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An interesting 2nd order non linear differential equation
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- čas přidán 14. 08. 2024
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NOTE: At the 1:02 mark, the equation is d^2y/dx^2=u(du/dy).
This equation can be easily reduced to first order by substitution u(y) = y'
y'' = u'(y)y'
y'' = u'(y)u(y)
yuu' - 1/2u^2 = -4y^2
Now we can solve it two ways as homogeneous or as Bernoulli
Wolfram alpha delivers for this ODE: y(x)=C2*cos^2(sqrt(2)*(C1+x))
This was awesome! Love your channel !
0:58 Isn't there a mistake here? If you substitute u=dy/dx into du/dy dy/dx you should get u du/dy instead of u dy/dx
Edit: okay I watched further and saw that you proceeded with the correct substitution in the next step
Thank you for your effort.
Very nice solution, thanks a lot👍 11:28
Thanks buddy
When can i apply the chain rule?, sometimes the chain rule has not been applied in y'=u
Ah it's been a while since I solved one of those bad boys, quick question: how do you know when to use the Exact differential method (ofc with the integrating factor cases)? what are the usual rules of thumb?
anyways great vid!
I just move stuff around. It helps when you have the equation in a differential form with one dx term and another dy term and from there you can use some intuitive guess work or the more formal rules for integrating factors (that I can't remember 😂).
I didnt use substitution and it worked so i am happy
What i essentially did was make y equal e^lambda x and then figure out the equation (making sure all the y’s are isolated and equalling 0) and then solving for e^{2lambda x}(…) =0 and then getting only a complex answer, plucking that in Eulers formula and then checking by calculating all the terms (4y^2),(y(y’’(x)),(y’(x))^2 and then plucking all those answers back into the first formula, which checked out
y for me eventually came out to be cos(sqrt(8)x) + isin(sqrt(8)x) and the equation was correct so I can’t complain
I found this one a bit simpler(or maybe your content has set the standards too high)
BTW i have hit more than a 100 subs:) which also has alot of contribution from your side.
I self learn math alot
Especially calcus
But the main reason I almost didn't do any searching about linear algebra deferential equation and complex analysis
Is because I don't like to learn anything that I didn't build my understanding of it from the very start
how did we even come with
Especially dealing with matrixs
I learned it in school and until now for me it's some random operations on some numbers
And now we are using them to solve equations?
I might as will just use Wolfram alpha to solve them
I love math (one of the reasons) because it pure logic
Basid on my (our) understanding and logical steps we can do things seemed impossible
So as a start to help me do you know an easy way to proof that any nth order linear deffertial equation have exactly n solutions
It's the reason I don't expect the solution of the integral cos(x)/(1+x²) form -∞ to ∞ with deffertial equations
What if there is more that one solution and only one work
Like solving for a side of a right angle and choosing the positive of the two answers because there is no thing as negative side
I think you should start with prof Leonard's series on DEs. That should set you up perfectly.
@@maths_505 when you said Leonard's series
I thought it's a mathematical series
Not a playlist 😅
@@illumexhisoka6181 😂😂😂
Please i would really like if you could tell me what do you use to draw, a tablet connect to your phone/tabled or a pen or something else. And if you could tell me the name of the product. ❤❤❤
It's an S6 tab and I use Samsung notes
And the pen that comes with it?@@maths_505
,👍...purtroppo i miei ricordi non mi consentono di risolvere queste equazioni... però ho capito...
Nice one. Why don't you solve some elliptic integral?
Ohh now that is a good idea🔥🔥🔥
y'' = y'dy'/dy
(1/2)y'^2 = 4y^2 + yy'dy'/dy
(1/2)y'^2 - yy'dy'/dy = 4y^2
y' = uy
(1/2)u^2y^2 - uy^2(u + ydu/dy) = 4y^2
u^2y^2 - 2u^2y^2 - 2uy^3du/dy = 8y^2
-u^2 - 2uydu/dy = 8
2uydu/dy = -u^2 - 8
2udu/(u^2 + 8) = dy/y
ln(u^2 + 8) = ln y + C
u^2 + 8 = ky
(y'/y)^2 = ky - 8
y'/y = √(ky - 8)
y' = y√(ky - 8)
dy/[y√(ky - 8)] = dx
atan(1/2√(ky/2 - 4))/√2 = x + C
y = Ksec^2((x + C)/√2)
You forgot a negative on line 9, and somehow, the wolframalpha solution is C_2cos^2(x*sqrt2+C_1) I checked with desmos, this is actually the right answer somehow! Also, how did atan turn into sec?
I love your solution development though, I also tried u=y'/y, but some silly mistakes took me out in the end