Your ability to break a complex problem down into simple and understandable terms is why thousands of students watch your videos when in dire need of mathematical help. One of the most annoying things about online teaching videos is when the video maker assumes that the consumer understands a step and jumps ahead. This can be frustrating, as often times I am simply misunderstanding one simple concept (missing a minus sign or something dumb). Thank you for working through each problem completely and allowing students like me to get a better understanding of differential equations.
honestly you may have just saved my life... DE is one of the most complicating subjects. plus my teacher explains it in ways that make no since and are very disorganized. Thank YOU thank YOU thank YOU
I started to watch your series of videos since I took my first calculus class in university. I will take ODE next semester and I am trying to preview for that class this summer. I have stuck at this point for a really long period of time. You have offered a really detailed and clear explanation about how to do such kind of problem. I am really grateful for your help!😁
I want to thank you for all of your videos. You've helped make being a nontraditional math major possible! With three kiddos and a husband, it is difficult to find time to study in-depth. If it wasn't for your videos, I would not have made it through calculus II. I've had a two-year break from calculus and I am now currently in differential equations... needless to say, I AM RUSTY with my calculus skills and am having to come back here to be saved. THANK YOU, THANK YOU, THANK YOU!!!!
Here we see two variables, x and y, both raised to the first power. Now if you had two x variables multiplied by one another, you know the result is x^2 and of the 2nd degree. The same holds true if you multiply x and y, as they are both variables but they happen to be different unknowns. Since x and y are both 1st degree variables, their multiplied resultant is xy where each single variable is of the 1st degree, but the combination are of the 2nd degree. Hope that helped.
just watched more than 5 videos to learn this rule! I have got nothing..now I am watching your video..this rule is now like a piece of cake..hats off 🤗
dude, you just saved me about 15 points in my exam this morning! I couldn't figure it out till i watched this and practiced before heading to school for the exam
I love differential equations right now. It is so crazy with the methods in the beginning, but when you can see the big picture it is interesting. These videos help reinforce and become more precise with DE, thanks!
My prof just reads pre-written math notes to us and I understand nothing Thank you for this video bc now I can actually learn the material. Wonderfully easy to understand
remember in calculus I when we are using limits and if we have a function lim x- infinity (x+1) we can factor out the x and change the equation to lim x - infinity (1 + 1/x) he is using the same concept
Thank you so much for these videos. I just wish you uploaded these a day earlier since my test on this subject is tomorrow and I'm already all studied out -_-
ur just saving my ass in eng'g since differential calculus, how the hell did my professors can't explain this very briefly like what u did, they're just making the problem worse, but all I can say thanks for making math an enjoying one.
If this was an initial value problem, would you solve for C, then plug it back into the equation? What if you don't have an explicit DE and you need to find y(1)=2/3 or something? Do you replace all y's with 2/3 and all x's with 1 and solve for C?
I know that when you multiply two variables, you add their exponents together. It still seems very odd to me that you are calling the product of two terms, and thus the sum of their exponents, the same as the numerator's exponents - where the x and y's exponents are left alone.
i need to see this when dy/dx are on different sides of the equation and cannot be combined. my book factors (x^2-u*x^2)(udx+xdu) into x^3(1-u)du and offers no explanation. is there a property of differentials in this form that I am missing?
Nathan Sowder This is probably too late, but I figured I'd throw it out there anyway. If I'm not mistaken, you're probably using Differential Equations by Zill, and I ran across a problem exactly as you described, and this very step tripped me up also. Remember, you're always looking for terms that cancel. ASSUMING you have the same problem I had, the equation looks something like this: ( x^2 + x^2u^2)dx + (x^2 - ux^2)[udx + x du]=0 First, FOIL the back half of the equation. Also, just to make the cancelling easier to see, we'll distribute the dx from the first half. x^2dx + x^2u^2dx + x^2udx + x^3du + -u^2x^2dx - ux^3du=0 Hopefully you now see that we have a positive and negative x^2u^2dx term, and they'll cancel. When you combine dx and du terms, you're left with: x^2dx + x^2udx + x^3du - ux^3du=0 No you can factor by grouping to get: x^2(1+u)dx + x^3(1-u)du=0 I hope this was helpful.
hi . what if the degree of the numerator and denominator (2nd-3rd step) is not the same? will the lcm/lcd be taken instead? or would the equation not be considered homogenous? ty
you're the man !! I've been looking for ..... keep it up and make more vids.... do you have Bernoulli ?? I can't find it out ...pls send link if it exists
but the example in my book makes it seem as though this would be degree zero since the lamdas from each term would factor out and cancel. i dont get it
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Seriously, if it wasn't for this guy I would not be studying engineering.
I'll second that!
pacrat90 I third that
Yup 💯
i'd quit Engineering right now if it were not for this guy!
PatrickJMT raised my GPA
Your ability to break a complex problem down into simple and understandable terms is why thousands of students watch your videos when in dire need of mathematical help. One of the most annoying things about online teaching videos is when the video maker assumes that the consumer understands a step and jumps ahead. This can be frustrating, as often times I am simply misunderstanding one simple concept (missing a minus sign or something dumb). Thank you for working through each problem completely and allowing students like me to get a better understanding of differential equations.
