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Separable Differential Equations 

Prime Newtons
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In this video, I solved a differential equation with initial value conditions

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18 дек 2023

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Комментарии : 24   
@afzalmehdi3269
@afzalmehdi3269 6 месяцев назад
I like your way of explaining. You talk soft and clear and more over use chalk and black board unlike others using digital pen and don't even talk English properly. Keep it up bro
@adw1z
@adw1z 6 месяцев назад
And now solving first order non-linear differential equations - I have never seen anyone cover as wide a range of questions as you do, and it's a joy to watch. Keep going and never stop learning 🫡
@holyshit922
@holyshit922 6 месяцев назад
There are three main types of first order ordinary differential equations 1. Separable ODE 2. Linear ODE 3. Exact ODE Other types of first order ordinary differential equations which can be seen in textbooks can be reduced to this three types by substitution, integrating factor , introducing parameter
@johnconrardy8486
@johnconrardy8486 2 дня назад
your fun to watch great teacher
@kyeburns1163
@kyeburns1163 6 месяцев назад
thank you serge ibaka 🙏
@michaelbaum6796
@michaelbaum6796 6 месяцев назад
Very nice example, thanks👍
@Spider70465
@Spider70465 6 месяцев назад
Amazing 🤩
@Private-se1gl
@Private-se1gl 6 месяцев назад
very good بسیار عالی حل کردید 🎉
@richardbraakman7469
@richardbraakman7469 6 месяцев назад
Those trig functions keep showing up where they weren't invited :)
@noblesleem1077
@noblesleem1077 6 месяцев назад
😂
@channelbuattv
@channelbuattv 6 месяцев назад
exponentials are even worse. you'll see lots of them in diff eq
@abhishankpaul
@abhishankpaul 2 месяца назад
Logarithms and exponents have a different opinion in this matter
@surajsk7315
@surajsk7315 6 месяцев назад
solve Jee Advanced problems of Mathematics
@user-yd4ky5vb3w
@user-yd4ky5vb3w 6 месяцев назад
Thanks for an other video
@antonionavarro1000
@antonionavarro1000 6 месяцев назад
Another solution: f(t)=(1-t)/(1+t) If you check it will be correct (with t not equals to -1).
@user-lg6fq1yt4g
@user-lg6fq1yt4g 6 месяцев назад
@holyshit922
@holyshit922 6 месяцев назад
For future video Let T_{n}(x) = cos(n*arccos(x)) Let t = arccos(x) y(t) = cos(n*t) y'(t) = -n*sin(nt) y''(t) = -n^2cos(nt) y''(t) = -n^2y(t) y''(t) + n^2y(t) = 0 Lets change of independent variable t = arccos(x) dt/dx= -1/sqrt(1-x^2) dy/dt = dy/dx * dx/dt dy/dt = dy/dx * (-sqrt(1-x^2)) d^2y/dt^2 = d/dt(dy/dt) d^2y/dt^2 = d/dx(dy/dx*dx/dt)*dx/dt d^2y/dt^2 = d/dx(dy/dx * (-sqrt(1-x^2)))*(-sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2) * d/dx(dy/dx * sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2) * (d^2y/dx^2 * sqrt(1-x^2) + dy/dx * (-x)/sqrt(1-x^2)) d^2y/dt^2 = (1-x^2)d^2y/dx^2 - xdy/dx y''(t) + n^2y(t) = 0 d^2y/dt^2 = (1-x^2)d^2y/dx^2 - xdy/dx + n^2y(x) = 0 y(1) = 1 Solve equation above with power series
@holyshit922
@holyshit922 6 месяцев назад
Problem is that i get factorial of negative number or division by zero while expanding fraction to factorial and later to binomial coefficient
@aaronmisquith9341
@aaronmisquith9341 6 месяцев назад
Why is it that when you took the integral of dy/y²+1, you didnt add a +c onto it like you did for 1/t²+1?
@carultch
@carultch 6 месяцев назад
I'm assuming you are talking about time stamp 3:36. There is a constant of integration in both integrals, so technically, you can have a +C1 on the left integral, and a +C2 on the right, as you are suggesting, which gives us: arctanh(y) + C1 = -arctanh(t) + C2 However, we also can see that these two integration constants are not independent of each other. We can subtract C1 from both sides and get: arctanh(y) = -arctanh(t) + C2 - C1 Since it doesn't matter how we set C2 and C1 relative to each other, we can just combine them to one constant of integration, and get: arctanh(y) = -arctanh(t) + C Because this step happens in separable differential equations all the time, it is common to just keep it simple, and only add a +C on one of the integrals, but not the other.
@carultch
@carultch 6 месяцев назад
Generally, you will only have an undetermined constant in the final general solution, for every order of differentiation involved in the highest derivative. This is how you can anticipate how many of the constants of integration to either absorb each other in intermediate steps, or ultimately cancel through other algebra as you post-process your integration results.
@himanshuhooda8762
@himanshuhooda8762 6 месяцев назад
Eary for iit jee students
@nicolascamargo8339
@nicolascamargo8339 6 месяцев назад
Genial
@honestadministrator
@honestadministrator 6 месяцев назад
f ' ( t) /[ (f( t)) ^2 + 1] + 1/( t^2 + 1) = 0 d ( arc tan ( f(t)) + arc tan (t) ] = 0 arc tan ( f(t)) + arc tan (t) = arc tan ( 2) + arc tan ( 3) = π/2 - arc tan (1/2) + π/2 - arc tan (1/3) = π - arc tan ((1/2 + 1/3) / (1 - 1/6)) = π - arc tan ( 1) = 3 π /4 Hereby f (t) = tan ( 3 π /4 -arc tan ( t)) = (1 - t) /( 1 + t)
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