g(x)!!! It's so beautiful.
I feel like you get a better feel for the general pattern with the third order version of this. Basically there are n-1 constraints and the last step takes care of the non-homogeneous thing. It's been a while but yeah, the Wronskian matrix and linear independence is huge here, "invertibility goes brrrrr" as the kids say.
I wonder if you'll read this but if you do, is there a good resource in the internet where I can find problems and detailed solutions in math, for free preferably. I feel like textbooks do questions that cover the topic but they don't push the knowledge like exam questions do. thanks
How is the integral of tan(x)sin(x) hard at all? That equals sin^2 / cos which is (1 - cos^2) / cos, or sec(x) - cos(x).
ln|secx + tanx| - sin(x) + C
I have my exams tomorrow and you're videos are just great. Reviewing with your lectures is just great. Keep rocking Professor!!
You read my mind. We are not covering Variation of Parameters in my Diff Eqs class and this will help a lot.
But you can't assume things are magically equal to zero!
Dr Trefor: "But i Can"
Me: "Understandable have nice day" 👍
Hello Dr. Trefor! You're videos have been inmensely useful for me all the past year, first in calculus and now in ODE's. So thank you!
I was thinking that if you could format the content of the videos into slides and put it in the description as a pdf it would be very helpful. Because I usually take a photo with my ipad of a moment in the video, paste it in my note taking app and then scribble some comments over it. So having a pdf with all the contents would be much faster. I know there is a textbook but it just isn't the same.
Fantastic video. I did not know it was possible to turn the inhomogeneous second order linear DE into two first order DEs like this. Very cool!
For the integral of sec(x) (which is similar to sin(x)tan(x) as you pointed out), I much prefer the form ln|sec(x)+tan(x)|. It's equivalent but imo looks nicer because 1) it doesn't have the 1/2 in front, and 2) it's much easier to differentiate and gives you an intuitive sense for "why" that is the integral of sec(x). You get a very nice fraction: (sec^2(x)+sec(x)tan(x))/(sec(x)+tan(x)), which cancels down nicely.
Klaatu to Professor Barnhardt: "With *variation of parameters* this is the solution." "The Day the Earth Stood Still" (1951)
tanxsinx = sin²x/cosx = (1-cos²x)/cosx = secx - cosx
cosx can be integrated separately
secx = (sec²x - secxtanx)/(secx - tanx)
The numerator is the negative of the derivative of the denominator. So integrating, we get -ln|secx-tanx| = (-1/2)ln((1-sinx)²/(cos²x)) = (-1/2)ln((1-sinx)²/(1-sin²x)) = (-1/2)ln((1-sinx)/(1+sinx))
Thank you for this method of solving DEs, I did not know that :). I'm actually amazed it worked so well. Definitely will be using it in my exams.
hi , how do you get = (-1/2)ln((1-sinx)²/(cos²x)) from -ln|secx-tanx| ? thank you in advance
@@sarataha3258 I can help you out if you still want to know.
You're the best techer ❤️❤️
Thank you!!
Variation of parameters: works every time with 5 sheets of work
How cool is it to learn from you!
Such a divine presentation of mathematics.
Great sir.
Glad you enjoyed!
There is always a guy who say youtube channels are better than my professor. I bet he sleeps at his class. Good video Dr. Trevor. 🌹💚
Subtracts tan(x) from both sides and makes it homogenous
Puts on shades
Leaves
At 6:40 integral of U2 comes negative of what you have mentioned.
True, good catch!!
"we found a u1 and youtube" 🤣
Thanks for liking this comment btw, I was mostly just messing around but I would like to say that your videos are the best maths videos around. It really gives off that maths professor oozing with knowledge vibe that I can't find in other UA-cam videos. I love how you linked it to other aspects of calculus taught and really made me understand it at a deeper level, and make the learning process a lot more interesting and fun
Опять говорю, что в уравнении второй степени с решением y = C1*cos(x) + C2*sin*(x) удобно использовать при решении системы метод Крамера.
Sir is there any absolute method to solve the problem as guessing a solution is ambiguous and tedious.
why is no one stating this ?? at 6:02 u2' should be -tanx.sinx (sir has written tanx.sinx) sign should be negative
in 2:56, you just sounded like when Vsauce said "even though I am" :D
ohhhhh I think I finally understand your tshirt.... took me a solid 10 mins 😅
Hmmm... is there a way to tell, to get the final solution of an ODE, that the integral is impossible to do vs when it is just hard?
Not ahead of time. And integrals are often ad box that it looks insane but then this one weird trick makes it doable.
@@DrTrefor ooohh... so you got to pull out all the tricks you have when facing an integral? sounds like a wild west, to know which one has a closed form. similar feelings with nonlinear DEs
and when it *is* impossible, are there other ways to look into it further without going straight to numerical computing? (series expansion comes to mind)
Yeah i think there's a theorem that tells you whether a integral has a closed form solution or not
2:52 This sound something that Michael from Vsauce would say!