Skyman12
You’re not misunderstanding a concept. You just need better precision. Discipline yourself to not make silly mistakes.
Lucas M your comment comes on time that i need it .. I will !
honestly you may have just saved my life... DE is one of the most complicating subjects. plus my teacher explains it in ways that make no since and are very disorganized. Thank YOU thank YOU thank YOU
Patrick just makes so math soothing, he could convince me to do anything with that beautiful innocent siren voice
:0 it's so accurate!
Anything? Even divide by zero? ;-)
gachiGASM
@@annasarahkramerfazendeiro8030
Yeye
Whats 2020/0
(2×0×2×0)/0=22
I started to watch your series of videos since I took my first calculus class in university. I will take ODE next semester and I am trying to preview for that class this summer. I have stuck at this point for a really long period of time. You have offered a really detailed and clear explanation about how to do such kind of problem. I am really grateful for your help!😁
You have a gift for making these topics 10x less intimidating. THANK YOU!
I want to thank you for all of your videos. You've helped make being a nontraditional math major possible! With three kiddos and a husband, it is difficult to find time to study in-depth. If it wasn't for your videos, I would not have made it through calculus II. I've had a two-year break from calculus and I am now currently in differential equations... needless to say, I AM RUSTY with my calculus skills and am having to come back here to be saved. THANK YOU, THANK YOU, THANK YOU!!!!
Suddenly I am enjoying DEs. Thanks alot
PatrickJMT, you are the best as always!
Here we see two variables, x and y, both raised to the first power. Now if you had two x variables multiplied by one another, you know the result is x^2 and of the 2nd degree. The same holds true if you multiply x and y, as they are both variables but they happen to be different unknowns. Since x and y are both 1st degree variables, their multiplied resultant is xy where each single variable is of the 1st degree, but the combination are of the 2nd degree. Hope that helped.
this method is 100000000% easier than the method my professor showed me. thanks, patrick!
just watched more than 5 videos to learn this rule! I have got nothing..now I am watching your video..this rule is now like a piece of cake..hats off 🤗
What would I do without you?😭thank youuuuu so much
I think you're the incarnation of either Newton or Leibniz. Thank you, so comprehensive
dude, you just saved me about 15 points in my exam this morning! I couldn't figure it out till i watched this and practiced before heading to school for the exam
this video (like most if not all) has been very helpful to countless persons. keep up the good work
Here I was thinking that patrickJMT stopped after Multivariable Calculus. So happy that you do Differential Equations too!!!
I would like to thank you
You just saved me from failing a course
You are a magician, i dont know why my teacher never taught it like this before.. so much easier!!
Well... time to drop the course.
This is ten months later, but I just wanted to say that I hope you didn't. Calculus is so pretty.
+Adam G this is almost two years after. but FYI this is DVQ
This is three years later. I wonder where you are now.
this is 4 years later and I'm going to become an engineer because of Patrick!
4 years later and I’m about to take my board exam to be a licensed engineer so I’m back again for these tutorials for review 😂😂
I love differential equations right now. It is so crazy with the methods in the beginning, but when you can see the big picture it is interesting. These videos help reinforce and become more precise with DE, thanks!
This dude is my savior. I’m going to build a shrine of him on my desk.
well, i would be very flattered
from one pat to another, many many thanks for your math videos!
this american accent with a touch of irish really makes this more soothing
I have a test tmr. this helped me to understand really fast. thx.
good luck!
@@patrickjmt thx. I still have an hour left for the test.
you know this thing take me a week and my book couldn't help me. wow this 7 minutes finally save me
this video explains it so much better than my textbook
SO thankful for all your videos!
thanks you so much man you made this vid in 2011 and here i am learning from you in 2017 YOU ARE A LEGEND
And here I am learning from him in 2023
My prof just reads pre-written math notes to us and I understand nothing
Thank you for this video bc now I can actually learn the material. Wonderfully easy to understand
come back any time :)
remember in calculus I when we are using limits and if we have a function
lim x- infinity (x+1) we can factor out the x and change the equation to lim x - infinity (1 + 1/x) he is using the same concept
Dude, thanks a whole heap! You're the best math teacher in the world. Your other video on triple integrals was AWESOME and so is this one!
You're a lifesaver man.
Patrick is the math god for us 🙌PREACH
do you have videos on Bernoulli. I can't seem to find them..
no, i don't at the moment
pls do one ASAP !!! We need it
Such an amazing resource. Chapeau!
Thank you so much for these videos. I just wish you uploaded these a day earlier since my test on this subject is tomorrow and I'm already all studied out -_-
Seriously, THANK YOU Patrick!!
We love you Patrick ❤️
PATRICK!! you are the BOMB.COM .... now i understand. with love from South Africa
Thank you again Patrick
explicit and extremely helpful. Thank you so much.
You don't know how much I love you right now. Thanks bro T^T
He multiplies the top and bottom by the same thing in order to preserve the value of the fraction. Essentially, you're multiplying it by 1.
