I checked it. Well, I checked y=exp(2Atan(Ax+B)) which is the same enough after a B transformation. It works. It's annoying that all these steps seem logical enough, but I'd blank on drawing some of them if I was solving it myself.
8:10 oh hell no! Lol. However you could expand the term inside the tangent function and call AB a new constant C, so then 2Atan(Ax+C) has a nice scale factor, frequency factor and phase shift factor which makes it nice to read and interpret :) Differentiation is left as an exercise to the reader. Or maybe I need to start a maths channel where I just do the inverses of your integral problems! 😊
You know; at a certain level; ODEs lose their charm and becomes really repetitive. Glad we have my boy kamal to make any ode a party hardy cheering up my weekend Anyways this is a second order non linear ode with no x so we apply the nice substitution z(y) =d/dx y So that y’’ = zdz/dy which makes it linear or exact call it whatever you want but it’s super solvable; what’s interesting though would be the integration part of the problem so I’ll watch the vid now. Keep up them vids my boii❤❤
One could abbreviate the calculation a little by setting y = exp(u). Then you get first u''/u' = u or u'' = 1/2 (u^2)' so that we have u'=1/2 u^2 +c ,which is separable and the rest is easy.
You kept trying to play with the constant of integration A to further simplify, but you could have instead played with the constant B. If you redefine B, you can remove a set of parentheses and simply the exponent to 2A tan (Ax+B)
I love non-linear equations. I hate it how Physics courses only teach you how to solve linear ODEs, linear algebra etc, but hardly spend any time on non-linear. Most of the physical world is non-linear! Why isn't more time spent on this?!
This is still incomplete because it's missing the cases A = 0 and A < 0. This is fun to solve and has an interesting symmetry with the original case if you use tanh^-1 (or coth^-1 where appropriate) instead of partial fractioning and writing the solution in terms of logs.
@@ninadmunshi2879 negative case should be fine. 1/ki arctan(x/ki)=-1/k artanh(x/k) And if you want an arcoth in there arcoth(1/x)=artanh(x) But yes. A=0 would instead give y=e^(-2/(x+B))
Ainda não entendo muito de equações diferenciais, mas elas parecem bem divertidas de se resolver (pricipalmente quando não sou eu que tenho que resolver :3 )
y=e^u...risulta 1+u"/u'-u'=u..u'=p...risulta 1+dp/du-p=u.. è un'equazione differenziale del primo ordine non omogenea del tipo y'+a(x)y=b(x)..che sappiamo risolvere ..=>p(u)=ce^u-u=du/dx...dx=du/(ce^u-u)...integro x+C2=int(1/(ce^u-u))du...uhm...
deepseek r1 solution is more correct than yours. (still missing some parts in conclusion, but after asking (and following though thinking process) it do adds them)
Oh yeah, tbh I should've handled those cases separately while solving the integral. But why use deepseek for something that wolfram alpha gets correct in the first attempt?😂
@@maths_505 to check how good it is, and wolfram alpha in free mode does not give intermediate steps, only QWQ32b and r1 does give intermediate steps in free mode.
I checked it. Well, I checked
y=exp(2Atan(Ax+B))
which is the same enough after a B transformation. It works. It's annoying that all these steps seem logical enough, but I'd blank on drawing some of them if I was solving it myself.
8:10 oh hell no! Lol. However you could expand the term inside the tangent function and call AB a new constant C, so then 2Atan(Ax+C) has a nice scale factor, frequency factor and phase shift factor which makes it nice to read and interpret :)
Differentiation is left as an exercise to the reader. Or maybe I need to start a maths channel where I just do the inverses of your integral problems! 😊
You know; at a certain level; ODEs lose their charm and becomes really repetitive. Glad we have my boy kamal to make any ode a party hardy cheering up my weekend
Anyways this is a second order non linear ode with no x so we apply the nice substitution z(y) =d/dx y
So that y’’ = zdz/dy which makes it linear or exact call it whatever you want but it’s super solvable; what’s interesting though would be the integration part of the problem so I’ll watch the vid now.
Keep up them vids my boii❤❤
As always (or at least as sometimes), you make it look so easy
Twi terribly sorry about that today guys, happy sunday
Hi,
"ok, cool" : 0:14 , 2:49 ,
"terribly sorry about that" : 4:36 , 5:17 , 6:25 .
One could abbreviate the calculation a little by setting y = exp(u). Then you get first u''/u' = u or u'' = 1/2 (u^2)' so that we have u'=1/2 u^2 +c ,which is separable and the rest is easy.
yeah this is what I did, the u’’ computation gets a little hairy but then it simplifies out really nicely and the rest is easy
You kept trying to play with the constant of integration A to further simplify, but you could have instead played with the constant B. If you redefine B, you can remove a set of parentheses and simply the exponent to 2A tan (Ax+B)
@@michaelihill3745 I agree, that would've been a better idea.
@ 1:55 you cancelled the variable u, which I believe caused a loss of a solution: dy/dx = u = 0. Then y = C is a solution.
Are you sure y=C works?
0/0-0/C=ln(C)
Not a safe insert into the original.
We have u≠0 because y' is in the denominator of the original equation
Excellent Effort. Thanks
I love non-linear equations. I hate it how Physics courses only teach you how to solve linear ODEs, linear algebra etc, but hardly spend any time on non-linear. Most of the physical world is non-linear! Why isn't more time spent on this?!
Exactly!
I wonder what physical system the equation in the video could model? @@maths_505
🤔 y = y'/y'' + y''/y''' + y'''/y'''' ...
This is still incomplete because it's missing the cases A = 0 and A < 0. This is fun to solve and has an interesting symmetry with the original case if you use tanh^-1 (or coth^-1 where appropriate) instead of partial fractioning and writing the solution in terms of logs.
if i understand correctly negative part is covered by complex numbers, but zero case needs separate expansion.
@@ninadmunshi2879 negative case should be fine.
1/ki arctan(x/ki)=-1/k artanh(x/k)
And if you want an arcoth in there
arcoth(1/x)=artanh(x)
But yes. A=0 would instead give
y=e^(-2/(x+B))
Ainda não entendo muito de equações diferenciais, mas elas parecem bem divertidas de se resolver (pricipalmente quando não sou eu que tenho que resolver :3 )
we’re do you even get your questions, do you just think of them on the spot for what conveniently yields the most exiting integrals.
@@sak6653 yeah that's one way
Hi Kamaal
case where A=0 is lost, so this is not full solution.
y=e^u...risulta 1+u"/u'-u'=u..u'=p...risulta 1+dp/du-p=u.. è un'equazione differenziale del primo ordine non omogenea del tipo y'+a(x)y=b(x)..che sappiamo risolvere ..=>p(u)=ce^u-u=du/dx...dx=du/(ce^u-u)...integro x+C2=int(1/(ce^u-u))du...uhm...
Sir please can you tell me which app you are using for writing.
deepseek r1 solution is more correct than yours. (still missing some parts in conclusion, but after asking (and following though thinking process) it do adds them)
Oh yeah, tbh I should've handled those cases separately while solving the integral. But why use deepseek for something that wolfram alpha gets correct in the first attempt?😂
@@maths_505 to check how good it is, and wolfram alpha in free mode does not give intermediate steps, only QWQ32b and r1 does give intermediate steps in free mode.