Thanks a lot Trefor! You explain waaay better than my Prof
Thanks. Great Video. I especially like your discussion of the constraint. It's always just stated in text discussions without any explanation of why it's safe to do that.
Wow, just in time! This is what I'm studying right now!
haha awesome!
Best explanation of the concept, thank you! so much!
your vids are of great help, thanks!
Thank you for the only video that gave me peace of mind about the justification of the constraint
Love coming to your videos to check my understanding or clear up any questions
Sir thank you so much for making the video public. I'm following your content past 3 months; extremely helpful for competitive exams.
Thank you once again :)
Shouldnt there be a minus sign in front of the tanxsinx at 5:47??
The solutions provided by this method are the only ones. Or maybe at least combining then linearly forming every solution possible?
It is unique up to linear combinations of the homogeneous solution, a consequence of existence and uniqueness theorem
8:12 “we solved the solution”. Succinct, sir.
Haha
Good timing! I saw this posted as members only about a week ago, do you post these members only ahead of time then open it up to everyone?
Ya exactly, basically post them about a week early for members. So everyone gets to watch eventually.
Omg I forgot to tie my shoes today
I'm not high enough to get this
I'll be right back
Your enthusiasm is so involving and reassuring, thanks a lot prof!
It is an interesting topic, and well explained, but why is the notation so 'weird'?
I am going to assume there must be a historical reason for it.
Do any of the videos explain the historical reason we prefer
y'' + y = tan(x)
to what to me intuitively seems cleaner
y'' + y = tan
or perhaps even better
y''(x) + y(x) = tan(x)
And prefer
y = u1*sin(x) + u2*cos(x)
to
y = u1*sin + u2*cos
Is it because it is trigonometric functions that their notation differs?
Also is the multiplication (and addition) involved meant as the product of the functions or a product for some particular x (and then repeated for all real x in the "zero to pi domain") ?
When I sat in mathematics classes many years ago we were taught that one of the most important things was to introduce/specify the ingredients (and I largely still try to adhere to that advice).
Ya I hear your point. I suppose I’m habituated to always write (x) explicitly for trig functions, but for generic ones I only write (x) explicitly if needed and omit it for shorthand convenience.
@@DrTrefor Cool. I will chalk it up to personal preference then. That is perfectly fine, I was worried there was some kind of established principle at play. 🙂
Wow that's amazing explanation ; things are crystal clear
Your explanation is brilliant. I searched for hours trying to understand this but only found methodologies w/o explanations. I could see clearly Cramer's rule being used, worked out what the matrix equation must be but could not understand where the matrix equation (Y.U' = G) was coming from. I now feel that I at last understand the method of variation of parameters completely.
btw in 8:02 I think your u2 has the wrong sign.
We used a different method when we solved this kind of d.e. some 40 years back. We would have operated on tan(x) with the operator 1/(D²+1)
This was a nice review. Thank you for the video.
Hy
is that important for homogenoeus DE it should first be linear?
Absolutely, just didn’t fit in the title!
this video's release date has synchronized with the time I had to study variation of parameters. thank you!
Perfect!
Thank you so much Professor Bazett for an absolutely fantastic review on variation of parameters method.
Glad it was helpful!
Thank you! best explanation on UA-cam!
This was very helpful thanks!
Thanks captain...love from india plz reply captain🙂
Thank you so much 😀
Professor Bazett, thank you for analyzing and solving Non-Homogeneous Differential Equations using Variation of Parameters. This topic is messy from start to finish.
Variation of parameters? More like "Very pertinent information for us!" Thank you so much for making all of these very high-quality videos for us! 👍
Are solutions fully generalized? If you drop the restriction that simplified your solution method, could you arrive at more generalized solution sets?
It is! We know 2nd order linear does have two linearly independent solutions and this up to a constant we have the general solution.
@@DrTrefor Do we know that though? I know its true for homogeneous equations. Non-homogeneous adds a complication and Ive never seen a proof that only two linearly independent solutions exist for this case. Dont get me wrong, I believe you. It just isnt as obvious to me as it is for linear homogeneous equations.
can this method be used to solve PDE?
Guys, I'm searching for a site or an app for graphing differential equations. Can you help me?
I want it to approximate or show the actual graph of the solution.
Thanks
Nice
Thanks for the videos you have been a huge help! What is an appropriate way to deal with online tests for proof heavy classes? I'm a proof class and enjoy it, but I don't even want to write closed book, no Google, tests because I feel like I'm at a disadvantage if I'm honest. I feel like it would be fair to allow open book, but test on the explanations rather than the answer. Google isn't always perfect and I find the people who break it down sometimes do it in a way that's hard to explain. Kudos to them, but annoying when you get stuck on the same part of an induction proof as a mathematician in our textbook though!