хаз кхетир!) баркал хьун)
Like the video, not the comments. Thanks PatrickJMT
ur just saving my ass in eng'g since differential calculus, how the hell did my professors can't explain this very briefly like what u did, they're just making the problem worse, but all I can say thanks for making math an enjoying one.
@rinwhr ha, yep!
u can put y =vx in the starting......so it becomes easy not so complicated
You're a life saver!
This helped soo much. Thanks a lot
Thank you Patrick!
thanks a lot you're always save me when i be in problem
Is there an intuitive reason why it's called homogeneous? Does it describe any specific phenomenon that might account for the name?
Homogeneous = the same, so everything in the equation has to have the same variable in it
Kameron Irvin not variable I think it's degree
Yes same degree.
This is very clear! Thank you!
You clarify things far more better then f* Wiley's book.
Hi your videos are amazing i just wanted to clarify as to why we do an implicit differentiation with y = xv
brilliant stuff sir....god bless you
very well explained
Thank you soo much. Love your videos and keep making more please :)
I suggest you review your laws of exponents.
good job that helped a lot !
Weird I've never learned this method in my ODEs class. I'm guessing it will be in ODEs 2 !
If this was an initial value problem, would you solve for C, then plug it back into the equation? What if you don't have an explicit DE and you need to find y(1)=2/3 or something? Do you replace all y's with 2/3 and all x's with 1 and solve for C?
why multiply with ( 1/x ) when you can directly put ( y=vx)
is possible to have some video which explain the interval of existence/ uniqueness about IVP? thank you
thanx you have cleared my mind
Hi Patrick,
Could you make a video on how to solve non exact equations?
Adopt me please
I'll think about it
+patrickJMT
"He Acknowledged it!" - Ted (Ted meets Flash Gordon)
please, me too! mathematician dad! :)
You are so good sir
Thanks for your amazing explain
THANK YOU !!! you helped me alot
why dont you multiply both sides by (1/x^2)?
bless this man
Couldn't you have also separated the numerator to make it (y/x) - (x/y)?
question: why would you add the degree of y and x in the numerator? are they supposed to be treated as one variable? sorry i'm so confused
I know that when you multiply two variables, you add their exponents together. It still seems very odd to me that you are calling the product of two terms, and thus the sum of their exponents, the same as the numerator's exponents - where the x and y's exponents are left alone.
i need to see this when dy/dx are on different sides of the equation and cannot be combined. my book factors (x^2-u*x^2)(udx+xdu) into x^3(1-u)du and offers no explanation. is there a property of differentials in this form that I am missing?
Nathan Sowder
This is probably too late, but I figured I'd throw it out there anyway. If I'm not mistaken, you're probably using Differential Equations by Zill, and I ran across a problem exactly as you described, and this very step tripped me up also. Remember, you're always looking for terms that cancel.
ASSUMING you have the same problem I had, the equation looks something like this:
( x^2 + x^2u^2)dx + (x^2 - ux^2)[udx + x du]=0
First, FOIL the back half of the equation. Also, just to make the cancelling easier to see, we'll distribute the dx from the first half.
x^2dx + x^2u^2dx + x^2udx + x^3du + -u^2x^2dx - ux^3du=0
Hopefully you now see that we have a positive and negative x^2u^2dx term, and they'll cancel. When you combine dx and du terms, you're left with:
x^2dx + x^2udx + x^3du - ux^3du=0
No you can factor by grouping to get:
x^2(1+u)dx + x^3(1-u)du=0
I hope this was helpful.
why did you use the product rule. there are 3 variables in y = vx. wouldn't that require partial differentiation?
What if you had dy/dx = (x+2y+1)/(2x+4y-1) would this be homogenous?
+Oreen Yousuf yup
Is the method you use here the same as variation of the parameter?
I need more questions about this topic
who cares how homogenous is pronounced... i can change variables now!! woohoo thanks heaps!!
Thank you man!
Why don't you just cross multiplied?
Can someone further explain why Patrick multiplied by 1/x ?
thanks a lot ....you saved me.
question! Why did you multiply the equation with 1/x^2? @ 1:06
I FUCKIN LOVE YOU PATRICK
hi . what if the degree of the numerator and denominator (2nd-3rd step) is not the same? will the lcm/lcd be taken instead? or would the equation not be considered homogenous?
ty
you're the man !! I've been looking for ..... keep it up and make more vids.... do you have Bernoulli ?? I can't find it out ...pls send link if it exists
i have a couple on bernoulli, just search on my channel or look at my recent videos.
how to found that you have to find dy/dx instead of dx/dy
but the example in my book makes it seem as though this would be degree zero since the lamdas from each term would factor out and cancel. i dont get it
In the book for homogenous it says y=0 at the end for some reason. Can you explain that?
Thanks patrick :)
it make a lot of sense now
In what world would you say XY has degree 2?
Thank you A LOT !! 💙
why did u multiply it with 1/x^2
why multiply by 1/x^2 but not 1/y^2