I got one question
What will be the probability of getting any number lets say 10 by subtracting two numbers?
Thank you Dr. Bazett that was really helpful. Can you please upload a video about the Cauchy Euler Method?
So grateful for your videos! 🙌
You ate with this video. Bless
Had used the method of variation of parameters before but never understood the intuition behind it , thanks prof
Is there a mistype at 9:38? U2 should not have a negative sign in front of the integral
Thanks a lot sir 🔥🔥🔥
I call it the method of "playing around with the equation until it fits"
you are amazing!!!!😃😁
Good explanation
let’s gooooooooooooo
Thanks
Thanks
Abel's identity can be helpful. Just my opinion
I think the - sign for u2 integral should be + instead.
Can you please do a video on actual examples of underdamping, overdamping, etc.? Like work out examples that we could see on an exam or homework.
Given a 2nd order linear homogeneous diffEQ in the form of:
a*y"(t) + b*y'(t) + c*y(t) = 0
The standard solution method is to assume an Ansatz of y(t) = e^(r*t). When you take derivatives, you'll accumulate r in front of e^(r*t) each time. After factoring, you end up with:
(a*r^2 + b*r + c)*e^(r*t) = 0
Since e^(r*t) can never permanently be zero for all t-values, the polynomial of the variable r in front of it, must be zero. So you are interested in solving for r, which is a quadratic formula.
If you get two distinct real solutions, you have overdamping. The solution will be a linear combination of e^(r1*t) and e^(r2*t). If both r's are negative, you have two exponential decays added together. If one or both of the r-values are positive, you'll have an unstable system that eventually grows indefinitely as t gets large. Example: y"(t) + 6*y'(t) + 5*y = 0.
In the special case that you get a single real solution, you have critical damping. Due to the overlap, the solution will be a linear combination of e^(r*t) and t*e^(r*t). This is the case that decays the fastest, without overshooting its target. Example: y"(t) + 6*y'(t) + 9*y = 0.
For a complex conjugate pair of solutions for r, you end up with underdamping. The imaginary component will be the frequency of the sine and cosine terms, and you'll have a linear combination of the two of them. Common to both terms, you'll have an exponential decay, with a decay constant equal to the real component of the solutions for r. It will have exponential decay when r is negative, and exponential growth when r is positive. Example: y"(t) + 6*y'(t) + 13*y = 0.
For purely imaginary solutions for r, you'll get no damping. The solution is a linear combination of sine and cosine.
Example: y"(t) + 9*y = 0.
so great
so great
7:17
Hi
i need proof of behavior of integral proof
Hello, Dr. Bazett! Do you have a video on how to solve a higher-order differential equation (say degree three) to get the particular solution using this method?
The way that it works for third order equations, and beyond is as follows.
Given a diffEQ in the form of:
a*y'" + b*y" + c*y' + d*y = g(t)
Step 1: find the homogeneous solution components. For 3rd order diffEQ's there will be 3 of them, and you'd solve a cubic equation to find them. I'll call them yh1(t), yh2(t), and yh3(t).
Step 2: find the first and second derivative of each one of them.
Step 3: calculate the Wronskian, with row 1 of the functions from part 1, row 2 for first derivatives, and row 3 for 2nd derivatives.
Step 4: calculate what I call, the Cramer Wronskians, because they come from Cramer's rule. There are 3 of them for a 3rd order system, which I'll call W1, W2, and W3. Replace column 1 in the original Wronskian, with a vertical column of , to get W1. Replace column 2 in the original Wronskian, with the same vertical column, to get W2. Likewise for W3.
Then we construct the particular solution:
yp = yh1*integral W1/W * g(t) dt + yh2*integral W2/W * g(t) dt + yh3*integral W3/W * g(t) dt
The complete solution will be:
y = A*yh1 + B*yh2 + C*yh3 + yp
It's easier to understand the 2nd order case knowing how it works beyond 2nd order equations, to see where the sign pattern comes from. I've tried to find examples that integrate in closed form to demonstrate this, that you couldn't otherwise do with undetermined coefficients or Laplace/Fourier transforms , but haven't had any success yet.
One example that looks promising, since it's as simple as I can make a cubic with two purely imaginary roots, so the homogeneous solutions are sin(sqrt(3)*t), cos(sqrt(3)*t), and e^(-3*t). It's great to work with sine and cosine, since they commonly reduce to a constant in the Wronskian.
y"' + 3*y" + 3*y' + 9*y = g(t)
But I've yet to find an example of g(t) that would be challenging enough to require variation of parameters.
How can proof behavior proof please?
Great video. Thank you
Sir, At 8:55, is there a negative sign in the expression for u2? Thanks.
Super well explained!
Concise explanations!
So basically the method using the Wronskian
mentioned around 8:35
Z=yَ